OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(c)

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S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(c)

Question 1.
Write the given statement in symbolic form using the letter in parentheses to represent the corresponding component.
(i) This is April (p) and income tax returns must be filed (q).
(ii) Accountancy is a required subject for Chartered Accountants (m) but not for engineers (n).
(iii) Mukesh Patel is a teacher (t) or a lawyer (u).
(iv) Jack went up the hill (c) and Jill went up the hill (d).
(v) I plan to take science (a) or commerce (c) in class 11.
(vi) I will not drive to Jaipur (~ d) but I shall go by train (i) or by plane (p).
Solution:
(i) p ∧ q
(ii) m ∧ ~ n
(iii) t ∨ u
(iv) c ∧ d
(v) Compound statement is the disjunction of a and c i.e. a ∨ c
(vi) The compound statement is the conjunction of ~ d and (disjunction of t and p) i.e. (~ d) A (t ∨ p)

Question 2.
Let p be “Shruti can type”, and let q be “Shruti takes shorthand.” Write the following statements in symbolic form :
(i) Shruti can type and take shorthand.
(ii) Shruti can type but she does not take shorthand.
(iii) Shruti can neither type nor take shorthand.
(iv) It is not true that Shruti can type and take shorthand.
Solution:
Given p : Shruti can type
q : Shruti takes shorthand
(i) Compound statement is the conjunction of p and q i.e. p ∧ q.
(ii) ~ q : she does not takes shorthand The given statement is the conjunction ofp and ~ q i.e. p ∧ q.
(iii) ~ p : Shruti can’t type
∴ given statement is the conjunction of ~p and ~ q i.e. ~p ∧ ~ q.
(iv) p ∧ q : Shruti can type and takes shorthand
∴ ~(p ∧ q): It is not true that Shruti can type and take shorthand.

Question 3.
Use p : Ramesh is rich ; q : Pradeep is poor. Think of “poor” as “not rich”, and write each of these statements in symbolic form.
(i) Ramesh is poor and Pradeep is rich.
(ii) Pradeep and Ramesh are both rich.
(iii) Neither Ramesh nor Pradeep is rich.
(iv) Ramesh is not rich and Pradeep is poor.
(v) It is not true that Ramesh and Pradeep both are rich.
(vi) Either Ramesh is poor or Pradeep is poor.
(vii) Either Ramesh or Pradeep is rich.
Solution:
Given statements are : p : Ramesh is rich q : Pradeep is poor ~p : Ramesh is not rich i.e. poor ~ q : Pradeep is not poor i.e. rich.
(i) (~P) ∧ (~ q)
(ii) (~ q) ∧ p
(iii) (~ p) ∧ q
(iv) ~ p ∧ q
(v) ~ [p ∧ (~ q)]
(vi) ~p ∨ q
(vii) p ∨ (~q)

OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(c)

Question 4.
Use p : I like this school ; q : I like Mr. Sexena. Express each of the following statements in words.
(i) p ∧ q
(ii) ~ q
(iii) ~p
(iv) (~p) ∧ (~ q)
(v) (~p) ∧ q
(vi) p ∨ q
(vii) ~ (p ∧ q)
(viii) ~ [(~ p) ∧ q]
Solution:
Given statements are ;
p : I like this school
q : I like Mr. Sexena.
(i) p ∧ q : I like this school and I like Mr. Sexena.
(ii) ~ q : I do not like Mr. Sexena.
(iii) ~p : I do not like this school.
(iv) ~p ∧ ~ q : I do not like this school and I do not like Mr. Sexena.
(v) ~ p ∧ q : I donot like this school but I like Mr. Sexena.
(vi) p v q : I like this school or I like Mr. Sexena.
(vii) ~ (p ∧ q) : It is not true that I like this school and I like Mr. Sexena.
(viii) ~ [(~ p) ∧ q] : It is not true that I donot like this school and I like Mr. Sexena.

Question 5.
Give the negation of each of the following statements.
(i) Either he is bald or he is tali.
(ii) Nobody does not like Madhuri.
(iii) It is not true that the set of prime numbers is finite.
(iv) All circles are round.
(v) Some students passed this course.
Solution:
(i) Let p : he is bald
q : he is tall.
∴ p ∨ q: either he is bald or he is tall.
Then ~ (p ∨ q) : He is not bald and he is not tall.
(ii) Negation of given statement is that somebody does not like Madhuri.
(iii) Let p : It is not true that set of prime numbers is finite.
∴ ~p : set of prime numbers is finite.
(iv) p : all circles are sound
~ p : all circles are not round or it is not the case that all circles are round.
(v) p : Some students passed this course
~ p : It is not true that, some students passed this course or No student passed this course.

OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(b)

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S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(b)

Question 1.
Identify the quantifier in the following statements.
(i) There exists a capital city for every state of India.
(ii) For every real numbers, x is less than x + 1.
(iii) At least one natural number is not a prime number.
Solution:
(i) There exists.
(ii) For every.
(iii) Atleast.

OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(b)

Question 2.
Symbolise the following statements.
(i) There is at least one number in the set of natural numbers which is equal to ‘its’ cube.
(ii) The square of every real number is positive.
(iii) There exists at least one number in A = {5, 7, 8, 9, 10} which is an even number.
(iv) For every real number x, x < x + 1.
(v) The square roots of all prime numbers are irrational numbers. (Let P denote the set of prime numbers and Q that of irrational numbers).
Solution:
(i) ∃ x ∈ N s.t x = x3
since 1 e N and 1 = 13
(ii) ∀ x ∈ R s.tx2 > 0
(iii) ∃ x ∈ A s.t x is even, since 8, 10 ∈ A
(iv) ∀ x ∈ R, x < x + 1, since successor of every real number is greater than itself.
(v) ∀ x ∈ P, √x ∈ Q

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1

Very short answer type questions :

Determine the order and the degree (when defined) of each of the following (1 to 20) differential equations:

Question 1.
(i) \(\frac{d y}{d x}\) – cos x = 0 (NCERT)
(ii) \(\left(\frac{d y}{d x}\right)^2+\frac{d y}{d x}\) – sin y = 0.
Solution:
Given differential equation be,
\(\frac{d y}{d x}\) – cos x = 0
Which is polynomial in derivatives.
The highest ordered derivative existing in the given diff. eqn. be \(\frac{d y}{d x}\) so its order is 1.
Also, the exponent of \(\frac{d y}{d x}\) is 1.
∴ degree of given differential equation be 1.

(ii) Given differential equation be,
\(\left(\frac{d y}{d x}\right)^2+\frac{d y}{d x}\) – sin y = 0
which is polynomial in derivatives.
The higher order derivative existing in given the given diff. eqn. be \(\frac{d y}{d x}\) so its order is 1.
and the highest power of \(\frac{d y}{d x}\) in given diff. eqn. be 2 so its degree is 2.

Question 2.
(i) y’ + y = ex
(ii) \(\frac{d^2 y}{d x^2}\) = sin 3x + cos 3x. (NCERT)
Solution:
Given differential eqn. be
y’ + y = ex i.e. \(\frac{d y}{d x}\) + y = ex
which is polynomial in derivatives.
The highest order derivative present in given differential eqn. be \(\frac{d y}{d x}\) and its exponent be 1.
So order of given diff. eqn. be 1 and its degree be also 1.

(ii) Given diff. eqn. be,
\(\frac{d^2 y}{d x^2}\) = sin 3x + cos 3x
which is polynomial in derivatives.
The highest order derivative existing in given diff. eqn. be \(\frac{d^2 y}{d x^2}\).
So its order be 2.
The highest power of \(\frac{d^2 y}{d x^2}\) be 1.
∴ degree of given differential eqn. be 1.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1

Question 3.
(i) (x2y – 3x) dy + (x3 – 3y2) dx = 0
(ii) \(\sqrt{1-y^2} d x+\sqrt{1-x^2} d y\) = 0
Solution:
(i) Given differential eqn. can be written as,
\(\frac{d y}{d x}+\frac{x^3-3 y^2}{x^2 y-3 x}\) = 0
which is polynomial in derivatives.
The highest order derivative present in given differential eqn. be \(\frac{d y}{d x}\).
so its order be 1.
Also the degree of given diff. eqn. be the highest exponent of \(\frac{d y}{d x}\) which is 1.
Thus its degree be 1.

(ii) Given differential eqn. can be written as,
\(\frac{d y}{d x}+\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\) = 0
which is polynomial in derivatives.
The highest order derivatives present in given diff. eqn. be \(\frac{d y}{d x}\) and its order be 1.
The degree of given diff. eqn. be the highest power of \(\frac{d y}{d x}\) which is 1.
So its degree be 1.

Question 4.
(i) \(\frac{1}{x} \cdot \frac{d^2 y}{d x^2}+5 x \frac{d y}{d x}\) = sin 2x
(ii) \(\left(\frac{d y}{d x}\right)^4+3 y \frac{d^2 y}{d x^2}\) = 0
Solution:
(i) Given differential eqn. be,
\(\frac{1}{x} \cdot \frac{d^2 y}{d x^2}+5 x \frac{d y}{d x}\) = sin 2x
which is polynomial in derivatives.
The highest ordered derivative present in given diff. eqn. be \(\frac{d^2 y}{d x^2}\)
So its order is 2.
The degree of given diff. eqn. be the highest exponent of \(\frac{d^2 y}{d x^2}\) which is 1.
So its degree be 1.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1

Question 5.
(i) (y’)2 + y2 – 1 = 0
(ii) y” + 5x (y’)2 – 6y = log x
Solution:
(i) Given diff. eqn. be,
y’2 + y2 – 1 = 0
which is polynomial in derivatives.
The highest ordered derivative present in given diff. eqn. be \(\frac{d y}{d x}\) i.e. y’.
so its order be 1.
The degree of the given diff. eqn. is the highest exponent of which is \(\frac{d y}{d x}\).
So its degree be 2.

(ii) Given diff. eqn. be,
y” + 5xy’2 – 6y = log x
which is polynomial in derivatives.
The highest ordered derivative in the given diff. eqn. be y”.
so its order be 2.
The degree of the given diff. eqn. be the highest exponent of \(\frac{d^2 y}{d x^2}\) which is 1.
Thus its degree be 1.

Question 6.
(i) x3 (\(\frac{d^2 y}{d x^2}\))2 + x (\(\frac{d y}{d x}\))4 = 0
(ii) y(iv) + sin y(i) = 0 (NCERT)
Solution:
(i) Here the highest ordered derivative existing in given differential eqn. be \(\frac{d^2 y}{d x^2}\)
so its order be 2.
Further given differential eqn. can be expressed as polynomial in derivatives.
So the exponent of highest ordered derivative \(\frac{d^2 y}{d x^2}\) which is 2 gives the degree of given differential eqn.

(ii) Here the highest ordered derivative existing in given differential equation be y(iv) so its order be 4.
Since the given differential eqn. can’t be expressed as polynomial in derivatives as it contains term like sin y(1) which contains infinite number of terms.
So degree of given differential eqn. is not defined.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1

Question 7.
(i) \(\left(\frac{d^2 y}{d x^2}\right)^3+\frac{d^2 y}{d x^2}+\sin \left(\frac{d y}{d x}\right)\) = 2x
(ii) \(\left(\frac{d^2 y}{d x^2}\right)^3+2 y \frac{d y}{d x}+\sin y=5 x^{\frac{2}{3}}+\log x\)
Solution:
(i) Here, the highest ordered derivative existing in given diff. eqn. be \(\frac{d^2 y}{d x^2}\) and its order be 2.
Clearly given diff. eqn. cannot be expressed as polynomial in \(\frac{d y}{d x}\)
so its degree is not defined [as it contains terms like sin (\(\frac{d y}{d x}\))]

(ii) Clearly the highest ordered derivative existing in given differential eqn. be \(\frac{d^2 y}{d x^2}\)
so its order is 2.
Since each term in derivatives is a polynomial.
So degree of given differential eqn. be highest exponent of \(\frac{d^2 y}{d x^2}\) i.e. 3.

Question 8.
(i) 2x2 \(\frac{d^2 y}{d x^2}\) – 3 \(\frac{d y}{d x}\) + y = 0 (NCERT)
(ii) (y”)3 + (y’)2 + sin y’ + 1 = 0. (NCERT)
Solution:
(i) Given differential eqn. be
2x2 \(\frac{d^2 y}{d x^2}\) – 3 \(\frac{d y}{d x}\) + y = 0,
which is polynomial in derivatives.
The highest ordered derivative present in \(\frac{d^2 y}{d x^2}\) given diff. eqn. be \(\frac{d^2 y}{d x^2}\).
So its order be 2.
The degree of given diff. eqn. is the highest power of \(\frac{d^2 y}{d x^2}\) which is 1.
Thus, its degree be 1.

(ii) Given differential eqn. be
(y”)3 + (y’)2 + sin y’ + 1 = 0
The highest order derivative existing in given diff. eqn. be y” and its order be 2.
Thus, the order of given diff. eqn. be 2.
Since the term sin y’ is not polynomial in y’.
∴ The degree of given diff. eqn. is not defined.

Question 9.
(i) x \(\frac{d^2 y}{d x^2}\) = (1 + (\(\frac{d y}{d x}\))2)4
(ii) \(\left(\frac{d^4 y}{d x^4}\right)^2=\left(x+\left(\frac{d y}{d x}\right)^2\right)^3\)
Solution:
(i) Clearly the highest ordered derivative existing in given diff. eqn. be \(\frac{d^2 y}{d x^2}\).
so its order be 2.
Clearly the given diff. eqn. can be expressed as polynomial in derivatives.
Hence degree be the highest exponent of \(\frac{d^2 y}{d x^2}\) which is 1.

(ii) Clearly the highest ordered derivative existing in given diff. eqn. be \(\frac{d^4 y}{d x^4}\).
so its order be 4.
Here given differential eqn. can be expressed as polynomial in derivatives.
Hence degree of given diff. eqn. be the highest exponent of \(\frac{d^4 y}{d x^4}\) which is 2.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1

Question 10.
(i) 3x \(\frac{d y}{d x}\) + \(\frac{5}{\frac{d y}{d x}}\) = y3
(ii) y = x \(\frac{d y}{d x}\) + a \(\sqrt{1+\left(\frac{d y}{d x}\right)^2}\)
Solution:
(i) Given differential eqn. be,
3x \(\frac{d y}{d x}\) + \(\frac{5}{\frac{d y}{d x}}\) = y3
⇒ 3x (\(\frac{d y}{d x}\))2 + 5 = y3 \(\frac{d y}{d x}\)
which is polynomial in derivatives.
The highest ordered derivative present in given diff. eqn. be \(\frac{d y}{d x}\).
so its order be 1.
The degree of given diff. eqn. be the highest exponent of \(\frac{d y}{d x}\) which is 2.
Thus its degree be 2.

(ii) Given differential eqn. be,
y = x \(\frac{d y}{d x}\) + a \(\sqrt{1+\left(\frac{d y}{d x}\right)^2}\)
⇒ y – x \(\frac{d y}{d x}\) = a \(\sqrt{1+\left(\frac{d y}{d x}\right)^2}\)
On squaring, we have
(y – x \(\frac{d y}{d x}\))2 = a2 [1 + (\(\frac{d y}{d x}\))2]
The highest ordered derivative exiosting in given diff. eqn. be \(\frac{d y}{d x}\) and its power 2.
∴ it is of order 1 and degree 2.
Clearly it is a non-linear differential equation.

Question 11.
(i) \(\sqrt{1+\left(\frac{d y}{d x}\right)^2}\) = 3x – \(\frac{d y}{d x}\)
(ii) \(5 \frac{d^2 y}{d x^2}=\left(1+\left(\frac{d y}{d x}\right)^2\right)^{\frac{1}{4}}\)
Solution:
(i) Given differential eqn. be,
\(\sqrt{1+\left(\frac{d y}{d x}\right)^2}\) = 3x – \(\frac{d y}{d x}\)
On squaring both sides ;
1 + (\(\frac{d y}{d x}\))2 = 9x2 + (\(\frac{d y}{d x}\))2 – 6x \(\frac{d y}{d x}\)
⇒ 6x \(\frac{d y}{d x}\) = 9x2 + 1 = 0
The highest ordered derivative existing in given diff. eqn. be \(\frac{d y}{d x}\) so its order be 1.
Further given duff. eqn. can be expressed as polynomial in derivative.
So degree be the highest exponent of which is 1.

(ii) Given differential eqn. can be written as
\(5 \frac{d^2 y}{d x^2}=\left(1+\left(\frac{d y}{d x}\right)^2\right)^{\frac{1}{4}}\)
Clearly the highest ordered derivative existing in given diff. eqn. be \(\frac{d^2 y}{d x^2}\) and
hence its order be 2.
Here, given diff. eqn. can be expressed as polynomial in derivatives.
So degree of given diff. eqn. be the highest power of which is 4.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1

Question 11 (old).
\(\left(\frac{d^2 x}{d t^2}\right)^4-7 t\left(\frac{d x}{d t}\right)^3\) = log t.
Solution:
Given differential eqn. be,
\(\left(\frac{d^2 x}{d t^2}\right)^4-7 t\left(\frac{d x}{d t}\right)^3\) = log t
which is polynomial in derivatives.
The highest ordered derivative present in given diff. eqn. be \(\frac{d^2 y}{d x^2}\) and its order be 2.
The degree of given diff. eqn. be the highest exponent of \(\frac{d^2 y}{d x^2}\) which is 4.
Thus its degree be 4.

Question 12.
(i) y” + (y’)2 + 2y = 0 (NCERT)
(ii) y” + 2y’ + sin y = 0
Solution:
(i) Given differential eqn. be,
y” + (y’)2 + 2y = 0
which is polynomial in derivatives.
The highest order derivative present in given diff. eqn. be y” and its order be 2.
The degree of given differential equation is the highest exponent of y” which is 1.
Thus its degree be 1.

(ii) Given differential eqn. be,
y” + 2y’ + sin y = 0
which is polynomial in derivatives.
The highest order derivative present in given diff. eqn. be y” and its order is 2.
The degree of given diff. eqn. be the highest exponent of y” which is 1.
Thus its degree be 1.

Question 12 (old).
y(iv) + sin y”’ = 0 (NCERT)
Solution:
Given diff. eqn. be,
y(iv) + sin y”’ = 0
The order of given diff. eqn. be the highest order derivative present in given diff. eqn. which is y(iv) and its order be 4.
Here the term sin y”’ is not polynomial in y”’.
Thus, the degree of given differential eqn. is not defined.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1

Question 13.
(i) (y”’)2 + (y”)3 + (y’)4 + y5 = 0 (NCERT)
(ii) (1 – (y’)2)3/2 = ky”
Solution:
(i) Given diff. eqn. be,
(y”’)2 + (y”)3 + (y’)4 + y5 = 0
which is polynomial in derivatives.
The highest order derivative present in given diff. eqn. be y”’ so its order be 3.
The degree of given diff. eqn. is the highest exponent of y”’, which is 2.
Thus degree of given diff. eqn. be 2.

(ii) Given diff. eqn. can be written as
(1 – (y’)2)3 = (ky”)2
Clearly the highest ordered derivative existing in given duff. eqn. be y” so its order 2.
Here the given duff. eqn. can be expressed as polynomial in derivatives.
Thus its degree be the highest exponent of y” which is 2.

Question 14.
Write the sum of the order and the degree of the differential equation \(\left(\frac{d y}{d x}\right)^5+3 x y\left(\frac{d^3 y}{d x^3}\right)^2+y\left(\frac{d^2 y}{d x^2}\right)^4\) = 0.
Solution:
Clearly the highest ordered derivative existing in given diff. eqn. be \(\frac{d^3 y}{d x^3}\).
so its order be 3.
Clearly the given duff. eqn. can be expressed as polynomial in derivatives.
So degree of given diff. eqn. be the highest exponent of \(\frac{d^3 y}{d x^3}\) which is 2.
∴ required sum = 3 + 2 = 5.

Question 15.
Find the product of the order and degree of the following differential equation :
x (\(\frac{d^2 y}{d x^2}\))2 + (\(\frac{d y}{d x}\))2 + y2 = 0.
Solution:
The given differential eqn. be
x (\(\frac{d^2 y}{d x^2}\))2 + (\(\frac{d y}{d x}\))2 + y2 = 0
which is polynomial in derivatives.
The highest order derivative present in given diff. eqn. be \(\frac{d^2 y}{d x^2}\).
So its order be 2.
The degree of given diff. eqn. be the highest exponent of \(\frac{d^2 y}{d x^2}\) which is 2.
Thus, its degree be 2.
∴ required product of order and degree of given differential equation 2 × 2 = 4.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1

Question 20 (old).
y = px + \(\sqrt{a^2 p^2+b^2}\), where p = \(\frac{d y}{d x}\).
Solution:
Given diff. eqn. can be written as
(y – px)2 = a2p2 + b2; where p = \(\frac{d y}{d x}\)
Clearly it is a polynomial in p.
The highest order derivative existing in diff. eqn. be p i.e. \(\frac{d y}{d x}\) and its power 2.
Thus given diff. eqn. is of order 1 and degree 2.
Clearly it is a non-linear differential equation.

Question 21 (old).
Write the sum of the order and the degree of the differential equation \(\left(\frac{d^2 y}{d x^2}\right)^2-\left(\frac{d y}{d x}\right)^3\) = y3.
Solution:
Given differential equation be,
\(\left(\frac{d^2 y}{d x^2}\right)^2-\left(\frac{d y}{d x}\right)^3\) = y3
which is polynomial in derivatives.
The highest order derivative present in given
diff. eqn. be \(\frac{d^2 y}{d x^2}\).
so its order be 2.
The degree of given diff. eqn. be the highest power of \(\frac{d^2 y}{d x^2}\) which is 2.
So its degree be 2 and required sum = 2 + 2 = 4.

OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(a)

Practicing ISC Mathematics Class 11 OP Malhotra Solutions Chapter 27 Mathematical Reasoning Ex 27(a) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(a)

Question 1.
Open the door.
Solution:
It is an imperative sentence so it does not represents a statement.

Question 2.
5 is a prime number.
Solution:
It is a statement.

Question 3.
Do you like mathematics ?
Solution:
It is not a statement as it is an Interrogative sentence.

OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(a)

Question 4.
Every rectangle is a square.
Solution:
It is a statement.

Question 5.
Today is Sunday and tomorrow is Monday.
Solution:
It is a compound statement as its component statement are;
(i) Today is Sunday
(ii) Tomorrow is Monday.

Question 6.
May you live long!
Solution:
It is not a statement. It is an exclamatory sentence.

Question 7.
Rekha is studying in class eleven and she has to offer 5 subjects.
Solution:
It is a compound statement and its components are ;
(i) Rekha is studying in class eleven.
(ii) She has to offer 5 subjects.

Question 8.
The earth revolves around the moon.
Solution: It is a simple statement.

Question 9.
New Delhi is a big city and it is the capital of India.
Solution:
It is a compound statement and its components are ;
(i) New Delhi is a big city.
(ii) It is the capital of India.

Question 10.
20 is a prime number and 20 is less than 21.
Solution:
It represents a compound statement and its components are ;
(i) 20 is a prime number
(ii) 20 is less than 21.
Write down whether the following statements are true (T) or false (F).

Question 11.
8 is a prime number.
Solution:
It is a false statement.

OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(a)

Question 12.
Every square is a rectangle.
Solution:
It is a true statement.

Question 13.
The earth revolves around the moon.
Solution:
It is a false statement.

Question 14.
The set of whole numbers is a finite set.
Solution:
It is a false statement as the set of whole numbers is an infinite set.

Question 15.
32 is a multiple of 8.
Solution:
It is a true statement.

Question 16.
3 + 4i is a complex number.
Solution:
It is a true statement.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Choose the correct answer from the given four options in questions (1 to 55) :

Question 1.
∫ sin \(\frac{x}{2}\) cos \(\frac{x}{2}\) cos x dx is equal to
(a) – \(\frac{1}{4}\) cos 2x + C
(b) – \(\frac{1}{8}\) cos 2x + C
(c) \(\frac{1}{8}\) cos 2x + C
(d) – \(\frac{1}{8}\) cos 2x + C
Solution:
(b) – \(\frac{1}{8}\) cos 2x + C

∫ sin \(\frac{x}{2}\) cos \(\frac{x}{2}\) cos x dx
= \(\frac{1}{2}\) ∫ (2 sin \(\frac{x}{2}\) cos \(\frac{x}{2}\)) cos x dx
= \(\frac{1}{2}\) ∫ sin x cos x dx
= \(\frac{1}{4}\) ∫ sin 2x dx
= \(\frac{1}{4}\left(-\frac{\cos 2 x}{2}\right)\) + C
= – \(\frac{1}{8}\) cos 2x + C

Question 2.
∫ \(\frac{d x}{\sin ^2 x \cos ^2 x}\) is equal to
(a) tan x + cot x + C
(b) (tan x + cot x)2 + C
(c) tan x – cot x + C
(d) (tan x – cot x)2 + C
Solution:
∫ \(\frac{d x}{\sin ^2 x \cos ^2 x}\)
= ∫ \(\frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x}\) dx
= ∫ \(\left[\frac{1}{\cos ^2 x}+\frac{1}{\sin ^2 x}\right]\) dx
= ∫ [sec2 x + cosec2 x] + C
= tan x – cot x + C

Question 3.
∫ \(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx is equal to
(a) 2 (sin x + x cos α) + C
(b) 2 (sin x – x cos α) + C
(c) 2 (sin x + 2x cos α) + C
(d) 2 (sin x – 2x cos α) + C
Solution:
(a) 2 (sin x + x cos α) + C

∫ \(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx
= 2 ∫ \(\frac{2 \cos ^2 x-1-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha}\) dx
= 2 ∫ \(\frac{(\cos x-\cos \alpha)(\cos x+\cos \alpha)}{\cos x-\cos \alpha}\) dx
= 2 [sin x + cos x] + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 4.
∫ \(\frac{x}{4+x^4}\) dx is equal to
(a) \(\frac{1}{4}\) tan-1 x2 + C
(b) \(\frac{1}{2}\) tan-1 \(\frac{x^2}{2}\) + C
(c) \(\frac{1}{4}\) tan-1 \(\left(\frac{x^2}{2}\right)\) + C
(d) \(\frac{1}{2}\) tan-1 2x2 + C
Solution:
(c) \(\frac{1}{4}\) tan-1 \(\left(\frac{x^2}{2}\right)\) + C

∫ \(\frac{x}{4+x^4}\) dx ;
put x2 = t
⇒ 2x dx = dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 1

Question 5.
∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) is equal to
(a) 2 cos √x + C
(b) 2 sin √x + C
(c) – 2 sin √x + C
(d) sin √x + C
Solution:
(b) 2 sin √x + C

put √x = t
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
∴ ∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx = ∫ cos t . 2t dt
= 2 sin t + C
= 2 sin √x + C

Question 6.
If ∫ x ekx2 dx = \(\frac{1}{4}\) e2x2 + C, then the value of k is,
(a) 4
(b) – 2
(c) 2
(d) 1
Solution:
(c) 2

Given ∫ x ekx2 dx = \(\frac{1}{4}\) e2x2 + C
put x2 = t
⇒ 2x dx = dt
⇒ ∫ ekt \(\frac{dt}{2}\) = \(\frac{1}{4}\) e2x2 + C
⇒ \(\frac{1}{2} \frac{e^{k t}}{k}\) = \(\frac{1}{4}\) e2x2 + C
⇒ \(\frac{1}{2 k}\) e2x2 + C = \(\frac{1}{4}\) e2x2 + C
∴ 2k = 4
⇒ k = 2

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 7.
If ∫ x6 sin (5x7) dx = \(\frac{k}{5}\) cos (5x7) + C, then the value of k is
(a) – \(\frac{1}{7}\)
(b) \(\frac{1}{7}\)
(c) \(\frac{1}{35}\)
(d) – \(\frac{1}{35}\)
Solution:
(a) – \(\frac{1}{7}\)

∫ x6 sin (5x7) dx = \(\frac{k}{5}\) cos (5x7) + C
put x7 = t
⇒ 7x6 dx = dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 2

Question 8.
If ∫ \(\frac{2^x}{\sqrt{1-4^x}}\) dx = k sin-1 (2x) + C, then the value of k is
(a) log 2
(b) \(\frac{1}{\log 2}\)
(c) \(\frac{2}{\log 2}\)
(d) \(\frac{\log 2}{2}\)
Solution:
(b) \(\frac{1}{\log 2}\)

Given ∫ \(\frac{2^x}{\sqrt{1-4^x}}\) dx = k sin-1 (2x) + C
put 2x = t
⇒ 2x log 2 dx = dt
⇒ ∫ \(\frac{d t}{\log 2 \sqrt{1-t^2}}\) = k sin-1 (2x) + C
⇒ \(\frac{1}{\log 2}\) sin-1 + C = k sin-1(2x + C
⇒ \(\frac{1}{\log 2}\) sin-1 + C = k sin-1 2x + C
∴ k = \(\frac{1}{\log 2}\)

Question 9.
If ∫ |x| dx = kx |x| + C, x ≠ 0, then the value of k is
(a) 2
(b) – 2
(c) – \(\frac{1}{2}\)
(d) \(\frac{1}{2}\)
Solution:
(d) \(\frac{1}{2}\)

Let I = ∫ |x| . 1 dx
= |x| . x – ∫ \(\frac{x}{|x|}\) . x dx
= x |x| – ∫ \(\frac{|x|^2}{|x|}\) dx
[∵ x2 = |x|2]
= x |x| – ∫ |x| dx
[∵ x ≠ 0]
⇒ I = x |x| – I
⇒ 2I = x |x|
⇒ I = \(\frac{x|x|}{2}\) + C
Now I = kx |x| + C
⇒ k = \(\frac{1}{2}\).

Question 10.
∫ cot x log (sin x) dx is equal to
(a) \(\frac{1}{2}\) (log (sin x)2 + C
(b) log (sin x) + C
(c) \(\frac{1}{2}\) (log (cos x))2 + C
(d) none of these
Solution:
(a) \(\frac{1}{2}\) (log (sin x)2 + C

Let I = ∫ cot x log (sin x) dx
put log sin x = t
⇒ \(\frac{1}{\sin x}\) cos x dx = dt
⇒ cot x dx = dt
= ∫ t dt
= \(\frac{t^2}{2}\) + C
= \(\frac{1}{2}\) [log (sin x)]2 + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 11.
∫ \(\frac{x+\sin x}{1+\cos x}\) dx is equal to
(a) log |1 + cos x| + C
(b) log |x + sin x| + C
(c) x – tan \(\frac{x}{2}\) + C
(d) x tan \(\frac{x}{2}\) + C
Solution:
(d) x tan \(\frac{x}{2}\) + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 3

Question 12.
∫ \(\left(\frac{1-x}{1+x^2}\right)^2\) ex dx is equal to
(a) \(\frac{e^x}{1+x^2}\) + C
(b) – \(\frac{e^x}{1+x^2}\) + C
(c) \(\frac{e^x}{\left(1+x^2\right)^2}\) + C
(d) – \(\frac{e^x}{\left(1+x^2\right)^2}\) + C
Solution:
(a) \(\frac{e^x}{1+x^2}\) + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 4

Question 13.
∫ (x – 1)e-x dx is equal to
(a) (x – 2) e-x + C
(b) xe-x + C
(c) – xe-x + C
(d) (x + 1)e-x + C
Solution:
(c) – xe-x + C

∫ (x – 1)e-x dx = (x – 1) \(\frac{e^{-x}}{-1}\) – ∫ 1 . \(\frac{e^{-x}}{-1}\) dx +C
= – (x – 1) e-x – e-x + C
= – (x – 1 + 1) e-x + C
= – x e-x + C

Question 14.
∫ ex (1 – cot x + cot2 x) dx is equal to
(a) ex cosec x + C
(b) – ex cosec x + C
(c) ex cot x + C
(d) – ex cot x + C
Solution:
(d) – ex cot x + C

Let I = ∫ ex (1 – cot x + cot2 x) dx
= ∫ ex (1 – cot x + cosec2 x – 1) dx
= ∫ ex (cosec2 x) dx – ∫ ex cot x dx
= ex (- cot x) – ∫ ex (- cot x) dx
= ex (- cot x) – ∫ ex (- cot x) dx – ∫ ex cot x dx + C
= – ex cot x + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 15.
If ∫ \(\frac{1+\cos 4 x}{\cot x-\tan x}\) dx = k cos 4x + C, then the value of k is
(a) \(\frac{1}{4}\)
(b) – \(\frac{1}{2}\)
(c) – \(\frac{1}{8}\)
(d) – \(\frac{1}{4}\)
Solution:
(c) – \(\frac{1}{8}\)

Let I = ∫ \(\frac{1+\cos 4 x}{\cot x-\tan x}\) dx
= ∫ \(\frac{2 \cos ^2 2 x}{\left(\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}\right)}\) dx
= ∫ \(\frac{\frac{2 \cos ^2 2 x d x}{\cos ^2 x-\sin ^2 x}}{\sin x \cos x}\) dx
= ∫ \(\frac{2 \cos ^2 2 x d x}{\frac{\cos 2 x}{\sin x \cos x}}\) dx
= ∫ cos 2x (2 sin x cos x) dx
= ∫ cos 2x sin 2x dx
= \(\frac{1}{2}\) ∫ sin 4x dx
= – \(\frac{1}{2} \frac{\cos 4 x}{4}\) + C
= – \(\frac{1}{8}\) cos 4x + C
Also, I = k cos 4x + C
∴ k = – \(\frac{1}{8}\)

Question 16.
∫ \(\frac{d x}{e^x+e^{-x}+2}\) is equal to
(a) \(\frac{1}{e^x+1}\) + C
(b) \(\frac{1}{1+e^{-x}}\) + C
(c) – \(\frac{1}{e^x+1}\) + C
(d) none of these
Solution:
(c) – \(\frac{1}{e^x+1}\) + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 5

Question 17.
∫ \(\frac{(\log x)^5}{x}\) dx is equal to
(a) \(\frac{\log x^6}{6}\) + C
(b) \(\frac{(\log x)^6}{6}\) + C
(c) \(\frac{(\log x)^6}{3 x^2}\) + C
(d) none of these
Solution:
(b) \(\frac{(\log x)^6}{6}\) + C

∫ \(\frac{(\log x)^5}{x}\) dx ;
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
= ∫ t5 dt
= \(\frac{t^6}{6}\) + C
= \(\frac{(\log x)^6}{6}\) + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 18.
∫ \(\frac{d x}{\sqrt{2 x-x^2}}\) is equal to
(a) sin-1 (x – 1) + C
(b) sin-1 (x – 1) + C
(c) – \(\sqrt{2 x-x^2}\) + C
(d) none of these
Solution:
(a) sin-1 (x – 1) + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 6

Question 19.
∫ \(\frac{x^2+1}{x^2-1}\) dx is equal to
(a) x + log \(\left|\frac{x+1}{x-1}\right|\) + C
(b) x + log \(\left|\frac{x-1}{x+1}\right|\) + C
(c) log |(x – 1) (x + 1)| + C
(d) log |x2 + 1| + C
Solution:
(b) x + log \(\left|\frac{x-1}{x+1}\right|\) + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 7

Question 20.
∫ (sin6 x + cos6 x + 3 sin2 x cos2 x) dx is equal to
(a) x + C
(b) \(\frac{3}{2}\) sin 2x + C
(c) – \(\frac{3}{2}\) cos 2x + C ‘
(d) none of these
Solution:
(a) x + C

∫ (sin6 x + cos6 x + 3 sin2 x cos2 x) dx
= ∫ [(sin2 x)3 + (cos2 x)3 + 3 sin2 x cos2 x (sin2 x + cos2 x)] dx
= ∫ (sin2 x+ cos2 x)3 dx
= ∫ dx
= x + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 21.
∫ \(\frac{d x}{x\left(x^7+1\right)}\) is equal to
(a) log \(\left|\frac{x^7}{x^7+1}\right|\) + C
(b) \(\frac{1}{7} \log \left|\frac{x^7}{x^7+1}\right|\) + C
(c) log \(\left|\frac{x^7+1}{x^7}\right|\) + C
(d) \(\frac{1}{7} \log \left|\frac{x^7+1}{x^7}\right|\) + C
Solution:
(b) \(\frac{1}{7} \log \left|\frac{x^7}{x^7+1}\right|\) + C

Let I = ∫ \(\frac{d x}{x\left(x^7+1\right)}\) ;
put x7 = t
⇒ 7x6 dx = dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 8

Question 22.
∫ (sin4 x – cos4 x) dx is equal to
(a) \(\frac{1}{2}\) cos 2x + C
(b) – \(\frac{1}{2}\) cos 2x + C
(c) \(\frac{1}{2}\) sin 2x + C
(d) – \(\frac{1}{2}\) sin 2x + C
Solution:
(d) – \(\frac{1}{2}\) sin 2x + C

∫ (sin4 x – cos4 x) dx = ∫ (sin2 x – cos2 x) (sin2 x + cos2 x) dx
= – ∫ cos 2x dx
= – \(\frac{\sin 2 x}{2}\) + C

Question 23.
∫ \(\frac{\left(\tan ^{-1} x\right)^3}{1+x^2}\) dx is equal to
(a) 3 (tan-1 x)2 + C
(b) (tan-1 x)4 + C
(c) (tan-1 x)4 + C
(d) none of these
Solution:
(b) (tan-1 x)4 + C

put tan-1 x = t
⇒ \(\frac{1}{1+x^2}\) dx = dt
∴ I = ∫ \(\frac{\left(\tan ^{-1} x\right)^3}{1+x^2}\) dx
= ∫ t3 dt
= \(\frac{t^4}{4}\) + C
= \(\frac{\left(\tan ^{-1} x\right)^4}{4}\) + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 24.
∫ e3 log x (x4 + 1)-1 dx is equal to
(a) \(\frac{1}{4}\) log (x4 + 1) + C
(b) – \(\frac{1}{4}\) (x4 + 1) + C
(c) log (x4 + 1) + C
(d) none of these
Solution:
(a) \(\frac{1}{4}\) log (x4 + 1) + C

∫ e3 log x (x4 + 1)-1 dx
= ∫ elog x3 \(\frac{1}{x^4+1}\) dx
= ∫ \(\frac{x^3 d x}{x^4+1}\) ;
put x4 + 1 = t
⇒ 4x3 dx = dt
= \(\frac{1}{4} \int \frac{d t}{t}\)
= \(\frac{1}{4}\) log |t| + C
= \(\frac{1}{4}\) log |x4 + 1| + C

Question 25.
∫ (sin (log x) + cos (log x)) dx is equal to
(a) sin (log x) + C
(b) cos (log x) + C
(c) x cos (log x) + C
(d) x sin (log x) + C
Solution:
(d) x sin (log x) + C

Let I = ∫ (sin (log x) + cos (log x)) dx
= sin (log x) . x – ∫ cos (log x) . \(\frac{1}{x}\) . x dx + ∫ cos (log x) dx
= x sin (log x) + C

Question 26.
∫ \(\frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}\) dx is equal to
(a) \(\frac{1}{2} \sqrt{1+x}\) + C
(b) \(\frac{1}{2}\) (1 + x)3/2 + C
(c) 2 (1 + x)3/2 + C
(d) \(\sqrt{1 + x}\) + C
Solution:
(b) \(\frac{1}{2}\) (1 + x)3/2 + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 9

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 27.
∫ \(\frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}\) dx is equal to
(a) f(x) log (f(x)) + C
(b) log (log (f(x)) + C
(c) \(\frac{f(x)}{\log (f(x))}\) + C
(d) \(\frac{1}{\log (\log (f(x)))}\) + C
Solution:
(b) log (log (f(x)) + C

Let I = ∫ \(\frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}\) dx
put log (f(x)) = t
⇒ \(\frac{f^{\prime}(x)}{f(x)}\) dx = dt
= ∫ \(\frac{1}{t}\) dt
= |log |t| + C
= log |log f(x)| + C

Question 28.
∫ x3 log x dx is equal to
(a) \(\frac{x^4 \log x}{4}\) + C
(b) \(\frac{x^4}{8}\left(\log x-\frac{4}{x^2}\right)\) + C
(c) \(\frac{x^4}{16}\) (4 log x – 1) + C
(d) \(\frac{x^4}{16}\) (4 log x + 1) + C
Solution:
(c) \(\frac{x^4}{16}\) (4 log x – 1) + C

∫ x3 log x dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 10

Question 29.
∫ ex log a ex dx is equal to
(a) \(\frac{a^x}{\log a e}\) + C
(b) \(\frac{e^x}{1+\log a}\)
(c) (ae)x + C
(d) \(\frac{(a e)^x}{\log a e}\) + C
Solution:
(d) \(\frac{(a e)^x}{\log a e}\) + C

∫ ex log a ex dx = ∫ elog ax . ex dx
= ∫ ax . ex dx
= ∫ (ae)x dx
= \(\frac{(a e)^x}{\log a e}\) + C

Question 30.
∫ \(\left(\frac{1-\sin x}{1-\cos x}\right)\) ex dx is equal to
(a) – ex tan \(\frac{x}{2}\) + C
(b) – ex cot \(\frac{x}{2}\) + C
(c) – \(\frac{1}{2}\) ex tan \(\frac{x}{2}\) + C
(d) – \(\frac{1}{2}\) ex cot \(\frac{x}{2}\) + C
Solution:
(b) – ex cot \(\frac{x}{2}\) + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 11

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 31.
∫ \(\left(\frac{1+x+x^2}{1+x^2}\right)\) etan-1 x dx is equal to
(a) x + etan-1 x + C
(b) etan-1 x – x + C
(c) etan-1 x + C
(d) x etan-1 x + C
Solution:
(d) x etan-1 x + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 12

Question 32.\(\)
If \(\int_0^{40} \frac{d x}{2 x+1}\) = log k, then value of k is
(a) 3
(b) \(\frac{9}{2}\)
(c) 9
(d) none of these
Solution:
(c) 9

Given \(\int_0^{40} \frac{d x}{2 x+1}\) = log k
⇒ \(\left.\frac{\log |2 x+1|}{2}\right]_0^{40}\) = log k
⇒ \(\frac{1}{2}\)[log 81 – log 1] = log k
⇒ \(\frac{1}{2}\) log 92 = log k
⇒ log 9 = log k
∴ k = 9

Question 33.
\(\int_0^{\sqrt{3}} \frac{d x}{1+x^2}\) is equal to
(a) \(\frac{\pi}{3}\)
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{6}\)
(d) \(\frac{\pi}{12}\)
Solution:
(d) \(\frac{\pi}{12}\)

\(\left.\int_0^{\sqrt{3}} \frac{d x}{1+x^2}=\tan ^{-1} x\right]_1^{\sqrt{3}}\)
= tan-1 √3 – tan-1 1
= \(\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 34.
If \(\int_0^k \frac{1}{9 x^2+1} d x=\frac{\pi}{12}\), then k is equal to
(a) \(\frac{\pi}{4}\)
(b) \(\frac{1}{3}\)
(c) 3
(d) none of these
Solution:
(b) \(\frac{1}{3}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 13

Question 35.
\(\int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin x}\) dx is equal to
(a) 3 – log 12
(b) 1 – log 2
(c) 1 + log 2
(d) none of these
Solution:
(b) 1 – log 2

\(\int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin x}\) dx ;
put sin x = t
⇒ cos x dx = dt
When x = 0 ⇒ t = 0 ;
When x = \(\frac{\pi}{2}\) ⇒ t = 1
= \(\int_0^1 \frac{t d t}{1+t}\)
= \(\int_0^1\left[\frac{1+t-1}{1+t}\right]\) dt
= \(\int_0^1\left[1-\frac{1}{1+t}\right]\) dt
= t – log |1 + t|\(\)
= 1 – log 2 – 0 – 0
= 1 – log 2

Question 36.
The value of \(\int_{\pi / 6}^{\pi / 3} \frac{1}{\sin 2 x}\) dx is
(a) log 3
(b) \(\frac{1}{2}\) log 2
(c) \(\frac{1}{2}\) log 3
(d) none of these
Solution:
(c) \(\frac{1}{2}\) log 3

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 14

Question 37.
\(\int_{-1}^0 \frac{d x}{x^2+2 x+2}\) is equal to
(a) 0
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{4}\)
(d) – \(\frac{\pi}{4}\)
Answer:
(c) \(\frac{\pi}{4}\)

\(\int_{-1}^0 \frac{d x}{x^2+2 x+2}\) = \(\int_{-1}^0 \frac{d x}{(x+1)^2+1^2}\)
= tan-1 x\(]_{-1}^0\)
= tan-1 1 – tan-1 0
= \(\frac{\pi}{4}\) – 0 = \(\frac{\pi}{4}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 38.
\(\int_0^1 \frac{\tan ^{-1} x}{1+x^2}\) dx is equal to
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi^2}{32}\)
(c) 1
(d) \(\frac{\pi^2}{16}\)
Solution:
(b) \(\frac{\pi^2}{32}\)

\(\int_0^1 \frac{\tan ^{-1} x}{1+x^2}\) dx ;
put tan-1 x = t
⇒ \(\frac{1}{1+x^2}\) dx = dt
When x = 0
⇒ t = tan-1 0 = 0
When x = 1
⇒ t = tan-1 1 = \(\frac{\pi}{4}\)
= \(\left.\int_0^{\pi / 4} t d t=\frac{t^2}{2}\right]_0^{\pi / 4}\)
= \(\frac{\pi^2}{32\)

Question 39.
\(\int_0^1 \frac{d x}{e^x+e^{-x}}\) is equal to
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{2}\)
(c) tan-1 \(\left(\frac{e-1}{e+1}\right)\)
(d) tan-1 \(\left(\frac{1-e}{1+e}\right)\)
Solution:
(c) tan-1 \(\left(\frac{e-1}{e+1}\right)\)

\(\int_0^1 \frac{d x}{e^x+e^{-x}}=\int_0^1 \frac{e^x d x}{e^{2 x}+1}\) ;
put ex = t
⇒ ex dx = dt
When x = 0 ⇒ t = 1
When x = 1 ⇒ t = e
= \(\int_1^e \frac{d t}{t^2+1}\)
= tan-1 t\(]_1^e\)
= tan-1 e – tan-1 1
= tan-1 e – \(\frac{\pi}{4}\)
= tan-1 \(\left(\frac{e-1}{e+1}\right)\)

Question 40.
\(\int_0^1 \frac{x^4+1}{x^2+1}\) dx is equal to
(a) \(\frac{1}{6}\) (3 – 4π)
(b) \(\frac{1}{6}\) (3π + 4)
(c) \(\frac{1}{6}\) (3 + 4π)
(d) \(\frac{1}{6}\) (3π – 4)
Solution:
(d) \(\frac{1}{6}\) (3π – 4)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 15

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 41.
\(\int_{a+c}^{b+c}\) f(x) dx is equal to
(a) \(\int_a^b\int_a^b\) f(x – c) dx
(b) \(\int_a^b\) f(x + c) dx
(c) \(\int_a^b\) f(x) dx
(d) \(\int_{a-c}^{b-c}\) f(x) dx
Solution:
(b) \(\int_a^b\) f(x + c) dx

put x = t + c
⇒ dx = dt
When x = a + c
⇒ t = a
When x = b + c
⇒ t = b
∴ I = \(\int_a^b\) f(t + c) dt
= \(\int_a^b\) f(x + c) dx
[∵ \(\int_a^b\) f(x) dx = \(\int_a^b\) f(t) dt]

Question 42.
\(\int_{-\pi / 2}^{\pi / 2}\) cos x dx is equal to
(a) 0
(b) 1
(c) 2
(d) 4
Solution:
(c) 2

\(\int_{-\pi / 2}^{\pi / 2}\) cos x dx = + sin x\(]_{-\pi / 2}^{\pi / 2}\)
= sin \(\frac{\pi}{2}\) – sin (- \(\frac{\pi}{2}\))
= 1 + sin \(\frac{\pi}{2}\)
= 1 + 1 = 2

Question 43.
\(\int_{-\pi / 4}^{\pi / 4} \frac{d x}{1+\cos 2 x}\) is equla to
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(a) 1

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 16

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Question 44.
\(\int_0^{\pi / 2}\) log |tan x| dx is equal to
(a) π – log 2
(b) – π log 2
(c) 0
(d) none of these
Solution:
(c) 0

Let I = \(\int_0^{\pi / 2}\) log |tan x| dx
I = \(\int_0^{\pi / 2}\) log tan x dx ………………(1)
[when 0 ≤ x < \(\frac{\pi}{2}\) ⇒ tan x > 0
⇒ |tan x| = tan x]
We know that
\(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx
∴ I = \(\int_0^{\pi / 2}\) log tan (\(\frac{\pi}{2}\) – x) dx
= \(\frac{\pi}{2}\) log cot x dx ………………(2)
On adding (1) and (2) ; we have
2I = \(\int_0^{\pi / 2}\) (log tan x + log cot x) dx
= \(\int_0^{\pi / 2}\) log (tan x . cot x) dx
= \(\int_0^{\pi / 2}\) log 1 dx = 0
⇒ I = 0

Question 45.
\(\int_{-\pi / 2}^{\pi / 2} \sqrt{\frac{1-\cos 2 x}{2}}\) dx is equal to
(a) 0
(b) \(\frac{1}{2}\)
(c) 1
(d) 2
Solution:
(d) 2

\(\int_{-\pi / 2}^{\pi / 2} \sqrt{\frac{1-\cos 2 x}{2}}\) dx
= \(\int_{-\pi / 2}^{\pi / 2} \sqrt{\frac{2 \sin ^2 x}{2}}\)
= \(\int_{-\pi / 2}^{\pi / 2}\) |sin x| dx
= \(\int_{-\pi / 2}^0\) |sin x| dx + \(\int_0^{\pi / 2}\) |sin x| dx
When – \(\frac{\pi}{2}\) ≤ x < 0
⇒ sin x ≥ 0
⇒ |sin x| = – sin x
When 0 ≤ x ≤ \(\frac{\pi}{2}\)
⇒ sin x ≥ 0
⇒ |sin x| = sin x
= \(\int_{-\pi / 2}^0\) – sin x dx + \(\int_0^{\pi / 2}\) sin x dx
= \(\left.\cos x]_{-\pi / 2}^0+(-\cos x)\right]_0^{\pi / 2}\)
= (1 – 0) + (- 0 + 1) = 2

Question 46.
\(\int_{-\pi}^\pi \frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}\) dx is equal to
(a) 2π
(b) π
(c) \(\frac{\pi}{2}\)
(d) \(\frac{\pi}{4}\)
Solution:
(b) π

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 17

Question 47.
\(\int_a^b \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}}\) dx is equal to
(a) \(\frac{\pi}{2}\)
(b) π
(c) \(\frac{1}{2}\) (b – a)
(d) b – a
Solution:
(c) \(\frac{1}{2}\) (b – a)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 18

Question 48.
\(\int_0^{\pi / 2} \frac{\cos x-\sin x}{2+\sin x \cos x}\) dx is equal to
(a) 0
(b) \(\frac{\pi}{6}\)
(c) \(\frac{\pi}{4}\)
(d) \(\frac{\pi}{2}\)
Solution:
(a) 0

Let I = \(\int_0^{\pi / 2} \frac{\cos x-\sin x}{2+\sin x \cos x}\) dx ……………..(1)
We know that
\(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx
∴ I = \(\int_0^{\pi / 2} \frac{\cos \left(\frac{\pi}{2}-x\right)-\sin \left(\frac{\pi}{2}-x\right)}{2+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}\)
I = \(\int_0^{\pi / 2} \frac{\sin x-\cos x}{2+\cos x \sin x}\) ………………….(2)
On adding (1) and(2) ; we have
2I = \(\int_0^{\pi / 2} \frac{\cos x-\sin x+\sin x-\cos x}{2+\cos x \sin x}\) dx = 0
⇒ I = 0

Question 49.
The value of \(\int_0^2\left|\cos \frac{\pi}{2} t\right|\) dt is equal to
(a) \(\frac{3}{4 \pi}\)
(b) \(\frac{4}{\pi}\)
(c) \(\frac{\pi}{2}\)
(d) 2π
Solution:
(b) \(\frac{4}{\pi}\)

Since, 0 ≤ t ≤ 2
⇒ 0 ≤ \(\frac{\pi}{2}\) t ≤ π
⇒ cos \(\frac{\pi}{2}\) t > 0
and 1 ≤ t ≤ 2
⇒ \(\frac{\pi}{2}\) ≤ \(\frac{\pi}{2}\) t ≤ π
⇒ cos \(\frac{\pi}{2}\) t < 0

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 19

Question 50.
\(\int_0^{\pi / 2} \sqrt{1-\sin 2 x}\) dx is equal to
(a) 2√2
(b) 2 (√2 + 1)
(c) 2
(d) 2 (√2 – 1)
Solution:
(d) 2 (√2 – 1)

Let I = \(\int_0^{\pi / 2} \sqrt{1-\sin 2 x}\) dx
= \(\int_0^{\pi / 2} \sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x}\)
= \(\int_0^{\pi / 2} \sqrt{(\sin x-\cos x)^2}\) dx
= \(\int_0^{\pi / 2}\) |sin x – cos x| dx
When 0 ≤ x < \(\frac{\pi}{4}\) ⇒ cos x > sin x
⇒ sin x – cos x < 0
∴ |sin x – cos x| = – (sin x – cos x)
and \(\frac{\pi}{4}\) ≤ x ≤ \(\frac{\pi}{2}\)
⇒ sin x > cos x
⇒ sin x – cos x > 0
|sin x – cos x| = sin x – cos x
= \(\int_0^{\pi / 4}\) – (sin x -cos x) dx + \(\int_{\pi / 4}^{\pi / 2}\) (sin x – cos x) dx
= – [- cos x – sin x\(]_0^{\pi / 4}\) + (- cos x- sin x)\(]_{\pi / 4}^{\pi / 2}\)
= \(\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-(1+0)\right]+\left(0-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)\)
= √2 – 1 – 1 + √2
= 2 (√2 – 1)

Question 51.
\(\int_0^{\pi / 2}\) sin 2x log cot x dx is equal to
(a) π
(b) 2π
(c) 0
(d) \(\frac{\pi}{2}\)
Solution:
(c) 0

Let I = \(\int_0^{\pi / 2}\) sin 2x log cot x dx ………..(1)
We know that
\(\int_0^a\) f(x) dx = \(\int_0^a\) f (a – x) dx
∴ I = \(\int_0^{\pi / 2}\) sin 2 (\(\frac{\pi}{2}\) – x) log cot (\(\frac{\pi}{2}\) – x) dx
= \(\int_0^{\pi / 2}\) sin (π – x) log cot (\(\frac{\pi}{2}\) – x) dx
∴ I = \(\int_0^{\pi / 2}\) sin x log tan x dx ……………..(2)
On adding (1) and (2) ; we have
2I = \(\int_0^{\pi / 2}\) sin 2x [log cot x + log tan x] dx
= \(\int_0^{\pi / 2}\) sin 2x log (cot x . tan x) dx
= \(\int_0^{\pi / 2}\) sin 2x log 1 dx = 0
[∵ log 1 = 0 ]
⇒ I = 0

Question 52.
The value of \(\int_{-\pi / 2}^{\pi / 2}\) (x5 + x sin2 x + 2 tan-1 x – 1) dx is
(a) π
(b) 2
(c) 1
(d) 0
Solution:
(a) π

Let I = \(\int_{-\pi / 2}^{\pi / 2}\) (x5 + x sin2 x + 2 tan-1 x – 1) dx
= \(\int_{-\pi / 2}^{\pi / 2}\) (x5 + x sin2 x + 2 tan-1 x) dx + \(\int_{-\pi / 2}^{\pi / 2}\) dx
= I1 + x\(]_{-\pi / 2}^{\pi / 2}\)
= I1 + \(\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\)
= I1 + π …………..(1)
Let f(x) = x5 + x sin2 x + 2 tan-1 x
∴ f(- x) = (- x)5 – x (sin (- x))2 + 2 tan-1 (- x)
= – x5 – x sin2 x – 2 tan-1 x
= – f(x)
Thus f(x) is an odd function.
∴ I1 = \(\int_{-\pi / 2}^{\pi / 2}\) f(x) = 0
∴ from (1) ;
I = 0 + π = π

Question 53.
If f (a + b – x) = f(x), then \(\int_a^b\) x f(x) dx is equal to
(a) \(\frac{a+b}{2} \int_a^b\) f (b – x) dx
(b) \(\frac{a+b}{2} \int_a^b\) f(a – x) dx
(c) \(\frac{b-a}{2} \int_a^b\) f(x) dx
(d) \(\frac{a+b}{2} \int_a^b\) f(x) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 20

Solution:
(d) \(\frac{a+b}{2} \int_a^b\) f(x) dx

Let I = \(\int_a^b\) x f(x) dx ………………(1)
We know that
\(\int_a^b\) f(x) dx = \(\int_a^b\) f(a + b – x) dx
I = \(\int_a^b\) (a + b – x) f(a + b – x) dx
I = \(\int_a^b\) (a + b – x) f(x) dx ………………(2)
[∵ f(a + b – x) = f(x)]
On adding (1) and (2) ; we get
2I = \(\int_a^b\) (a + b – x + x) f(x) dx
⇒ I = \(\frac{a+b}{2} \int_a^b\) f(x) dx

Question 54.
If a is a real number such that \(\int_0^a\) x dx ≤ a + 4, then
(a) 0 ≤ a ≤ 4
(b) – 2 ≤ a ≤ 0
(c) a ≤ – 2 or a ≥ 4
(d) – 2 ≤ a ≤ 4
Solution:
(d) – 2 ≤ a ≤ 4

\(\int_0^a\) x dx ≤ a + 4
⇒ \(\left.\frac{x^2}{2}\right]_0^a\) ≤ a + 4
⇒ \(\frac{a^2}{2}\) ≤ a + 4
⇒ a2 ≤ 2a + 8
⇒ (a2 – 2a – 8) ≤ 0
⇒ (a + 2) (a – 2) ≤ 0
⇒ – 2 ≤ a ≤ 4
[if (x – a) (x – b) ≤ 0 and a < b. Then a ≤ x ≤ b]

Question 55.
\(\int_0^{\pi / 8}\) tan2 2x dx is equal to
(a) \(\frac{4-\pi}{8}\)
(b) \(\frac{4+\pi}{8}\)
(c) \(\frac{4-\pi}{4}\)
(d) \(\frac{4-\pi}{2}\)
Solution:
(a) \(\frac{4-\pi}{8}\)

Let I = \(\int_0^{\pi / 8}\) tan2 2x dx
put 2x = t
⇒ 2 dx = dt
When x = 0 ⇒ t = 0 ;
When x = \(\frac{\pi}{8}\)
⇒ t = \(\frac{\pi}{4}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs 21

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Peer review of ML Aggarwal Class 12 Solutions can encourage collaborative learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Very Short answer type questions (1 to 7):

Evaluate the following (1 to 4) :

Question 1.
If f'(x) = √x and f(1) = 2, then find the f(x).
Solution:
Since f(x) = ∫ f'(x) dx + C
⇒ f(x) = ∫ √x dx + C
[∵ f'(x) = √x]
⇒ f(x) = \(\frac{2}{3}\) x3/2 + C ………..(1)
Since f(1) = 2
⇒ When x = 1 ; f(x) = 2
∴ from (1);
2 = \(\frac{2}{3}\) + C
⇒ C = \(\frac{4}{3}\)
∴ f(x) = \(\frac{2}{3}\) x3/2 + \(\frac{4}{3}\)

Question 1 (old).
(i) ∫ \(\frac{x}{\sqrt{1-x^2}}\) dx
(ii) ∫ x2 ex3 dx
(iii) ∫ eex ex dx
Solution:
(i) Let I = ∫ \(\frac{x}{\sqrt{1-x^2}}\) dx
put x2 = t
⇒ 2x dx = dt
= ∫ \(\frac{d t}{2 \sqrt{1-t}}\)
= \(\frac{1}{2}\) ∫ (1 – t)– \(\frac{1}{2}\)
= \(\frac{1}{2} \frac{(1-t)^{-\frac{1}{2}+1}}{(-1)\left(-\frac{1}{2}+1\right)}\) + C
= – \(\sqrt{1-x^2}\) + C

(ii) Let I = ∫ x2 ex3 dx
put x3 = t
⇒ 3x2 dx = dt
= ∫ et \(\frac{d t}{3}\)
= \(\frac{e^{x^3}}{3}\) + C

(iii) put ex = t
ex dx = dt
∴ I = ∫ eex ex dx
= ∫ et dt
= et + C
= eex + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 2.
If f'(x) = 4x3 – \(\frac{3}{x^4}\) and f(- 1) = 0, find f(x).
Solution:
Given f'(x) = 4x3 – \(\frac{3}{x^4}\)
On integrating both sides w.r.t x, we have
∫ f'(x) dx = ∫ (4x3 – \(\frac{3}{x^4}\)) dx + C
⇒ f(x) = \(\frac{4 x^4}{4}-3 \frac{x^{-4+1}}{-4+1}\) + C
⇒ f(x) = x4 + \(\frac{1}{x^3}\) + C …………….(1)
Given f(- 1) = 0 i.e.
When x = – 1 ;
f(x) = 0
0 = 1 – 1 + C
⇒ C = 0
∴ from (1) ;
f(x) = x4 + \(\frac{1}{x^3}\)

Question 2 (old).
(i) ∫ \(\frac{d x}{\left(1+x^2\right) \tan ^{-1} x}\)
(ii) ∫ \(\sqrt{2+\sin 3 x}\) cos 3x dx
(iii) ∫ \(\frac{d x}{4 \cos ^3 x-3 \cos x}\)
Solution:
(i) put tan-1 x = t
⇒ \(\frac{1}{1+x^2}\) dx = dt
∴ ∫ \(\frac{d x}{\left(1+x^2\right) \tan ^{-1} x}\) = ∫ \(\frac{d t}{t}\)
= log |t| + C
= log |tan-1 x| + C

(ii) Let I = ∫ \(\sqrt{2+\sin 3 x}\) cos 3x dx
put sin 3x = t
⇒ 3 cos 3x dx = dt
∴ I = ∫ \(\sqrt{2+t} \frac{d t}{3}\)
= \(\frac{1}{3} \frac{(2+t)^{3 / 2}}{3 / 2}\) + C
= \(\frac{2}{9}\) (2 + sin 3x)3/2 + C

(iii) Let I = ∫ \(\frac{d x}{4 \cos ^3 x-3 \cos x}\)
= ∫ \(\frac{d x}{\cos 3 x}\)
= ∫ sec 3x dx
= \(\frac{1}{3}\) log |sec 3x + tan 3x| + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 3.
(i) ∫ sec x \(\sqrt{\frac{1-\sin x}{1+\sin x}}\) dx
(ii) ∫ \(\frac{1}{\ {cosec} x-1}\) dx
Solution:
(i) ∫ sec x \(\sqrt{\frac{1-\sin x}{1+\sin x}}\) dx
= ∫ sec x \(\sqrt{\frac{1-\sin x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}}\) dx
= ∫ \(\frac{\sec x(1-\sin x)}{\cos x}\) dx
= ∫ \(\frac{(1-\sin x)}{\cos ^2 x}\) dx
= ∫ sec2 x dx – ∫ tan x sec x dx + C
= tan x – sec x + C

(ii) ∫ \(\frac{1}{\ {cosec} x-1}\) dx
= ∫ \(\frac{\sin x d x}{1-\sin x}\) dx
= ∫ \(\frac{\sin x(1+\sin x)}{\cos ^2 x}\) dx
= ∫ tan x sec x dx + ∫ (sec2 x – 1) dx
= sec x + tan x – x + C

Question 3 (old).
(i) \(\int_1^2 \frac{x}{x^2+1}\) dx
(ii) \(\int_1^e \frac{1+\log x}{2 x}\) dx
(iii) \(\int_0^{\pi / 4} \frac{\cos 3 x}{\cos x}\) dx
Solution:
(i) Let I = \(\int_1^2 \frac{x}{x^2+1}\) dx
put x2 + 1 = t
⇒ 2x dx = dt
When x = 1 ⇒ t = 2 ;
When x = 2 ⇒ t = 5
∴ I = \(\int_2^5 \frac{d t}{2 t}\)
= \(\left.\frac{1}{2} \log |t|\right]_2^5\)
= \(\frac{1}{2}\) (log 5 – log 2)
= \(\frac{1}{2}\) log \(\frac{5}{2}\).

(ii) put 1 + log x = t
⇒ \(\frac{1}{x}\) dx = dt
When x = 1
⇒ t = 1 + log 1 = 1
When x = e
⇒ t = 1 + log e = 1 + 1 = 2
∴ I = \(\int_1^2 \frac{1+\log x}{2 x}\) dx
= \(\frac{1}{2} \int_1^2\) t dt
= \(\left.\frac{t^2}{4}\right]_1^2\)
= \(\frac{1}{4}\) [4 – 1]
= \(\frac{3}{4}\)

(iii) \(\int_0^{\pi / 4} \frac{\cos 3 x}{\cos x}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 1

Question 4.
(i) ∫ \(\frac{1+\sin ^2 x}{1+\cos x}\) dx
(ii) ∫ (4 cot x – 5 tan x)2 dx
Solution:
(i) Let I = ∫ \(\frac{1+\sin ^2 x}{1+\cos x}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 3

(ii) Let I = ∫ (4 cot x – 5 tan x)2 dx
= ∫ (16 cot2 x + 25 tan2 x – 40) dx
= ∫ [16 (cosec2 x – 1) + 25 (sec2 x – 1) – 40] dx
= ∫ 16 cosec2 x dx + 25 ∫ sec2 x dx – 81 ∫ dx
= – cot x + 25 tan x – 81 x + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 4 (old).
(ii) \(\int_{0}^{\pi / 2} \frac{\sin ^n x}{\sin ^n x+\cos ^n x}\) dx
Solution:
Let I = \(\int_{0}^{\pi / 2} \frac{\sin ^n x}{\sin ^n x+\cos ^n x}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 2

Question 5.
(i) ∫ \(\frac{d x}{\sqrt{1-3 x}-\sqrt{5-3 x}}\)
(ii) ∫ (1 – x) \(\sqrt{1+x}\) dx
Solution:
(i) Let I = \(\frac{d x}{\sqrt{1-3 x}-\sqrt{5-3 x}}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 4

(ii) Let I = ∫ (1 – x) \(\sqrt{1+x}\) dx
= ∫ – (x + 1 – 2) \(\sqrt{1+x}\) dx
= – ∫ [(x + 1)3/2 – 2 (1 + x)1/2] dx
= – \(\frac{2}{5}\) (x + 1)5/2 + \(\frac{4}{3}\) (1 + x)3/2 + C

Question 5 (old).
If ∫ |x| dx = k x |x| + C, then what is the value of k ?
Solution:
Let I = ∫ |x| dx
= ∫ |x| . 1 dx
= |x| . x – ∫ \(\frac{x}{|x|}\) . x dx
∴ I = x |x| – ∫ \(\frac{|x|^2}{|x|}\) dx + 2C
⇒ I = x |x| – ∫ |x| dx + 2C
⇒ 2I = x |x| + 2C
⇒ I = \(\frac{x|x|}{2}\) + C ………….(1)
Also I = kx |x| + C …………….(2)
∴ from (1) and (2) ;
we have k = \(\frac{1}{2}\).

Question 6.
(i) ∫ (x + 1) (2x – 1)3/2 dx
(ii) ∫ sec2 2x cos 4x dx
Solution:
(i) Let I = ∫ (x + 1) (2x – 1)3/2 dx
= \(\frac{1}{2}\) ∫ (2x – 1 + 3) (2x – 1)3/2 dx
= \(\frac{1}{2} \int(2 x-1)^{5 / 2} d x+\frac{3}{2} \int(2 x-1)^{3 / 2}\) dx
= \(\frac{1}{2} \frac{(2 x-1)^{7 / 2}}{\frac{7}{2} \times 2}+\frac{3}{2} \frac{(2 x-1)^{5 / 2}}{\frac{5}{2} \times 2}\) + C
= \(\frac{1}{14}\) (2x – 1)3/2 + \(\frac{3}{10}\) (2x – 1)5/2 + C

(ii) ∫ sec2 2x cos 4x dx
= ∫ sec2 2x cos (2 × 2x) dx
= ∫ sec2 2x (2 cos2 2x – 1) dx
= ∫ [2 – sec2 2x] dx
= 2x – \(\frac{\tan 2 x}{2}\) + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 7.
(i) ∫ \(\frac{10 x^9+10^x \log 10}{x^{10}+10^x}\) dx (NCERT)
(ii) ∫ \(\frac{d x}{x \sqrt{a x-x^2}}\) dx (NCERT)
Solution:
(i) put x10 + 10x = t
⇒ (10x9 + 10x log 10) dx = dt
∴ ∫ \(\frac{10 x^9+10^x \log 10}{x^{10}+10^x}\) dx = ∫ \(\frac{d t}{t}\)
= log |t| + C
= log |x10 + 10x| + C

(ii) Let I = ∫ \(\frac{d x}{x \sqrt{a x-x^2}}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 5

Question 8.
(i) ∫ \(\frac{d x}{\cos (x+a) \cos (x+b)}\)
(ii) ∫ \(\frac{d x}{3 \cos x+4 \sin x}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{\cos (x+a) \cos (x+b)}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 6

(ii) Let I = ∫ \(\frac{d x}{3 \cos x+4 \sin x}\)
put 3 = r cos α …………..(1)
and 4 = r sin α …………..(2)
On squaring (1) and (2) ; we have
r = \(\sqrt{9+16}\) = 5
On dividing (2) by (1) ; we have
tan α = \(\frac{4}{3}\)
⇒ α = tan-1 \(\frac{4}{3}\)
∴ I = ∫ \(\frac{d x}{r(\cos \alpha \cos x+\sin \alpha \sin x)}\)
= ∫ \(\frac{d x}{5 \cos (x-\alpha)}\)
= \(\frac{1}{5}\) log |sec (x – α) + tan (x – α)| + C
where α = tan-1 \(\frac{4}{3}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 9.
(i) ∫ \(\frac{\sin 3 x}{\sin x}\) dx
(ii) ∫ \(\frac{\sec x}{\sec 2 x}\) dx
Solution:
(i) ∫ \(\frac{\sin 3 x}{\sin x}\) dx
= ∫ \(\left[\frac{3 \sin x-4 \sin ^3 x}{\sin x}\right]\) dx
= ∫ [3 – 4 sin2 x] dx
= ∫ [3 – 4 \(\left(\frac{1-\cos 2 x}{2}\right)\)] dx
= \(\frac{1}{2}\) ∫ [2 + 4 cos 2x] dx
= ∫ (1 + 2 cos 2x) dx
= x + \(\frac{2 \sin 2 x}{2}\) + C
= x + sin 2x + C

(ii) ∫ \(\frac{\sec x}{\sec 2 x}\) dx
= ∫ \(\frac{\cos 2 x}{\cos x}\) dx
= ∫ \(\left(\frac{2 \cos ^2 x-1}{\cos x}\right)\) dx
= 2 ∫ cos x dx – ∫ sec x dx
= 2 sin x – log |sec x + tan x| + C

Question 10.
(i) ∫ \(\frac{e^x}{\sqrt{e^{2 x}-4}}\) dx
(ii) ∫ ecot x cosec2 x dx
Solution:
(i) put ex = t
⇒ ex dx = dt
∴ I = ∫ \(\frac{e^x}{\sqrt{e^{2 x}-4}}\) dx
= ∫ \(\frac{d t}{\sqrt{t^2-2^2}}\)
= log |t + \(\sqrt{t^2-2^2}\)| + C
= log |ex + \(\sqrt{e^{2 x}-4}\)| + C

(ii) Let I = ∫ ecot x cosec2 x dx
put cos x = t
⇒ – cosec2 x dx = dt
∴ I = ∫ et (- dt)
= – et + C
= – ecot x + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 11.
(i) ∫ \(\frac{d x}{\sqrt{\sin ^3 x \cos x}}\)
(ii) ∫ \(\frac{\sin 2 x}{(a+b \cos 2 x)^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{\sin ^3 x \cos x}}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 7

(ii) Let I = ∫ \(\frac{\sin 2 x}{(a+b \cos 2 x)^2}\) dx
put cos 2x = t
⇒ – 2 sin 2x dx = dt
= ∫ \(\frac{d t}{-2(a+b t)^2}\)
= – \(\frac{1}{2} \frac{(a+b t)^{-2+1}}{(-2+1) b}\) + C
= \(\frac{1}{2 b} \frac{1}{(a+b \cos 2 x)}\) + C

Question 12.
(i) ∫ \(\frac{x+\left(\cos ^{-1} 3 x\right)^2}{\sqrt{1-9 x^2}}\) dx
(ii) ∫ \(\frac{x^3}{\sqrt{1+x^2}}\) dx
Solution:
(i) Let I = ∫ \(\frac{x+\left(\cos ^{-1} 3 x\right)^2}{\sqrt{1-9 x^2}}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 8

(ii) put x2 = t
⇒ 2x dx = dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 9

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 13.
(i) ∫ \(\frac{\sec x}{\log (\sec x+\tan x)}\) dx
(ii) ∫ \(\frac{\cos x-\sin x}{1-\sin 2 x}\) dx
Solution:
(i) put log (sec x + tan x) = t
⇒ \(\frac{1}{\sec x+\tan x}\) (sec x tan x + sec2 x) dx = dt
⇒ \(\frac{\sec x(\tan x+\sec x)}{\sec x+\tan x}\) dx = dt
⇒ sec x dx = dt
∴ I = ∫ \(\frac{\sec x d x}{\log (\sec x+\tan x)}\)
= ∫ \(\frac{d t}{t}\)
= log |t| + C

(ii) Let I = ∫ \(\frac{\cos x-\sin x}{1-\sin 2 x}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 10

Question 14.
(i) ∫ \(\sqrt{1+2 \tan x(\tan x+\sec x)}\) dx
(ii) ∫ \(\frac{4 x+1}{\sqrt{2 x^2+x-3}}\) dx
Solution:
(i) Let I = ∫ \(\sqrt{1+2 \tan x(\tan x+\sec x)}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 11

(ii) put 2x2 + x – 3 = t
⇒ (4x + 1) dx = dt
∴ I = ∫ \(\frac{(4 x+1) d x}{\sqrt{2 x^2+x-3}}\)
= ∫ \(\frac{d t}{\sqrt{t}}\)
= ∫ t-1/2 dt
= \(\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= 2 \(\sqrt{2 x^2+x-3}\) + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 15.
∫ \(\frac{\sin 2 x}{\sqrt{a^2 \sin ^2 x+b^2 \cos ^2 x}}\) dx, a ≠ ± b. What happens if a = ± b ?
Solution:
Case – I: When a ≠ ± b
Let I = ∫ \(\frac{\sin 2 x}{\sqrt{a^2 \sin ^2 x+b^2 \cos ^2 x}}\) dx
put a2 sin2 x + b2 cos2 x = t
⇒ [2a2 sin x cos x + 2b2 cos x (- sin x)] dx = dt
⇒ (a2 – b2) sin 2x dx = dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 12

Question 16.
(i) ∫ \(\frac{x d x}{\sqrt{x^2+a^2}+\sqrt{x^2-a^2}}\)
(ii) ∫ \(\frac{x e^{\sqrt{x^2+3}}}{\sqrt{x^2+3}}\) dx
Solution:
(i) Let I = ∫ \(\frac{x d x}{\sqrt{x^2+a^2}+\sqrt{x^2-a^2}}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 14

(ii) Let I = ∫ \(\frac{x e^{\sqrt{x^2+3}}}{\sqrt{x^2+3}}\) dx
put \(\sqrt{x^2+3}\) = t
⇒ \(\frac{1}{2}\left(x^2+3\right)^{-\frac{1}{2}}\) 2x dx = dt
⇒ \(\frac{x}{\sqrt{x^2+3}}\) dx = dt
⇒ I = ∫ et dt
= et + C
Thus, I = \(e^{\sqrt{x^2+3}}\) + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 17.
(i) ∫ tan x sec2 x \(\sqrt{1-\tan ^2 x}\) dx
(ii) ∫ sin3 x cos5 x dx
Solution:
(i) Let I = ∫ tan x sec2 x \(\sqrt{1-\tan ^2 x}\) dx
put tan2 x dx = t
⇒ 2 tan x sec2 x dx = dt
∴ I = \(\frac{1}{2}\) ∫ \(\sqrt{1-t}\) dt
= \(\frac{1}{2} \frac{(1-t)^{3 / 2}}{\frac{3}{2}(-1)}\) + C
= – \(\frac{1}{3}\) (1 – tan2 x)3/2 + C

(ii) Let I = ∫ sin3 x cos5 x dx
= ∫ sin2 x cos5 x sin x dx
= ∫ (1 – cos2 x) cos5 x (sin x) dx
put cos x = t
⇒ – sin x dx = dt
= ∫ (1 – t2) t5 (- dt)
= – \(\left[\frac{t^6}{6}-\frac{t^8}{8}\right]\) + C
= \(-\frac{\cos ^6 x}{6}+\frac{\cos ^8 x}{8}\) + C

Question 18.
(i) ∫ \(\frac{(a+\sqrt{x})^n}{\sqrt{x}}\) dx, n ≠ – 1
(ii) ∫ \(\frac{d x}{\cos ^2 x \sqrt{\tan x-1}}\)
Solution:
(i) Let I = ∫ \(\frac{(a+\sqrt{x})^n}{\sqrt{x}}\) dx, n ≠ – 1
put √x = t
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
∴ I = ∫ (a + t)n (2 dt)
= \(\frac{2(a+t)^{n+1}}{n+1}\) + C
= \(\frac{2}{n+1}\) (a + √x)n + 1 + C ; n ≠ – 1

(ii) Let I = ∫ \(\frac{d x}{\cos ^2 x \sqrt{\tan x-1}}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 15

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 19.
(i) ∫ \(\frac{d x}{(2 \sin x+3 \cos x)^2}\)
(ii) ∫ \(\frac{d x}{x^{2 / 3} \sqrt{x^{2 / 3}-4}}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{(2 \sin x+3 \cos x)^2}\)
Divide Numerator and deno. by cos2 x ; we have
= ∫ \(\frac{\sec ^2 x d x}{(2 \tan x+3)^2}\)
put tan x = t
⇒ sec2 x dx = dt
∴ I = ∫ \(\frac{d t}{(2 t+3)^2}\)
= ∫ (2t + 3)-2 dt
= \(\frac{(2 t+3)^{-2+1}}{2 \cdot(-2+1)}\) + C
= \(\frac{1}{2(2 t+3)}\) + C
= – \(\frac{1}{2(2 \tan x+3)}\) + C

(ii) Let I = ∫ \(\frac{d x}{x^{2 / 3} \sqrt{x^{2 / 3}-4}}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 16

Question 20.
(i) ∫ x \(\sqrt[3]{2 x+1}\) dx
(ii) ∫ \(\frac{2 x}{x^2+3 x+2}\) dx (NCERT)
Solution:
(i) Let I = ∫ x \(\sqrt[3]{2 x+1}\) dx
put \(\sqrt[3]{2 x+1}\) = t
⇒ 2x + 1 = t3
⇒ 2 dx = 3t2 dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 17

(ii) Let I = ∫ \(\frac{2 x}{x^2+3 x+2}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 18

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 21.
(i) ∫ \(\frac{\sqrt{x}}{\sqrt{x}+2}\) dx
(ii) ∫ √x (log x)2 dx
Solution:
(i) put √x + 2 = t
⇒ √x = t – 2
⇒ x = (t – 2)2
⇒ dx = 2 (t – 2) dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 19

= t2 – 8t + 8 log |t| + C
= (√x + 2)2 – 8 (√x + 2) + 8 log |t| + C
= x – 4√x + 8 log |√x + 2| + C’

(ii) Let I = ∫ √x (log x)2 dx
put log x = t
⇒ x = et
⇒ dx = et dt
∴ I = ∫ et/2 t2 et dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 20

Question 22.
(i) ∫ \(\frac{1}{x-x^3}\) dx (NCERT)
(ii) ∫ \(\frac{\cos x}{2+\cos ^2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{1}{x-x^3}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 21

(ii) Let I = ∫ \(\frac{\cos x}{2+\cos ^2 x}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 22

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 23.
(i) ∫ \(\frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{e^x}{e^{2 x}+6 e^x+5}\) dx
Solution:
(i) put x2 = y

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 23

(ii) Let I = ∫ \(\frac{e^x}{e^{2 x}+6 e^x+5}\) dx ;
put ex = t
⇒ ex dx = dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 24

Question 24.
(i) ∫ \(\frac{\sin x}{(2+\cos x)(5+\cos x)}\) dx
(ii) ∫ \(\frac{x^4}{(x-1)\left(x^2+1\right)}\) dx
Solution:
(i) I = ∫ \(\frac{\sin x}{(2+\cos x)(5+\cos x)}\) dx
put cos x = t
⇒ – sin x dx = dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 25

(ii) Let I = ∫ \(\frac{x^4}{(x-1)\left(x^2+1\right)}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 26

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 27

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 25
(i) ∫ cos 2x log (1 + tan x) dx
(ii) ∫ \(\frac{e^{\log \left(1+\frac{1}{x^2}\right)}}{x^2+\frac{1}{x^2}}\) dx
Solution:
(i) Let I = ∫ cos 2x log (1 + tan x) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 28

sin x = l (sin x + cos x) + m \(\frac{d}{d x}\) (sin x + cos x)
⇒ sin x = l (sin x + cos x) + m (cos x – sin x)
Coeff. of sin x ;
1 = l – m ;
Coeff. of cos x ;
0 = l + m
On solving these eqn.’s ;
l = \(\frac{1}{2}\) ; m = – \(\frac{1}{2}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 29

(ii) Let I = ∫ \(\frac{e^{\log \left(1+\frac{1}{x^2}\right)}}{x^2+\frac{1}{x^2}}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 31

Question 25 (old).
(ii) ∫ \(\frac{1-\cos x}{\cos x(1+\cos x)}\) dx
Solution:
Let I = ∫ \(\frac{1-\cos x}{\cos x(1+\cos x)}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 30

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 26.
(i) ∫ \(\frac{x^3 \sin ^{-1}\left(x^4\right)}{\sqrt{1-x^4}}\) dx
(ii) ∫ sin4 x cos5 x dx
Solution:
(i) Let I = ∫ \(\frac{x^3 \sin ^{-1}\left(x^4\right)}{\sqrt{1-x^4}}\) dx ;
put x4 = t
⇒ 4x3 dx = dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 32

(ii) Let I = ∫ sin4 x cos5 x dx
= ∫ sin4 x cos4 x cos x dx
= ∫ sin4 x (1 – sin2 x)4 cos x dx
put sin x = t
⇒ cos x dx = dt
∴ I = ∫ t4 (1 – t2)2 dt
= ∫ t4 (t4 – 2t2 + 1) dt
= \(\frac{t^9}{9}-2 \frac{t^7}{7}+\frac{t^5}{5}\) + C
= \(\frac{\sin ^9 x}{9}-\frac{2}{7} \sin ^7 x+\frac{1}{5} \sin ^5 x\) + C

Question 27.
(i) ∫ \(\frac{d x}{x^{1 / 2}-x^{1 / 4}}\)
(ii) ∫ tan-1 \(\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\) dx
Solution:
(i) Let I = \(\frac{d x}{x^{1 / 2}-x^{1 / 4}}\)
put x1/4 = t
⇒ x = t4
⇒ dx = 4t3 dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 33

= 2t2 + 4t + 4 log |t – 1| + C
= 2√x + 4x1/4 + 4 log |x1/4 – 1| + C

(ii) Let I = ∫ tan-1 \(\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\) dx
put x = cos 2t
⇒ dx = – 2 sin 2t dt

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 34

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 28.
(i) ∫ \(\frac{\log (x+1)-\log x}{x(x+1)}\) dx
(ii) ∫ \(\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}\) ex dx
Solution:
(i) Let I = ∫ \(\frac{\log (x+1)-\log x}{x(x+1)}\) dx ;
put log (x + 1) – log x = t

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 35

(ii) Let I = ∫ \(\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}\) ex dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 36

Question 29.
(i) ∫ x3 e– x2 dx
(ii) ∫ sin-1 \(\left(\frac{2 x+2}{\sqrt{4 x^2+8 x+13}}\right)\) dx
Solution:
(i) Let I = ∫ x3 e– x2 dx ;
put x2 = t
⇒ 2x dx = dt
∴ I = ∫ t e-t \(\frac{d t}{2}\)
= \(\frac{1}{2}\left[t \frac{e^{-t}}{(-1)}-\int 1 \times \frac{e^{-t}}{(-1)} d t\right]\) + C
∴ I = \(\frac{1}{2}\) [- t e-t – e-t] + C
= – \(\frac{1}{2}\) (t + 1) e-t + C
= – \(\frac{1}{2}\) (x2 + 1) e– x2 + C

(ii) Let I = ∫ sin-1 \(\left(\frac{2 x+2}{\sqrt{4 x^2+8 x+13}}\right)\) dx
first of all, we convert sin-1 to tan-1 with p = 2x + 2;
h = \(\sqrt{4 x^2+8 x+3}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 37

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 38

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 30.
(i) ∫ f'(ax + b) (f (ax + b))n dx (NCERT)
(ii) ∫ \(\frac{\sqrt{1-x}}{x^{7 / 2}}\) dx
Solution:
(i) Case – I :
When n ≠ – 1
Let I = ∫ f'(ax + b) (f (ax + b))n dx
put f (ax + b) = t
⇒ f’ (ax + b) . a dx = dt
∴ I = ∫ \(\frac{1}{a}\) tn dt
= \(\frac{1}{a} \frac{t^{n+1}}{n+1}\) + C ; n ≠ – 1
∴ I = \(\frac{(f(a x+b))^{n+1}}{(n+1) a}\) + C ; n ≠ – 1

Case – II :
When n = – 1
I = ∫ \(\frac{f^{\prime}(a x+b) d x}{f(a x+b)}\)
put f (ax + b) = t
⇒ f’ (ax + b) . a dx = dt
= ∫ \(\frac{d t}{a t}\)
= \(\frac{1}{a}\) log |t| + C
= \(\frac{1}{a}\) log |f (ax + b)| + C

(ii) Let I = ∫ \(\frac{\sqrt{1-x}}{x^{7 / 2}}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 39

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 31.
(i) ∫ \(\sqrt{4-x+x^2}\) dx
(ii) ∫ \(\frac{\tan ^{-1} x}{x^2}\) dx
(iii) ∫ \(\frac{\log |x|}{(x+1)^3}\) dx
Solution:
(i) Let I = ∫ \(\sqrt{4-x+x^2}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 40

(ii) Let I = ∫ \(\frac{\tan ^{-1} x}{x^2}\) dx
put tan-1 x = θ
⇒ x = tan θ
⇒ dx = sec2 θ
= ∫ \(\frac{\theta}{\tan ^2 \theta}\) sec2 θ dθ
= ∫ θ cosec2 θ dθ
= θ (- cot θ) – ∫ (- cot θ) dθ + C
= – θ cot θ + log |sin θ| + C
= \(-\frac{\tan ^{-1} x}{x}+\log \left|\frac{x}{\sqrt{1+x^2}}\right|\) + C

(iii) Let I = ∫ \(\frac{\log |x|}{(x+1)^3}\) dx
= ∫ log |x| . \(\frac{1}{(x+1)^3}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 41

Multiplying both sides of eqn. (2) by x(x + 1)2; we get
1 = A(x + 1)2 + Bx (x + 1) + Cx ………….(3)
putting x = 0, – 1 successively in eqn. (3); we have
1 = A
and 1= – C
⇒ C = – 1
Coeff. of x2;
0 = A + B,
⇒ B = – 1
∴ I1 = ∫ \(\left[\frac{1}{x}-\frac{1}{x+1}-\frac{1}{(x+1)^2}\right]\) dx
= log |x| – log |x + 1| + \(\frac{1}{x+1}\)
∴ from (1); we have
= \(-\frac{\log |x|}{2(x+1)^2}+\frac{1}{2} \log \left|\frac{x}{x+1}\right|+\frac{1}{2(x+1)}\) + C

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 32.
Evaluate the following definite integrals as the limit of sums :
(i) \(\int_1^3\) (x2 + 5x) dx
(ii) \(\int_0^1\) e2 – 3x dx (NCERT)
Solution:
(i) On comparing \(\int_1^3\) (x2 + 5x) dx with \(\int_a^b\) f(x) dx
Here f(x) = x2 + 5x ;
a = 1 ; b = 3
∴ nh = b – a
= 3 – 1 = 2
∴ f(a) = f(1)
= 12 + 5
f(a + h) = f(1 + h)
= (1 + h)2 + 5 (1 + h)
f(a + 2h) = f(1 + 2h)
= (1 + 2h)2 + 5 (1 + 2h)
……………………………………
……………………………………

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 42

(ii) On comparing\(\int_0^1\) e2 – 3x dx with \(\int_a^b\) f(x) dx
Here, f(x) = e2-3x;
a = 0;
b = 1;
nh = b – a = 1 – 0 = 1
∴ f(a) = f(0) = e2
f(a + h) = f(h) = e2 – 3h
f(a + 2h) = f(2h) = e2 – 6h
……………………………
……………………………
\(f(a+\overline{n-1} h)\) = f((n – 1)h)
= e2 – 3 (n – 1) h
∴ \(\int_0^1\) e2 – 3h dx = \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h[f(a) + f(a + h) + f(a + 2h) ………..+ \(f(a+\overline{n-1} h)\)]
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h [e2 + e2 – 3h + e2 – 6h + ……. + e2 – 3(n – 1) h]
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h[1 + e– 3h + e– 6h ……… n terms]

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 43

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Evaluate the following (33 to 36) definite integrals:

Question 33.
(i) \(\int_0^{\pi / 4}\) tan2 x dx
(ii) \(\int_0^2 \frac{5 x+1}{x^2+4}\) dx
Solution:
(i) \(\int_0^{\pi / 4}\) tan2 x dx
= \(\int_0^{\pi / 4}\) (sec2 – 1) dx
= tan x – x\(]_0^{\pi / 4}\)
= (tan \(\frac{\pi}{4}\) – \(\frac{\pi}{4}\)) – (0 – 0)
= 1 – \(\frac{\pi}{4}\)

(ii) \(\int_0^2 \frac{5 x+1}{x^2+4}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 44

Question 34.
(i) \(\int_0^{\pi / 6}\) (2 + 3x2) cos 3x dx
(ii) \(\int_0^1\) (cos-1 x)2 dx
Solution:
(i) \(\int_0^{\pi / 6}\) (2 + 3x2) cos 3x dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 45

(ii) Let I = \(\int_0^1\) (cos-1 x)2 dx
put cos-1 x = t
⇒ – \(\frac{1}{\sqrt{1-x^2}}\) dx = dt
⇒ x = cos t
⇒ dx = – sin t dt
When x = 0
⇒ t = \(\frac{\pi}{2}\) ;
When x = 1
⇒ t = 0
∴ I = \(\int_{\pi / 2}^0\) t2 (- sin t) dt
= – [- t2 cos t\(\}_{\pi / 2}^0\) + \(\int_{\pi / 2}^0\) t cos t dt]
∴ I = (0 × 1 – \(\frac{\pi^2}{4}\) × 0) – 2 [t sin t\(\}_{\pi / 2}^0\) – \(\int_{\pi / 2}^0\) sin t dt]
∴ I = – 2 [t sin t + cos t\(]_{\pi / 2}^0\)
= + 2 [t sin t + cos t\(t]_0^{\pi / 2}\)
= + 2 [\(\frac{\pi}{2}\) × 1 + 0 – 0 – 1]
= + 2 (\(\frac{\pi}{2}\) – 1)
= π – 2

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 35.
(i) \(\int_0^\pi\) x sin x cos2 x dx (NCERT Exemplar)
(ii) \(\int_0^{\pi / 2}\) \(\sqrt{sin x}\) cos5 x dx (NCERT)
Solution:
(i) Let I = \(\int_0^\pi\) x sin x cos2 x dx ……………(1)
∴ I = \(\int_0^\pi\) (π – x) sin (π – x) cos2 (π – x) dx
I = \(\int_0^\pi\) (π – x) sin x cos2 x dx ……………(2)
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx]
On adding (1) and (2) ; we have
2I = \(\int_0^\pi\) π sin x cos2 x dx
put cos x = t
⇒ – sin x dx = dt
When x = 0 ⇒ t = 1 ;
When x = π ⇒ t = – 1
⇒ 2I = π \(\int_1^{-1}\) t2 (- dt)
= – π \(\left.\frac{t^3}{3}\right]_1^{-1}\)
= – \(\frac{\pi}{3}\) (- 1 – 1)
= \(\frac{2 \pi}{3}\)
Thus, I = \(\frac{\pi}{3}\)

(ii) Let I = \(\int_0^{\pi / 2}\) \(\sqrt{sin x}\) cos5 x dx
= \(\int_0^{\pi / 2}\) \(\sqrt{sin x}\) (1 – sin2 x)2 cos x dx
put sin x = t ⇒ cos x dx = dt
When x = 0 ⇒ t = 0 ;
When x = \(\frac{\pi}{2}\) ⇒ t = 1
∴ I = \(\int_0^1\) √t (1 – t2)2 dt
= \(\int_0^1\) √t (t4 – 2t2 + 1) dt
= \(\left.\frac{2 t^{11 / 2}}{11}-\frac{2 t^{7 / 2}}{7 / 2}+\frac{t^{3 / 2}}{3 / 2}\right]_0^1\)
= \(\left[\frac{2}{11}-\frac{4}{7}+\frac{2}{3}\right]\)
= \(\frac{42-132+154}{231}=\frac{64}{231}\).

Question 36.
(i) \(\int_0^{\pi / 4} \frac{\sin x \cos x}{\cos ^2 x+\sin ^4 x}\) dx
(ii) \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{\cos ^2 x+\sin ^4 x}\) dx
Solution:
(i) I = \(\int_0^{\pi / 4} \frac{\sin x \cos x}{\cos ^2 x+\sin ^4 x}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 46

(ii) Let I = \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{\cos ^2 x+\sin ^4 x}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 47

put sin x – cos x = t
⇒ (cos x + sin x) dx = dt
On squaring; we have
(sin x – cos x)2 = t2
⇒ sin2 x + cos2 x – sin 2x = t2
⇒ 1 – sin 2x = t2
⇒ sin 2x = 1 – t2
When x = 0 ⇒ t = – 1 ;
When x = \(\frac{\pi}{4}\) ⇒ t = 0
∴ I = 4 \(\int_{-1}^0 \frac{d t}{4-\left(1-t^2\right)^2}\) ………………(1)
= 4 \(\int_{-1}^0 \frac{d t}{\left(2+1-t^2\right)\left(2-1+t^2\right)}\)
⇒ I = 4 \(\int_{-1}^0 \frac{d t}{\left(3-t^2\right)\left(1+t^2\right)}\)
put t2 = y
Then \(\frac{1}{\left(3-t^2\right)\left(1+t^2\right)}=\frac{1}{(3-y)(1+y)}=\frac{\mathrm{A}}{3-y}+\frac{\mathrm{B}}{1+y}\) …………….(2)
Multiply both sides of eqn. (2) by (3 – y) (1 + y) ; we get
1 = A (1 + y) + B (3 – y) ………….(3)
putting y = – 1, 3 successively in eqn. (3) ; we have
1 = 4B
⇒ B = \(\frac{1}{4}\)
and 1 = 4A
⇒ A = \(\frac{1}{4}\)
∴ from (2) ; we get ;

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 48

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

By using properties of definite integrals, evaluate the following (37 to 39) :

Question 37.
\(\int_0^4\) (|x| + |x – 2| + |x – 4|
Solution:
Let f(x) = |x| + |x – 2| + |x – 4|
The critical points of f(x) are given by x = 0,2, 4
When 0 ≤ x < 2
⇒ |x| = x ;
x – 2 < 0
⇒ |x – 2| = – (x – 2)
and x – 4< 0
⇒ |x – 4| = – (x – 4)
∴ f(x) = x – (x – 2) – (x – 4) = – x + 6
When 2 ≤ x < 4
Then x – 2 ≥ 0;
x – 4 < 0; x > 0
∴ f(x) = x + x – 2 – (x – 4) = x+ 2
∴ \(\int_0^4\) f(x) dx = \(\int_0^2\) f(x) dx + \(\int_2^4\) f(x) dx
= \(\int_0^2\) (- x + 6) dx + \(\int_2^4\) (x + 2) dx
= \(\left.\left[-\frac{x^2}{2}+6 x\right]_0^2+\frac{(x+2)^2}{2}\right]_2^4\)
= [(- 2 + 12) – (0 + 0)] + \(\frac{1}{2}\) [36 – 16]
= 10 + 10 = 20.

Question 38.
(i) \(\int_{1 / e}^e\) |log x| dx
(ii) \(\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}\)
(iii) \(\int_0^{\pi / 2} \frac{\sqrt{\cot x}}{\sqrt{1+\cot x}}\) dx
Solution:
(i) When \(\frac{1}{e}\) ≤ x ≤ 1 ;
log x ≤ 0
∴ |log x| = – log x
When 1 ≤ x ≤ e ;
log x ≥ 0
∴ |log x| = + log x

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 49

(ii) Let I = \(\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}\) …………….(1)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 50

(iii) Let I = \(\int_0^{\pi / 2} \frac{\sqrt{\cot x}}{\sqrt{1+\cot x}}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 51

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 39.
\(\int_0^1\) x (1 – x)5 dx
Solution:
Let I = \(\int_0^1\) x (1 – x)5 dx
∴ I = \(\int_0^1\) (1 – x) (1 – (1 – x))5 dx
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f (a – x) dx]
I = \(\int_0^1\) (1 – x) x5 dx
= \(\left[\frac{x^6}{6}-\frac{x^7}{7}\right]_0^1\)
= \(\left(\frac{1}{6}-\frac{1}{7}-0-0\right)\)
= \(\frac{1}{42}\)

(ii) Let I = \(\int_0^{\pi / 2} \frac{\sin ^2 x}{1+\sin x \cos x}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 52

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 53

Question 40.
(i) \(\int_{-\pi / 2}^{\pi / 2}\) (x3 + x cos x + tan5 x + 1) dx
(ii) \(\int_0^{\pi / 2} \frac{\sin ^2 x}{1+\sin x \cos x}\) dx
Solution:
(i) Here f(x) = x3 + x cos x + tan5 x + 1
∴ f (- x) = (- x)3 + (- x) cos (- x) + [tan (- x)5] + 1
= – x3 – x cos x – tan5 x + 1
So f(- x) is neither equal to f(x) nor equal to – f(x)
Let I = \(\int_{-\pi / 2}^{\pi / 2}\) [x3 + x cos x + tan5 x + 1] dx
= \(\int_{-\pi / 2}^{\pi / 2} x^3 d x+\int_{-\pi / 2}^{\pi / 2} x \cos x+\int_{-\pi / 2}^{\pi / 2} \tan ^5 x+\int_{-\pi / 2}^{\pi / 2} 1 d x\) …………..(1)
Let f(x) = x3 ;
f(- x) = (- x)3
= – x3
= – f(x)
∴ \(\int_{-\pi / 2}^{\pi / 2}\) f(x) = \(\int_{-\pi / 2}^{\pi / 2}\) x3 = 0 ……………….(2)
[∵ \(\int_{-a}^a\) f(x) = 0 if f(- x) = – f(x)]
Let g(x) = x cos x
∴ g(- x) = – x cos (- x)
= – x cos x
= – g(x)
∴ g(x) be an odd function.
\(\int_{-\pi / 2}^{\pi / 2}\) g(x) dx = 0
⇒ \(\int_{-\pi / 2}^{\pi / 2}\) x cos x dx = 0 ……………(3)
Let h(x) = tan5 x
⇒ h (- x) = [tan (- x)]5
= – tan5 x
= – h(x)
∴ \(\int_{-\pi / 2}^{\pi / 2}\) h(x) dx = 0
⇒ \(\int_{-\pi / 2}^{\pi / 2}\) tan5 x dx = 0 ……………….(4)
Using eqn. (2), (3) and (4) in eqn. (1) ; we get
I = 0 + 0 + 0 + x\(]_{-\pi / 2}^{\pi / 2}\)
= \(\frac{\pi}{2}+\frac{\pi}{2}\)
= π

(ii) Let I = \(\int_0^{\pi / 2} \frac{\sin ^2 x}{1+\sin x \cos x}\) dx …………….(1)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test 54

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Question 45 (old).
(i) \(\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right)\) dx
Solution:
Let I = \(\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right)\) dx
= \(\int_0^1 \tan ^{-1}\left(\frac{x+(x-1)}{1-x(x-1)}\right)\) dx
= \(\int_0^1\) [tan-1 x + tan-1 (x – 1)] dx
I = \(\int_0^1\) [tan-1 x + tan-1 (x – 1)] dx …………….(1)
∴ I = \(\int_0^1\) [tan-1 (1 – x) + tan-1 (1 – x – 1)] dx
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f (a – x) dx]
⇒ I = \(\int_0^1\) [tan-1 (1 – x) + tan-1 (- x)] dx
⇒ I = \(\int_0^1\) [- tan-1 (x – 1) – tan-1 x] dx …………….(2)
[∵ tan-1 (- x) = – tan-1 x]
On adding (1) and (2) ; we have
2I = 0
⇒ I = 0.

OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test

The availability of step-by-step ISC Mathematics Class 11 OP Malhotra Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test

Question 1.
Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).
Solution:
Let P be any point on y-axis and its coordinates are (0, y, 0) and given points are A(3, 1, 2) and B (5, 5, 2).
According to given condition; we have
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test Img 1
On squaring both sides; we have
9 + (1 – y)2 + 4 = 25 + (5 – y)2 + 4
⇒ 14 + y2 – 2y = 54 + y2 – 10y
⇒ 8y = 40
⇒ y = 5
Thus required point on y-axis be P(0, 5, 0).

Question 2.
Show that the points (a, b, c),(b, c, a) and (c, a, b) are the vertices of an equilateral trinagle.
Solution:
Let (a, b, c); (b, c, a) and (c, a, b) are the given vertices A, B and C of △ABC.
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test Img 2
Thus, △ABC be an equilateral triangle.

Question 3.
Find out whether the points (0, 7, -10), (1, 6, 6) and (4, 9, -6) are the vertices of a right angled triangle.
Solution:
Let A(0,7,-10) ; B(1, 6, 6) and C(4, 9, -6) are the vertices of △ABC respectively.
AB2 = (1 – 0)2 + (6 – 7)2 + (6 + 10)2 = 1 + 1 +256 = 258
BC2 = (4 – 1)2 + (9 – 6)2 + (- 6 – 6)2 = 9 + 9 + 144 = 162
CA2 = (4 – 0)2 + (9 – 7)2 + (- 6 + 10)2 = 16 + 4 + 16 = 36
Thus,
AB2 ≠ BC2 + AC2
BC2 ≠ AB2 + AC2
AC2 ≠ AB2 + BC2
Thus, pythagoras theorem does not holds good. ∴ △ABC is not a right angled Δ.

OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test

Question 4.
Show that the points A(-2, 3, 5), B(1, 2, 3) and C(7, 0, -1) are collinear.
Solution:
Given points are A(-2, 3, 5); B(1, 2, 3) and C(7, 0, -1)
Here AB = \(\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}\) = \(\sqrt{9+1+4}\) = \(\sqrt{14}\)
BC = \(\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}\) = \(\sqrt{36+4+16}\) = \(\sqrt{56}\)
and CA = \(\sqrt{(-2-7)^2+(3-0)^2+(5+1)^2}\) = \(\sqrt{81+9+36}\) = \(\sqrt{126}\) = \(3 \sqrt{14}\)
Thus AB + BC + CA
Therefore all the given points A, B and C lies on same straight line and hence given points are collinear.

Question 5.
Find the lengths of the medians of the tri- angle A(0, 0, 6), B(0, 4, 0) and C(6, 0, 0)
Solution:
Let D, E and F are the mid-points of sides BC, CA and AB of △ABC.
Thus coordinates of D, E and F using mid-point formula be given by D(3, 2, 0); E(3, 0, 3) and F(0, 2, 3).
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test Img 3

Question 6.
Using section formula, show that the points (2, -3, 4), (-1, 2, 1) and (0, \(\frac { 1 }{ 3 }\), 2) are collinear.
Solution:
Let A, B and C are the given points (2, -3, 4), (-1, 2, 1) and (0, \(\frac { 1 }{ 3 }\), 2) respectively.
Thus the coordinates of the point divides the straight line joining the points B and C in the ratio k : 1 internally are
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test Img 4
∴ 2 = –\(\frac{1}{k+1}\) ⇒ k + 1 = –\(\frac{1}{2}\) ⇒ k = –\(\frac{3}{2}\)
\(\frac{\frac{k}{3}+2}{k+1}\) = – 3 ⇒ \(\frac{k}{3}\) + 2 = – 3k – 3
⇒ \(\frac{10k}{3}\) = -5 ⇒ k = –\(\frac{3}{2}\)
and \(\frac{2 k+1}{k+1}\) ⇒ 2k + 1 = 4k + 4
⇒ 2k = – 3 ⇒ k = –\(\frac{3}{2}\)
Thus from all equations, we get same value of k.
Hence, the point A lies on the straight line joining B and C. Thus, points A, B and C are collinear.

Question 7.
Find the ratio in which the yz-plane divides the line segment formed by joining the point (-2, 4, 7) and (3, -5, 8).
Solution:
Let the point P divides the line segment joining A(-2, 4, 7) and B(3, -5, 8) in the ratio k : 1.
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test Img 5
Clearly the line segment AB is divided by yz-plane so x-coordinate of point P is 0.
∴ \(\frac{3 k-2}{k+1}\) = 0 ⇒ k = \(\frac{2}{3}\)
Thus, required ratio be k : 1 i.e. 2 : 3

OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b)

Students often turn to ISC Mathematics Class 11 OP Malhotra Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) to clarify doubts and improve problem-solving skills.

S Chand Class 11 ICSE Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b)

Question 1.
Find the coordinates of the points which divide the join of the points (2, -1, 3) and (4, 3, 1) in the ratio 3 : 4 internally.
Solution:
Let R be the point which divides the join of points A(2, -1, 3) and B(4, 3, 1) in the ratio 3 : 4 internally.
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 1

Question 2.
Find the coordinates of the points which divide the line joining the points (2, -4, 3), (-4, 5, -6) in the ratio
(i) 1 : -4
(ii) 2 : 1
Solution:
(i) Let the point P divides the line segment AB in ratio 1 :-4
Then coordinates of P are
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 2

Question 3.
Find the ratio in which the line joining the points (2, 4, 5),(3, 5, -4) is divided by the yz-plane.
Solution:
Let the point P divides the line segment joining A(2, 4, 5) and B(3, 5, -4) in the ratio k : 1
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 3
Since the line joining AB is divided by yz – plane
i.e. x-coordinates of point P is 0 .
\(\frac{3 k+2}{k+1}\) = 0
⇒ k = –\(\frac{2}{3}\)
Thus required ratio be k : 1 i.e. –\(\frac{2}{3}\) : 1 i.e. – 2 : 3.

OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b)

Question 4.
The three points A(0, 0, 0), B(2, -3, 3), C(-2, 3, -3) are collinear. Find in what ratio each point divides the segment joining the other two.
Solution:
Let the point B (2, -3, 3) divides the line segment AC in the ratio k : 1 internally.
Then by section formula, we have
The coordinates of B are
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 4
∴ \( \frac{-2 k}{k+1}\) = 2
⇒ – 2k = 2k + 2
⇒ 4k = – 2
⇒ k = \(\frac{-1}{2}\)
and \(\frac{3 k}{k+1}\) = – 3
⇒ 3k = -3 k – 3
⇒ 6 k = – 3
⇒ k =-\(\frac{1}{2}\)
and \(\frac{-3 k}{k+1}\) = 3
⇒ – 3k = 3k + 3
⇒ k = –\(\frac{1}{2}\)
Thus the required ratio be k : 1 i.e. – 1 : 2.
Let the point C(-2, 3, -3) divides AB in the ratio λ : 1.
Then coordinates of C are
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 5
\(\frac{2 \lambda}{\lambda+1}\) = – 2 ⇒ 2λ = – 2λ – 2 ⇒ λ = –\(\frac { 1 }{ 2 }\)
\(\frac{-3 \lambda}{\lambda+1}\) = 3 ⇒ -3λ = 3λ + 3 ⇒ λ = –\(\frac { 1 }{ 2 }\)
and \(\frac{3 \lambda}{\lambda+1}\) = – 3 ⇒ 6λ = – 3
Thus required ratio be λ : 1 i.e. -1 : 2. Let the point A(0, 0, 0) divides line segment BC in the ratio p : 1.
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 6

Question 5.
Find the coordinates of the points which trisect AB given that A(2, 1, -3) and B (5, -8, 3).
Solution:
Let P and Q be the point of trisection of line segment AB.
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 7
Thus point P divides the line segment AB in the ratio 1 : 2.
Then coordinates of P are
\(\left(\frac{5+4}{1+2}, \frac{-8+2}{1+2}, \frac{3-6}{1+2}\right)\) i.e. P(3, -2, -1)
Also, point Q divides the line segment AB in the ratio 2 : 1.
Then coordinates of $\mathrm{Q}$ are
\(\left(\frac{10+2}{2+1}, \frac{-16+1}{2+1}, \frac{6-3}{2+1}\right)\) i.e. Q(4, -5, 1).

Question 6.
Find the coordinates of the point which is three-fifths of the way from (3, 4, 5) to (-2, -1, 0).
Solution:
Let the given points are A(3, 4, 5) and E(-2, -1, 0) and let point P is at \(\frac{3}{5}\)th of the way from A.
∴ P is at a \(\frac{2}{5}\)th of the way from B i.e. p divides line segment AB in the ratio 3 : 2.
Then by section formula, we have
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 8
Thus, the coordinates of point P are (0, 1, 2).

Question 7.
Show that the point (1, -1, 2) is common to the lines which join (6, -7), 0) to (16, -19, -4) and (0, 3, -6) to (2, -5, 10).
Solution:
Any point on line segment AB be
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 9
If AB and CD have a common point. Then P and Q coincide for some values of k and k’.
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 10
Thus eqns. (1), (2) and (3) are satisfied or consistent for k = \(\frac{-1}{3}\) and k’ = 1
putting k = \(\frac{-1}{3}\) in coordinates of P
we get, the required point be (1, -1, 2). Hence, the point P(1, -1, 2) is common to lines which join A(6, – 7, 0)
and B(16, -19, -4) and C(0, 3, -6) and D(2, -5, 10).

OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b)

Question 8.
Find the lengths of the medians of the triangle whose vertices are A(2, -3, 1), B (-6, 5, 3), C (8, 7, – 7).
Solution:
Let D, E and F are the mid-points of sides BC, CA and AB of △ABC.
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 11

Question 9.
Find the point of intersection of the medians of the triangle with vertices (-1, -3, -4), (4, -2, -7), (2, 3, -8).
Solution:
Let the vertices of △ABC are A(-1, -3, -4); B(4, -2, -7) and C(2, 3, -8)
We know that the point of intersection of all medians of a triangle is called centroid of triangle.
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 12

Question 10.
Find the ratio in which the join of A(2, 1, 5) and B(3, 4, 3) is divided by the plane 2x + 2y – 2z = 1. Also, find the coordinates of the point of division.
Solution:
Let the point R divides the line segment PQ in the ratio k : 1 internally.
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 13
Then by section formula, we have coordinates of R are
\(\left(\frac{3 k+2}{k+1}, \frac{4 k+1}{k+1}, \frac{3 k+5}{k+1}\right)\)
Clearly it is given that, line segment PQ is divided by the plane
2x + 2y – 2z = 1 …(1)
Thus the point R lies on eqn. (1); we have
\(2\left(\frac{3 k+2}{k+1}\right)+2\left(\frac{4 k+1}{k+1}\right)-2\left(\frac{3 k+5}{k+1}\right)\) = 1
⇒ 6k + 4 + 8k + 2 – 6k – 10 = k + 1
⇒ 7k = 5 ⇒ k = \(\frac{5}{7}\)
Thus the required ratio be k : 1
i.e. \(\frac{5}{7}\) : 1 i.e. 5 : 7
Thus, required point of division be
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 14

Question 11.
The mid-points of the sides of astriangle are (1, 5, -1),(0, 4, -2) and (2, 3, 4). Find its vertices.
Solution:
Let the vertices of △ABC are A (x1, y1, z1);
B (x2, y2, z2) and
C (x3, y3, z3).
It is given that D (1, 5, -1) ; E(0, 4, – 2) and F(2, 3, 4) are the mid-points of sides BC, CA and AB of △ABC.
Now D be the mid-point of BC.
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 15
On adding eqn. (1), (4) and (7); we have
x1 + x2 + x3 = 3 …(10)
From eqn. (1) and eqn. (10); x1 = 1
From (4) and (10); x2 = 3
and From eqn. (7) and (10); x3 = -1
On adding eqn. (2), (5) and (8); we have
y1 + y2 + y3 = 12 …(11)
From (2) and (11); y1 = 2
From (5) and (11); y2 = 4
From (8) and (11); y3 = 6
On adding eqn. (3), (6) and (9); we have
z1 + z2 + z3 = 1 …(12)
From (3) and (12); z1 = 3
From (6) and (12); z2 = 5
From (9) and (12); z3 = -7
Thus the required vertices of △ABC are (1, 2, 3) ;(3, 4, 5) and (-1, 6, -7).

OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b)

Question 12.
Three vertices of a parallelogram ABCD are A(3, -1, 2), B(1, 2, -4) and C(-1, 1, 2). Find the coordinates of the fourth vertex D.
Solution:
Given vertices of parallelogram ABCD are A(3, -1, 2); B(1, 2, -4); C(-1, 1, 2) and let the coordinates of fourth vertex D are (α, β ,γ).
Mid-point of AC = \(\left(\frac{3-1}{2}, \frac{-1+1}{2}, \frac{2+2}{2}\right)\)
and Mid-point of BD = \(\left(\frac{1+\alpha}{2}, \frac{2+\beta}{2}, \frac{-4+\gamma}{2}\right)\)
Since ABCD be a parallelogram.
∴ diagonals of || gm ABCD bisect each other.
Thus, mid-point of AC = mid-point of BD
OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) Img 16
Thus the coordinates of fourth vertex D are $(1,-2,8)$.

Question 13.
What is the locus of a point for which
(i) x = 0
(ii) y = 0
(iii) z = 0
(iv) x =a
(v) y = b
(vi) z = c?
Solution:
(i) The point for which x = 0 is of the form (0, y, z) and lies in yoz plane. Thus required locus of point be yz-plane.

(ii) The point for which y = 0 is of the form (x, 0, z) and lies in the xoz plane.
Thus, required locus of point be xz-plane.

(iii) The point for which z = 0 is of the form (x, y, 0) and lies in xoy plane. Thus required locus of a point be xy-plane.

(iv) Since x = a be the plane parallel to x = 0 i.e. yz plane. Thus locus of a point for which x = a be a plane parallel to yz plane at a distance of a units from it.

(v) Since y = b be the plane parallel to y = 0 i.e. xz-plane. Thus, locus of a point for which y = b be a plane parallel to xz plane at a distance b units from it.

(vi) Since z = c be a plane parallel to plane z = 0 i.e. xy plane. Thus locus of a point for which z = c be a plane || to xy plane at a distance c units from it.

Question 14.
What is the locus of a point for which
(i) x = 0, y = 0
(ii) y = 0, z = 0
(iii) z = 0, x = 0
(iv) x = a, y = b
(v) y = b, z = c
(vi) z = c, x = a?
Solution:
(i) We know that on z-axis, x = 0 = y Thus required locus be z-axis.
(ii) We know that on x-axis, we have y = 0 = z Thus required locus be x-axis.
(iii) We know that on y-axis, we have x = 0 = z Thus, required locus be y-axis.
(iv) x = a be the line || to y-axis and y = b be the line || to x-axis
Thus locus of a point for which x = a, y = b is the line of intersection of given planes x = a and y = b
(v) Clearly the locus of a point for which y = b, z = c is the line of intersection of given planes y = b and z = c
(vi) Given planes are z = c and x = a
Required locus is the line of intersection of planes z = c and x = a.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19

Continuous practice using Class 12 ISC Maths Solutions can lead to a stronger grasp of mathematical concepts.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19

Evaluate the following (1 to 9) integrals :

Question 1.
(i) ∫ \(\frac{x^5}{\sqrt{1+x^3}}\) dx
(ii) ∫ \(\frac{d x}{\sqrt{2 e^x-1}}\)
Solution:
(i) Let I = ∫ \(\frac{x^5}{\sqrt{1+x^3}}\) dx
put \(\sqrt{1+x^3}\) = t
⇒ 1 + x3 = t2
⇒ x3 = t2 – 1
⇒ 3x2 dx = 2t dt
∴ I = ∫ \(\frac{\left(t^2-1\right) 2 t d t}{3 t}\)
= \(\frac{2}{3}\) ∫ (t2 – 1) dt
= \(\frac{2}{3}\left[\frac{t^3}{3}-t\right]\)
= \(\frac{2}{9}\) (1 + x3)3/2 – \(\frac{2}{3} \sqrt{1+x^3}\) + C

(ii) Let I = ∫ \(\frac{d x}{\sqrt{2 e^x-1}}\)
put \(\sqrt{2 e^x-1}\) = t
⇒ 2 ex – 1 = t2
⇒ 2 ex dx = 2t dt
∴ I = ∫ \(\frac{t d t}{e^x \cdot t}\)
= ∫ \(\frac{\frac{d t}{t^2+1}}{\frac{2}{2}}\)
= 2 ∫ \(\frac{d t}{t^2+1^2}\)
= 2 tan-1 t + C
= 2 tan-1 \(\left(\sqrt{2 e^x-1}\right)\) + C

Question 2.
(i) ∫ \(\frac{d x}{\tan x+\cot x+\sec x+\ {cosec} x}\)
(ii) ∫ \(\frac{d x}{\sec x+\ {cosec} x}\)
Solution:
(i) ∫ \(\frac{d x}{\tan x+\cot x+\sec x+\ {cosec} x}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 1

(ii) ∫ \(\frac{d x}{\sec x+\ {cosec} x}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 2

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19

Question 3.
(i) ∫ \(\frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x}\) dx
(ii) ∫ \(\frac{\cos 5 x+\cos 4 x}{1-2 \cos 3 x}\) dx (NCERT Exemplar)
Solution:
(i) Let I = ∫ \(\frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 3

(ii) Let I = ∫ \(\frac{\cos 5 x+\cos 4 x}{1-2 \cos 3 x}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 4

Question 4.
(i) ∫ \(\frac{d x}{2 \sin x+3 \sec x}\)
(ii) ∫ \(\frac{d x}{\sin ^3 x+\cos ^3 x}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{2 \sin x+3 \sec x}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 5

(ii) Let I = ∫ \(\frac{d x}{\sin ^3 x+\cos ^3 x}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 6

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19

Question 5.
(i) ∫ \(\frac{x^2-3}{x^3-2 x^2-x+2}\) dx
(ii) ∫ x tan-1 (2x + 3) dx
Solution:
(i) Let I = ∫ \(\frac{x^2-3}{x^3-2 x^2-x+2}\) dx
= ∫ \(\frac{\left(x^2-3\right) d x}{(x-1)\left(x^2-x-2\right)}\)
= ∫ \(\frac{\left(x^2-3\right) d x}{(x-1)(x+1)(x-2)}\)
Let \(\frac{x^2-3}{(x-1)(x+1)(x-2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x-2}\) …………..(1)
Multiply both sides ofeqn. (1) by (x- 1) (x + 1) (x – 2); we get
x2 – 3 = A (x + 1) (x – 2) + B (x – 1) (x – 2) + C (x – 1) (x + 1) …………..(2)
putting x = – 1, 1, 2 successively in eqn. (2) ; we have
– 2 = B (- 2) (- 3)
⇒ B = – \(\frac{1}{3}\)
– 2 = A (2) (- 1)
⇒ A = 1
and1 = 3C
⇒ C = \(\frac{1}{3}\)
∴ from (1) ; we get

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 7

(ii) Let I = ∫ x tan-1 (2x + 3) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 8

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 9

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19

Question 6.
(i) ∫ \(\frac{\tan ^{-1} x}{x^2}\) dx
(ii) ∫ \(\frac{\log |x|}{(x+1)^3}\) dx
Solution:
(i) Let I = ∫ tan-1 x . \(\frac{1}{x^2}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 10

(ii) Let I = ∫ \(\frac{\log |x|}{(x+1)^3}\) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 11

Multiplying eqn.(2) by x (x + 1)2;
we have 1 = A (x + 1)2 + B x (x + 1) + Cx ………….(3)
puuing x = 0, – 1 successively in eqn.(3);
we have 1 = A and 1 = – C
⇒ C = – 1
Coeff. of x2 ;
0 = A + B
⇒ B = – 1
∴ from eqn. (2); we get

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 12

Question 7.
(i) ∫ ex (log x + \(\frac{1}{x^2}\)) dx
(ii) ∫ cos 2x log \(\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\) dx
Solution:
(i) Let I = ∫ ex (log x + \(\frac{1}{x^2}\)) dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 13

(ii) Let I = ∫ cos 2x log \(\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\) dx [by parts]

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 14

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19

Question 8.
(i) ∫ \(\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}\) ex dx
(ii) ∫ \(\frac{x^2-1}{\left(x^4+3 x^2+1\right) \tan ^{-1}\left(x+\frac{1}{x}\right)}\) dx
Solution:
(i) Let I = ∫ \(\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}\) ex dx

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 15

(ii) Let I = ∫ \(\frac{x^2-1}{\left(x^4+3 x^2+1\right) \tan ^{-1}\left(x+\frac{1}{x}\right)}\) dx
Divide numerator and denominator by x2 ; we have

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 16

Question 9.
\(\int_0^\pi\) |cos x – sin x| dx
Solution:
Let I = \(\int_0^\pi\) |cos x – sin x| dx
When 0 ≤ x ≤ \(\frac{\pi}{4}\) ;
cos x ≥ sin x
⇒ cos x – sin x ≥ 0
∴ |cos x – sin x| = cos x – sin x
When \(\frac{\pi}{4}\) ≤ x ≤ 7 ;
sin x ≥ cos x
⇒ cos x – sin x ≤ 0
∴ |cos x – sin x| = – (cos x – sin x)
∴ I = \(\int_0^{\pi / 4}\) |cos x – sin x| dx + \(\int_{\pi / 4}^\pi\) |cos x – sin x| dx
= \(\int_0^{\pi / 4}\) (cos x – sin x) dx + \(\int_{\pi / 4}^\pi\) – (cos x – sin x) dx
= [sin x + cos x\(]_0^{\pi / 4}\) – [sin x + cos x\(]_{\pi / 4}^\pi\)
= \(\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1\right]-\left[0-1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right]\)
= (√2 – 1) – (- 1 – √2)
= 2√2

Question 10.
Prove that \(\int_0^\pi \frac{x}{1-\cos \alpha \sin x} d x=\frac{\pi(\pi-\alpha)}{\sin \alpha}\).
Solution:
Let I = \(\int_0^\pi \frac{x}{1-\cos \alpha \sin x}\) dx ……………(1)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 17

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 19

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19

Question 11.
Prove that \(\int_0^{\pi / 2} \frac{f(\sin x)}{f(\sin x)+f(\cos x)} d x=\frac{\pi}{4}\).
Solution:
Let I = \(\int_0^{\pi / 2} \frac{f(\sin x)}{f(\sin x)+f(\cos x)} d x=\frac{\pi}{4}\)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 19

Question 12.
Evaluate \(\int_0^2 \frac{d x}{\left(17+8 x-4 x^2\right)\left(e^{6(1-x)}+1\right)}\).
Solution:
Let I = \(\int_0^2 \frac{d x}{\left(17+8 x-4 x^2\right)\left(e^{6(1-x)}+1\right)}\) ……………(1)

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 20

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19 21

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19

Question 13.
If p (x) is a polynomial of least degree that has maximum value equal to 6 at x = 1, and a minimum value equal to 2 at x = 3, then show that \(\int_0^1\) p(x) dx = \(\frac{19}{4}\).
Solution:
Since p (x) be a polynomial of least degree
and it has maximum and minimum values.
∴ p’ (x) must be atleast polynomial of degree 2.
Thus p (x) must be a polynomial of degree 3.
Let p(x) = ax3 + bx2 + cx + d ……………….(1)
∴ p’(x) = 3ax2 + 2bx + c
Since p (x) has maximum values equal to 6 at x = 1
and a minimum value equal to 2 at x = 3.
∴ p'(1) = 0 – p'(3)
⇒ 3a + 2b + c = 0 …………..(2)
and 27a + 6b + c = 0 ……………(3)
also p(1) 6
⇒ a + b + c + d= 6 …………(4)
and p (3) = 2
⇒ 27a + 9b + 3c + d = 2 …………….(5)
eqn. (5) – eqn. (4) gives:
26a + 8b + 2c = – 4
⇒ 13a + 4b + c = – 2 ……………..(6)
eqn. (3) – eqn. (2) gives;
24a + 4b = 0
⇒ 6a + b = 0 …………….(7)
eqn. (3) – eqn. (6) gives
14a + 2b = 2
⇒ 7a + b = 1 ……………(8)
eqn. (8) – eqn. (7) gives;
a = 1
∴ from (7);
b = – 6
∴ from eqn.(6);
13 – 24 + c = – 2
⇒ c = 9
from eqn (4) ;
1 – 6 + 9 + d = 6
⇒ d = 2
∴ from (1) ;
p(x) = x3 – 6x2 + 9x + 2
∴ \(\int_0^1\) p(x) dx = \(\int_0^1\) (x3 – 6x2 + 9x + 2) dx
= \(\left[\frac{x^4}{4}-2 x^3+\frac{9 x^2}{2}+2 x\right]_0^1\)
= \(\left[\frac{1}{4}-2+\frac{9}{2}+2\right]=\frac{19}{4}\)

OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(a)

The availability of step-by-step ISC Mathematics Class 11 OP Malhotra Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(a) can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(a)

Question 1.
Find the distance from the origin to each of the points :
(i) (2, 2, 3)
(ii) (4, -1, 2)
(iii) (0, 4, -4)
(iv) (-4, -3, -2)
Solution:
(i) Required distance of origin O(0, 0, 0) from P(2, 2, 3) = \(\sqrt{(2-0)^2+(2-0)^2+(3-0)^2}\) = \(\sqrt{4+4+9}\) = \(\sqrt{17}\)
(ii) Required distance =\(\sqrt{(4-0)^2+(-1-0)^2+(2-0)^2}\) = \(\sqrt{16+1+4}\) = \(\sqrt{21}\)
(iii) Required distance = \(\sqrt{(0-0)^2+(4-0)^2+(-4-0)^2}\) = \(\sqrt{16+16}\) = \(4 \sqrt{2}\)
(iv) Required distance = \(\sqrt{(-4-0)^2+(-3-0)^2+(-2-0)^2}\) = \(\sqrt{16+9+4}\) = \(\sqrt{29}\)

Question 2.
Find the distance between each of the following pairs of points :
(i) (2, 5, 3) and (-3, 2, 1);
(ii)(0, 3, 0) and (6, 0, 2);
(iii) (-4, -2, 0) and (3, 3, 5).
Solution:
(i) Required distance = \(\sqrt{(-3-2)^2+(2-5)^2+(1-3)^2}\) = \(\sqrt{25+9+4}\) = \(\sqrt{38}\)
(ii) Required distance = \(\sqrt{(6-0)^2+(0-3)^2+(2-0)^2}\) = \(\sqrt{36+9+4}\) = \(\sqrt{49}\) = 7
(iii) Required distance = \(\sqrt{(3+4)^2+(3+2)^2+(5-0)^2}\) = \(\sqrt{49+25+25}\) = \(\sqrt{99}\) = \(3 \sqrt{11}\)

OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(a)

Question 3.
Show that the triangle with vertices (6, 10, 10),(1, 0, -5),(6, -10, 0) is a right-angled triangle, and find its axes.
Solution:
Let the given vertices of triangle are A(6, 10, 10); B(1, 0, -5) and C (6, -10, 0)
∴ AB = \(\sqrt{(1-6)^2+(0-10)^2+(-5-10)^2}\) = \(\sqrt{25+100+225}\) = \(\sqrt{350}\)
BC = \(\sqrt{(6-1)^2+(-10-0)^2+(0+5)^2}\) = \(\sqrt{25+100+25}\) = \(\sqrt{150}\)
CA = \(\sqrt{(6-6)^2+(-10-10)^2+(0-10)^2}\) = \(\sqrt{400+100}\) = \(\sqrt{500}\)
∴ AB2 + BC2 =350 + 150 = 500 = CA2
Thus △ABC be right angled △ at B.
Therefore A, B and C are the vertices of right angled triangle.
∴ area of △ABC = \(\frac { 1 }{ 2 }\) (AB) × (BC) = \(\frac { 1 }{ 2 }\) × \(\sqrt{350} \sqrt{150}\) = \(\frac { 1 }{ 2 }\) \(\sqrt{35 \times 15 \times 100}\)
= \(\frac { 1 }{ 2 }\) × 50√21 = 25√21 sq. units

Question 4.
Show that the triangle with vertices A(3, 5, -4), B(-1, 1, 2), C(-5, -5, -2) is isosceles.
Solution:
Given vertices of triangle are A (3, 5, -4), B(-1, 1, 2) and C(-5, -5, -2).
AB = \(\sqrt{(-1-3)^2+(1-3)^2+(2+4)^2}\) = \(\sqrt{16+16+36}\) = \(\sqrt{68}\) = \(2 \sqrt{17}\)
BC = \(\sqrt{(-5+1)^2+(-5-1)^2+(-2-2)^2}\) = \(\sqrt{16+36+16}\) = \(\sqrt{68}\)
CA = \(\sqrt{(3+5)^2+(5+5)^2+(-4+2)^2}\) = \(\sqrt{64+100+4}\) = \(\sqrt{168}\)
Thus AB = BC
∴ △ ABC be an isosceles triangle.

Question 5.
Show that (4, 2, 4),(10, 2, -2) and (2, 0, -4) are the vertices of an equilateral triangle.
Solution:
Let the vertices of triangle are A(4, 2, 4), B(10, 2, -2) and C(2, 0, -4).
AB = \(\sqrt{(10-4)^2+(2-2)^2+(-2-4)^2}\) = \(\sqrt{36+36}\) = \(6 \sqrt{2}\) = \(\sqrt{72}\)
BC = \(\sqrt{(2-10)^2+(0-2)^2+(-4+2)^2}\) = \(\sqrt{64+4+4}\) = \(\sqrt{72}\)
and CA = \(\sqrt{(4-2)^2+(2-0)^2+(4+4)^2}\) = \(\sqrt{4+4+64}\) = \(\sqrt{72}\)
Thus AB = BC = CA
∴ △ ABC is an equilateral triangle.

Question 6.
Show that the points (1, -1, 3),(2, -4, 5) and (5, -13, 11) are collinear.
Solution:
Given points are A(1, -1, 3); B(2, -4, 5) and C(5, -13, 11)
Here AB = \(\sqrt{(2-1)^2+(-4+1)^2+(5-3)^2}\) = \(\sqrt{1+9+4}\) = \(\sqrt{14}\)
BC = \(\sqrt{(5-2)^2+(-13+4)^2+(11-5)^2}\) = \(\sqrt{9+81+36}\) = \(\sqrt{126}\) = \(3 \sqrt{14}\)
CA = \(\sqrt{(1-5)^2+(-1+13)^2+(3-11)^2}\) = \(\sqrt{16+144+64}\) = \(\sqrt{224}\) = \(4 \sqrt{14}\)
Clearly AB + BC = CA
Thus the points A, B and C are collinear.

Question 7.
Derive the equation of the locus of a point equidistant from the points (1, -2, 3) and (-3, 4, 2).
Solution:
Let P(x, y, z) be any point on locus and A(1, -2, 3), B(-3, 4, 2) are given points.
According to given condition, we have |PA| = | PB |
\(\sqrt{(x-1)^2+(y+2)^2+(z-3)^2}\) = \(\sqrt{(x+3)^2+(y-4)^2+(z-2)^2}\)
On squaring both sides; we have
(x – 1)2 + (y + 2)2 + (z – 3)2 =(x + 3)2 + (y – 4)2 + (z – 2)2
⇒ -2x +4y – 6z + 14 = 6x – 8y – 4z + 29
⇒ 8x – 12y + 2z + 15 = 0
which is the required locus of a point.

Question 8.
Derive the equation of the locus of a point twice as far from (-2, 3, 4) as from (3, -1, -2).
Solution:
Let P(x, y, z) be any point on the locus
such that PA = 2 PB
where A be the point (-2, 3, 4) and B be the point (+3, -1, -2).
\(\sqrt{(x+2)^2+(y-3)^2+(z-4)^2}\) = \(2 \sqrt{(x-3)^2+(y+1)^2+(z+2)^2}\)
On squaring both sides; we have
(x + 2)2 + (y – 3)2 + (z – 4)2 =4[(x – 3)2 + (y + 1)2 + (z + 2)2]
⇒ 3x2 + 3y2 + 3z2 – 28x + 14y + 24z + 27 = 0
which is the required locus.

Question 9.
Find the equation of the locus of a point whose distance from the y-axis is equal to its distance from (2, 1, -1).
Solution:
Let P(x, y, z) be any point on locus and A(0, y, 0) be any point on y-axis and B(2, 1, -1) be the given point.
According to given condition, PA = PB
\(\sqrt{(x-0)^2+(y-y)^2+(z-0)^2}\) = \(\sqrt{(x-2)^2+(y-1)^2+(z+1)^2}\)
On squaring both sides; we have
x2 + z2 = x2 + y2 + z2 – 4x – 2y + 2z + 6
⇒ y2 – 4x – 2y + 2z + 6 = 0 be the required eqn. of locus.

OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(a)

Question 10.
Find the equation of the locus of a point whose distance from the xy-plane is equal to distance from the point (-1, 2, -3).
Solution:
Let P(x, y, z) be any point on locus and A(x, y, 0) be any point in xy plane and B(-1, 2, -3) b the given point.
Then according to given condition; we have PA = PB
\(\sqrt{(x-x)^2+(y-y)^2+(z-0)^2}\) = \(\sqrt{(x+1)^2+(y-2)^2+(z+3)^2}\)
On squaring both sides; we have
0 + 0 + z2 = x2 + y2 + z2 + 2x – 4y + 6z + 14
⇒ x2 + y2 + 2x – 4y + 6z + 14 = 0
which is the required locus.

Question 11.
A point moves so that the difference of the squares of its distances from the x-axis and the y-axis is constant. Find the equatiol of its locus.
Solution:
Let P(x, y, z) be any point on locus and let Q(x, 0, 0) and R(0, y, 0) be any two points on x-axis and y-axis.
Then according to given condition; we have
PQ2 – PR2 = constant = k
⇒ \(\left[\sqrt{(x-x)^2+(y-0)^2+(z-0)^2}\right]^2\) – \(\left[\sqrt{(x-0)^2+(y-y)^2+(z-0)^2}\right]^2\) = k
⇒ y2 + z2 – x2 – z2 = k
⇒ y2 – x2 = k
which is the required locus.

Question 12.
Find the equation of the locus of a point whose distance from the z-axis is equal to its distance from the xy-plane.
Solution:
Let P(x, y, z) be any point on locus and let Q(0, 0, z) be any point on z-axis and R(x, y, 0) be any point in xy-plane.
Then PQ = PR
⇒ \(\sqrt{(x-0)^2+(y-0)^2+(z-z)^2}\) = \(\sqrt{(x-x)^2+(y-y)^2+(z-0)^2}\)
On squaring both sides; we have
x2 + y2 = z2
⇒ x2 + y2 – z2 = 0
which is the required locus.

OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Chapter Test

The availability of step-by-step ISC Mathematics Class 11 OP Malhotra Solutions Chapter 25 Hyperbola Chapter Test can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 25 Hyperbola Chapter Test

Question 1.
Find the eccentricity and the coordinate of foci of the hyperbola 25x2 – 9y2 = 225.
Solution:
Given eqn. of hyperbola can be written as ;
\(\frac{x^2}{9}\) – \(\frac{y^2}{25}\) = 1 …(1)
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
we have a2 = 9 and b2 = 25
We know that b2 = a2 (e2 – 1)
⇒ 25 = 9(e2 – 1)
⇒ \(\frac{25}{9}\) + 1 = e2
⇒ e = \(\frac{\sqrt{34}}{3}\) (∵ e > 0)
Thus the required eccentricity of given eqn. (1) be \(\frac{\sqrt{34}}{3}\)
The coordinates of foci are ( ± ae, 0) i.e. \(\left( \pm 3 \times \frac{\sqrt{34}}{3}, 0\right)\) i.e. \(( \pm \sqrt{34}, 0)\).

OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Chapter Test

Question 2.
Find the value (s) of k so that the line 2x + y + k = 0 may touch the hyperbola 3x2 – y2 = 3.
Solution:
eqn. of given hyperbola be 3x2 – y2 = 3 ⇒ \(\frac{x^2}{1}\) – \(\frac{y^2}{3}\) = 1
On comparing eqn. (1) with\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
we have a2 = 1; b2 = 3
eqn. of given line be
2x + y + k = 0 ⇒ y = – 2x – k
We know that the line y = mx + c touches hyperbola (1) if c = ±\(\sqrt{a^2 m^2-b^2}\)
Here m = – 2 and c = – k
Thus, eqn. (2) touches hyperbola (1)
if – k = ± \(\sqrt{1 \times(-2)^2-3}\)
⇒ – k = ± 1
⇒ k = ± 1

Question 3.
From the following information, find the equation of the hyperbola and the equation of the transverse axis.
Focus (-2, 1), Directrix : 2x – 3y + 1 = 0, e = \(\frac{2}{\sqrt{3}}\)
Solution:
Let P (x, y) be any point on the hyperbola.
Then by definition, we have | PF | = | PM |
OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Chapter Test Img 1
On squaring both sides ; we have
(x + 2)2 + (y – 1)2 =\frac{4}{39}(2x – 3y + 1)2
⇒ 39 [x2 + 4x + 4 + y2 – 2y + 1]
= 4 [4x2 + 9y2 – 12xy – 6y + 1+ 4x]
which is the required eqn. of hyperbola.
Axis of hyperbola is a line ⊥ to the directrix and pass through the focus (- 2, 1).
Thus eqn. of line ⊥ to 2x – 3y + 1 = 0 be given by
3x + 2y + k = 0 …(1)
eqn. (1) pass through the point (- 2, 1).
-6 + 2 + k = 0
⇒ k = 4
Thus eqn. (1) reduces to ; 3x + 2y + 4 = 0 be the required eqn. of axis of hyperbola.

Question 4.
Find the equation of the hyperbola whose eccentricity is √5 and the sum of whose semi-axes is 9.
Solution:
Let the required eqn. of hyperbola be
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
given eccentricity of hyperbola be √5
∴ e = √5
Let a be length of semi-major and b be the length of semi-minor axes of hyperbola.
According to given condition, we have
a + b = 9 …(2)
We know that b2 = a2 (e2 – 1)
⇒ b2 = a2(5 – 1) = 4a2
⇒ (9 – a)2 = 4a2 [using eqn. (2)]
⇒ 3a2 + 18a – 81 = 0
⇒ (a – 3)(3a + 27) = 0
⇒ a = 3 (∵ a > 0)
∴ from (2); b = 9 – 3 = 6
Thus eqn. (1) reduces to ;
\(\frac{x^2}{9}\) – \(\frac{y^2}{36}\) = 1
which is the required eqn. of hyperbola.

Question 5.
Find the equation of the hyperbola whose foci are (4, 1),(8, 1) and whose eccentricity is 2.
Solution:
We know that centre is the mid-point of line joining the two foci (4, 1) and (8, 1).
∴ Coordinates of centre of hyperbola are \(\left(\frac{4+8}{2}, \frac{1+1}{2}\right)\) i.e. (6, 1).
Since ordinate of both foci are identical
∴ Transverse axis of the hyperbola is parallel to x-axis.
Thus eqn. of hyperbola can be taken as
\(\frac{(x-6)^2}{a^2}\) – \(\frac{(y-1)^2}{b^2}\) = 1
Given eccentricity of eqn. (1) be 2 i.e. e = 2
∴ distance between foci
= \(\sqrt{(8-4)^2+(1-1)^2}\) = 4
⇒ 2ae = 4 ⇒ ae = 2 ⇒ a = 1
∴ b2 = a2 (e2 – 1) = 12(4 – 1) = 3
Thus, eqn. (1) reduces to ;
\(\frac{(x-6)^2}{1}\) – \(\frac{(y-1)^2}{3}\) = 1
⇒ 3 (x – 6)2 – (y – 1)2 = 3
which is the required eqn. of hyperbola.

Question 6.
Show that the line y = x + √7 touches the hyperbola 9x2 – 16y2 = 144.
Solution:
Given eqn. of hyperbola be
9x2 – 16y2 = 144
and eqn. of given line be
y = x + √7 …(2)
From (2); y = x + √7, putting in eqn. (1); we get
9x2 – 16 (x + √7)2 = 144
⇒ 9x2 – 16 (x2 + 7 + 2√7 x) – 144 = 0
⇒ -7x2 – 32√7x – 256 = 0
⇒ 7x2 + 32√7x + 256 = 0
⇒ (√7x +1 6)2 = 0
eqn. (3) is quadratic in x and have equal roots
∴ from (3) ; x = \(\frac{-16}{\sqrt{7}}\)
Thus line (2) touches hyperbola eqn. (1).

OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Chapter Test

Question 7.
Find the equation of the hyperbola whose foci are (0, ± 13) and the length of the conjugate axis is 20.
Solution:
Coordinates of foci are (0, ± 13) which are lies on y-axis and y-axis be the transverse axis of hyperbola.
Let the eqn. of hyperbola be,
\(\frac{y^2}{a^2}\) – \(\frac{x^2}{b^2}\) = 1
Here ae = 13
and 2b = 20 and b = 10
We know that, b2 = a2 (e2 – 1)
⇒ 100 = 132 – a2
⇒ a2 = 69
Thus eqn. (1) reduces to ; \(\frac{y^2}{69}\) – \(\frac{x^2}{100}\) = 1,
which is the required eqn. of hyperbola.

Question 8.
Find the equation of the hyperbola whose transverse and conjugate axes are the x and y axes respectively, given that the length of conjugate axis is 5 and distance between the foci is 13 .
Solution:
Let the eqn. of hyperbola be
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 …(1)
Given length of conjugate axis be 5
∴ 2b = 5 ⇒ b = \(\frac{5}{2}\)
and distance between foci = 13
⇒ 2 ae = 13 …(2)
We know that b2 = a2 (e2 – 1)
⇒ \(\left(\frac{5}{2}\right)^2\) = \(\left(\frac{13}{2}\right)^2\) – a2
⇒ a2 = \(\frac{169}{4}\) – \(\frac{25}{4}\) = \(\frac{144}{4}\) = 36
Thus, eqn. (1) reduces to;
\(\frac{x^2}{36}\) – \(\frac{y^2}{25 / 4}\) = 1
⇒ \(\frac{x^2}{36}\) – \(\frac{4 y^2}{25}\) = 1
⇒ 25x2 – 144y2 = 900
which is the required eqn. of hyperbola.

Question 9.
Find the equation to the conic whose focus is (1, -1), eccentricity is \(\frac{1}{2}\) and the directrix is the line x – y = 3. Is the conic section an ellipse?
Solution:
Given focus of ellipse be F (1, -1) and eqn. of directrix be x – y -3 = 0 and e = \(\frac{1}{2}\)
Let P (x, y) be any point on ellipse. Then by def. of ellipse, we have | PF | = e | PM |
where PM be the ⊥ drawn from P on given directrix.
OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Chapter Test Img 2
On squaring both sides, we have
8[(x – 1)2 + (y + 1)2] = (x – y – 3)2
⇒ 8 (x2 – 2x + y2 + 2y + 2) = x2 + y2 + 9 – 2xy + 6y – 6x
⇒ 7x2 + 7y2 + 2xy – 10x + 10y + 7 = 0
which is the required eqn. of ellipse.

OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Ex 25(b)

Utilizing ISC Mathematics Class 11 OP Malhotra Solutions Chapter 25 Hyperbola Ex 25(b) as a study aid can enhance exam preparation.

S Chand Class 11 ICSE Maths Solutions Chapter 25 Hyperbola Ex 25(b)

Question 1.
Find the tangent to the parabola y2 = 16x, making an angle of 45° with the x-axis.
Solution:
The eqn. of tangent to parabola be
y = mx + \(\frac{a}{m}\) …(1)
Here m = tan 45° = 1
On comparing y2 = 16x with y2 = 4ax ;
we have, 16 = 4a
⇒ a = 4
Thus eqn. (1) reduces to ; y = x + 4
which is the required eqn. of tangent to given parabola.

Question 2.
A tangent to the parabola y2 = 16x makes an angle of 60° with the x-axis. Find its point of contact.
Solution:
We know that, the line y = mx + c may touch the parabola y2 =4ax then the point of contact be given by \(\left(\frac{a}{m^2}, \frac{2 a}{m}\right)\),
On comparing y2 = 16x with y2 = 4ax
we have, 16 = 4a ⇒ a = 4
and m = tan 60° = √3
∴ required point of contact be \(\left(\frac{4}{3}, \frac{2 \times 4}{\sqrt{3}}\right)\) i.e. \(\left(\frac{4}{\sqrt{3}}, \frac{8}{\sqrt{3}}\right)\)

OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Ex 25(b)

Question 3.
(i) Find the equations of the tangents to the parabola y2 = 6x which pass through the point \(\left(\frac{3}{2}, 5\right)\).
(ii) Find the equations of the tangents to the parabola y2 + 12x = 0 from the point (3, 8).
Solution:
(i) On comparing y2 = 6x with y2 = 4ax; we have
4a = 6 ⇒ a = \(\frac{3}{2}\)
We know that, eqn. of any tangent to parabola y2 = 4ax be given by
y = mx + \(\frac{a}{m}\) …(1)
Thus required eqn. of tangent to parabola y2 = 6x be given by
y = mx + \(\frac{3}{2 m}\) …(2)
Now eqn. (2) passes through the point \(\left(\frac{3}{2}, 5\right)\).
5 = \(\frac{3}{2}\) m + \(\frac{3}{2 m}\)
⇒ 10m = 3m2 + 3
⇒ 3m2 – 10m + 3 = 0
⇒ (m – 3) (3m – 1) = 0
⇒ m = 3, \(\frac{1}{3}\)
putting the values of m in eqn. (2); we have
y = 3x + \(\frac{3}{2 \times 3}\)
⇒ y = 3x + \(\frac{1}{2}\)
⇒ 2y = 6x + 1 …(3)
and y = \(\frac{1}{3}\)x + \(\frac{3 \times 3}{2 \times 1}\)
⇒ y = \(\frac{x}{3}\) + \(\frac{9}{2}\)
⇒ 6y = 2x + 27 …(4)
Thus eqn. (3) and eqn. (4) are the required eqns. of tangents to given parabola.

(ii) Given eqn. of parabola be
y2 = -12x …(1)
On comparing eqn. (1) with y2 = 4ax
we have, 4a = – 12 ⇒ a = – 3
We know that, eqn. of any tangent to parabola y2 = 4ax be given by
y = mx + \(\frac{a}{m}\)
Thus eqn. of any tangent to given parabola (1) be given by
y = mx – \(\frac{3}{m}\)
Now eqn. (2) passes through the point (3, 8).
8 = 3m – \(\frac{3}{m}\) ⇒ 3m2 – 8m – 3 = 0
⇒ m = \(\frac{8 \pm \sqrt{64+36}}{6}\) = \(\frac{8 \pm 10}{6}\) = 3, \(\frac{1}{3}\)
putting m = 3 in eqn. (2) ; we have
y = 3x – 1 …(3)
putting m = –\(\frac{1}{3}\) in eqn. (2) ; we have
y = –\(\frac{1}{3}\) + 9 ⇒ 3y = -x + 27 …(4)
Thus, eqn. (3) and eqn. (4) gives the required tangents to given parabola.

Question 4.
Show that the line 12y – 20x – 9 = 0 touches the parabola y2 = 5x.
Solution:
Given eqn. of parabola be
y2 = 5x …(1)
On comparing eqn. (1) with y2 = 4ax ; we have
4a = 5
⇒ a = \(\frac{5}{4}\)
Given eqn. of line can be written as ;
y = \(\frac{20 x}{12}\) + \(\frac{9}{12}\)
⇒ y = \(\frac{5 x}{3}\) + \(\frac{3}{4}\) …(2)
Comparing eqn. (2) with y = mx + c
we have, m = \(\frac{5}{3}\) and c = \(\frac{3}{4}\)
Here, \(\frac{a}{m}\) = \(\frac{\frac{5}{4}}{\frac{5}{3}}\) = \(\frac{3}{4}\) = c
Thus line (2) touches the parabola (1).

Question 5.
Show that the line x + y = 1 touches the parabola y = x – x2.
Solution:
Given eqn. of line be
x + y = 1 …(1)
and eqn. of parabola be y = x – x2 …(2)
From (1); y = 1 – x, putting in eqn. (2); we have
1 – x = x – x2
⇒ x2 – 2x + 1 = 0
⇒ (x – 1)2 = 0 …(3)
i.e. eqn. (3) have equal roots.
Thus eqn. (1) touches eqn. (2).

Question 6.
Show that the line x + ny + an2 = 0 touches the parabola y2 = 4ax and find the point of contact.
Solution:
Given eqn. of line be
x + ny + an2 = 0 …(1)
y2 = 4ax …(2)
From (1); y = \(\frac{-x-a n^2}{n}\)
putting the value of y in eqn. (2) ; we have
OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Ex 25(b) Img 1
which is quadratic in x and have equal roots. Thus line (1) touches parabola (2).
∴ from (3); x = an2
from (1); ny + 2an2 = 0 ⇒ y = – 2an
Hence the required point of contact be (an2, – 2an).

OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Ex 25(b)

Question 7.
Find the tangents to the ellipse x2 + 9y2 = 3, which are (i) parallel (ii) perpendicular to the line 3x + 4y = 9.
Solution:
Given eqn. of ellipse be
x2 + 9y2 = 3 ⇒ \(\frac{x^2}{3}\) + \(\frac{y^2}{\frac{1}{3}}\) = 1 …(1)
which is a horizontal ellipse.
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
we have a2 = 3 and b2 = \(\frac{1}{3}\)
eqn. of given line be
3x + 4y – 9 = 0 …(2)
∴ slope of line (2) = –\(\frac{3}{4}\)
∴ slope of line || to line (2) = –\(\frac{3}{4}\) = m
The eqns. of tangents to ellipse (1) be
OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Ex 25(b) Img 2

(ii) slope of line ⊥ to line (2)
= \(-\frac{1}{-\frac{3}{4}}\) = \(\frac{4}{3}\) = m
∴ required eqns. of tangents to ellipse (1) be given by
OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Ex 25(b) Img 3

Question 8.
Find the equations of the tangents to the ellipse \(\frac{x^2}{2}\) + \(\frac{y^2}{7}\) = 1 that make an angle of 45° with the x-axis.
Solution:
Given eqn. of ellipse be
\(\frac{x^2}{2}\) + \(\frac{y^2}{7}\) = 1 …(1)
which is a vertical ellipse.
On comparing eqn. (1) with
\(\frac{x^2}{b^2}\) + \(\frac{y^2}{a^2}\) = 1, a > b > o
We have a2 = 7 ; b2 = 2
Here m = tan 45° = 1
The eqns. of tangents to given ellipse (1) be given by
y = mx ± \(\sqrt{b^2 m^2+a^2}\)
⇒ y = x ± \(\sqrt{2+7}\)
⇒ y = x ± 3

Question 9.
Find the equation of the tangents to the ellipse \(\frac{x^2}{16}\) + \(\frac{y^2}{9}\) = 1, which make equal intercepts on the axes.
Solution:
Given eqn. of ellipse be,
\(\frac{x^2}{16}\) + \(\frac{y^2}{9}\) = 1 …(1)
which is a horizontal ellipse.
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
where a > b > 0
we have a2 = 16 and b2 = 9
Since tangents makes equal intercepts on axes
∴ slope of tangents = m = ± 1
Thus required eqns. of tangents to eqn. (1) be given by
OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Ex 25(b) Img 4

Question 10.
Find the value of ‘c’ so that 2x -y + c = 0 may touch the ellipse x2 + 2y2 = 2.
Solution:
Given eqn. of line be 2x -y + c = 0
⇒ y = 2x + c …(1)
and eqn. of given ellipse be x2 + 2y2 = 2
⇒ \(\frac{x^2}{2}\) + \(\frac{y^2}{1}\) = 1 …(2)
We know that the line y = mx + c touches the ellipse \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
if c = ± \(\sqrt{a^2 m^2+b^2}\)
if c = ± \(\sqrt{2 \times 2^2+1}\)
⇒ c = ± 3
[Here m = 2 ; a2 = 2 ; b2 = 1]

Question 11.
Show that the line lx + my = 1 will touch the ellipse \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 if a2l2 + b2m2 = 1.
Solution:
eqn. of given line be
lx + my = 1 …(1)
and eqn. of ellipse be \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 … (2)
From (1) ; y = \(\frac{1-l x}{m}\)
putting the value of y in eqn. (2); we have
\(\frac{x^2}{a^2}\) + \(\frac{1}{b^2}\) \(\left(\frac{1-l x}{m}\right)^2\) = 1
⇒ m2b2x2 + a2 (1 – lx)2 = a2b2m2
⇒ x2 [m2b2 + a2l2] – 2a2lx + a2 – a2b2m2 = 0 …(3)
Now eqn. (1) touches eqn. (2)
if roots of quadratic eqn. (3) are equal
if Discriminant = 0
if (- 2a2l)2 – 4 (b2m2 + a2l2) (a2 – a2b2m2) = 0
if 4a4l2 – 4a2b2m2 + 4a2b4m4 – 4a4l2 + 4a4b2l2m2 = 0
if 4a2b2m2 (b2m2 + a2l2 – 1) = 0
if b2m2 + a2l2 = 1 which is the required condition.

Question 12.
Show that the following lines are tangents to the given hyperbola and determine the points of contact.
(i) x + 1 = 0, 4x2 – 3y2 = 4
(ii) x – 2y + 1 = 0, x2 – 6y2 = 3
Solution:
Given eqn. of line be x + 1 = 0 ⇒ x = – 1 …(1)
and eqn. of given hyperbola be 4x2 – 3y2 = 4 i.e. \(\frac{x^2}{1}\) – \(\frac{y^2}{\frac{4}{3}}\) = 1
putting eqn. (1) in eqn. (2) ; we have
4 – 3x2 = 4 ⇒ y2 = 0 which is quadratic in y and gives equal roots and each root be 0.
Thus line (1) touches hyperbola (2).
Putting y = 0 in eqn. (1); x = – 1
Hence the required point of contact be (- 1, 0)

(ii) Given eqn. of line be x- 2y + 1 = 0 …(1)
and eqn. of hyperbola be x2 – 6y2 = 3 …(2)
From (1); y = \(\frac{x+1}{2}\), putting the value of y in eqn. (2); we have
x2 – 6\(\left(\frac{x+1}{2}\right)^2\) = 3
⇒ x2 – \(\frac{3}{2}\) (x + 1)2 = 3
⇒ 2x2 – 3 (x2 + 2x + 1) = 6
⇒ -x2 – 6x – 9 = 0
⇒ (x – 3)2 = 0
Clearly eqn. (3) have equal roots.
Thus line (1) touches given hyperbola (2).
∴ from (3); x + 3 = 0 ⇒ x = – 3
∴ from (1); y = – 1
Thus, the required point of contact be (- 3, – 1).

OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Ex 25(b)

Question 13.
Find the equations of the tangents to the hyperbola 2x2 – 3y2 = 6, which re parallel to the Iine x + y – 2 = 0.
Solution:
eqn. of given hyperbola be, 2x2 – 3y2 = 6 ⇒ \(\frac{x^2}{3}\) – \(\frac{y^2}{2}\) = 1
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
we have a2 = 3 and b2 = 2
and eqn. of given line be x + y – 2 = 0
i.e. y = – x + 2 …(2)
On comparing eqn. (2) with y = mx + c
we have m = – 1 and c = 2
Thus the required eqns. of tangents to eqn. (1) be given by
y = mx ± \(\sqrt{a^2 m^2-b^2}\)
⇒ y =- x ± \(\sqrt{3(-1)^2-2}\)
⇒ y = – x ± 1
⇒ x + y ± 1 = 0

Question 14.
The tangents from P to the hyperbola \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 are mutually perpendicular, show that the locus of P is the circle x2 + y2 = a2 – b2.
Solution:
Given eqn. of hyperbola be
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 …(1)
Thus, eqn. of tangents to hyperbola (1) at P (x, y) be given by
y = mx ± \(\sqrt{a^2 m^2-b^2}\)
⇒ y – mx = ± \(\sqrt{a^2 m^2-b^2}\)
On squaring both sides ; we have
(y – mx)2 = a2m2 – b2
⇒ m2x2 +y2 – 2myx + b2 – a2m2 = 0
⇒ m2x2 – 2myx + (b2 – a2m2 + y2) = 0
⇒ m2(x2 – a2) – 2xym + b2 + y2 = 0 …(2)
eqn. (2) is a quadratic in m and hence two roots say m1 and m2
∴ product of roots = m1 m2 = – 1
⇒ \(\frac{y^2+b^2}{x^2-a^2}\) = -1
y2 + b2 = -x2 + a2
⇒ x2 +y2 = a2 – b2 which is the required locus of point P.

Question 15.
Show that the straight line x + y = 1 touches the hyperbola 2x2 – 3y2 = 6. Also find the coordinates of the point of contact.
SoLUTION:
Given eqn. of line be
x + y =1 …(1)
and eqn. of given hyperbola be
2x2 – 3y2 = 6 …(2)
∴ from (1); y = 1 – x, putting in eqn. (3); we get
2x2 – 3 (1 – x)2 = 6
⇒ 2x2 – 3 (x2 + 1 – 2x) = 6
⇒ -x2 + 6x – 9 = 0
⇒ x2 – 6x + 9 = 0
⇒ (x – 3)2 = 0 …(3)
Thus eqn. (3) is a quadratic in x and have two equal roots i.e. x = + 3, + 3.
∴ eqn. (1) touches hyperbola (2).
putting x = + 3 in eqn. (1); we have y = -2.
∴ required point of contact be (+ 3, – 2).

Question 16.
Find the equations of the tangents to the hyperbola 4x2 – 9y2 = 144, which are perpendicular to the line 6x + 5y = 21. Solution:
Given eqn. of hyperbola be
4x2 – 9y2 = 144 ⇒ \(\frac{x^2}{36}\) – \(\frac{y^2}{16}\) = 1 …(1)
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
we have, a2 = 36 and b2 = 16
slope of given line 6x + 5y – 21 = 0 be \(\frac{-6}{5}\)
∴ slope of line ⊥ to given line = \(\frac{-1}{-6/5}\) = \(\frac{5}{6}\) = m
Thus, eqns. of tangents to hyperbola (1) be given by
y = mx ± \(\sqrt{a^2 m^2-b^2}\)
⇒ y = \(\frac{5}{6}\) x ± \(\sqrt{36 \times \frac{25}{36}-16}\)
⇒ y = \(\frac{5}{6}\) x ± 3
⇒ 6y = 5x ± 18
⇒ 5x – 6y ± 18 = 0