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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 = \(\sqrt{(3+4)^2+(3+2)^2+(5-0)^2}\) = \(\sqrt{49+25+25}\) = \(\sqrt{99}\) = \(3 \sqrt{11}\)

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.

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.

Practicing S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Chapter Test is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Chapter Test

Question 1.
In rolling two fair dice, what is the probability of obtaining a sum greater than 3 but not exceeding 6 ?
Solution:
In rolling two dice, total no. of outcomes = n (S) = 6² = 36
Let A : event of obtaining a sum greater than 3 but not exceeding 6 = {(1,3), (2, 2), (3, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1,5), (2, 4), (4, 2), (3,3), (5,1)}
∴ n( A) =12
Thus reqd/ probability = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{12}{36}\) = \(\frac{1}{3}\)

Question 2.
Find the probability of obtaining a sum 8 in a single throw of two dice.
Solution:
In a single throw of two dice
Then total no. of outcomes = n (S) = 6² = 36
Let A : obtaining a sum 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
∴ n (A) = 5
Thus required probability
= P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{5}{36}\)

Question 3.
A bag contains 4 red, 6 white and 5 black balls. 2 balls are drawn at random. Find the probability of getting one red and one Whitehall.
Solution:
Given no. of red balls = 4, no. of white balls = 6 and no. of black balls = 5
Total no. of balls = 4 + 6 + 5 = 15
∴ n (S) = Total no. of ways of drawing 2 balls out of 15 = ¹⁵C₂
Let A : getting one red and one white ball
∴ n (A) = Total no. of ways of drawing one red ball out of 4 and 1 white ball out of
6 = ⁴C₁ × ⁶C₁
Thus required probability
= P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{{ }^4 C_1 \times{ }^6 C_1}{{ }^{15} C_2}\) = \(\frac{\frac{4 \times 6}{15 \times 14}}{2}\) = \(\frac{24}{105}\) = \(\frac{8}{35}\)

Question 4.
Out of 26 cards numbered from 1 to 26, one card is chosen. Find the probability that it is not divisible by 4.
Solution:
Total no. of outcomes = n (S) = 26
Let A : number on card is divisible by 4 = {4, 8, 12, 16, 20, 24}
∴ n (A) = 6
Thus, P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{6}{26}\) = \(\frac{3}{13}\)
∴ required probability that chosen card bear a number which is not divisible by 4
= \(P(\bar{A})\) = 1 – P (A) = 1 – \(\frac{3}{13}\) = \(\frac{10}{13}\)

Question 5.
If A and B are mutually exclusive events with \(\frac{1}{2}\) = P (B) and A ∪ B = S, the sample space. Find P (A).
Solution:
Given A and B are mutually exclusive events
∴ A ∩ B = Φ ⇒ P (A ∩ B) = 0
Given \(\frac{1}{2}\) = P (B)
Also S = A ∪ B ⇒ P(S) = P (A ∪ B) ⇒ P (A ∪ B) = 1 = P (A) + P (B) – P (A ∩ B)
⇒ 1 = P (A) + \(\frac{1}{2}\) – 0 ⇒ P (A) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)

Question 6.
A and B are two events, such that
P (A) = 0.42, P (B) = 0.48, P (A and B) = 0.16 Determine (i) P (not A), (ii) P (not B), (iii) P (A or B)
Solution:
Given P (A) = 0.42 ; P (B) = 0.48 ; P (A and B) = P (A ∩ B) = 0.16
(i) P(not A) = \(P(\bar{A})\) = 1 – P (A) = 1 – 0.42 = 0.58
(ii) P (not B) = \(P(\bar{B})\) = l – P (B) = 1 – 0.48 = 0.52
(iii) P(A or B) = P (A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.42 + 0.48 – 0.16 = 0.9 – 0.16 = 0.74

Students can cross-reference their work with OP Malhotra Maths Class 11 Solutions Chapter 10 Quadratic Equations Chapter Test to ensure accuracy.

S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Chapter Test

Solve the following equations :

Question 1.
5^x+1 + 5^2-x = 5³ + 1
Solution:
Given equation be
5^x+1 + 5^2-x = 5³ + 1 …(1)
putting 5^x = t in equation (1); we have
t x 5 + \(\frac { 25 }{ t }\) = 126
⇒ 5t² – 126t + 25 = 0
⇒ 5t² – 125t – t + 25 = 0
⇒ 5t (t – 25) – 1 (t – 25) = 0
⇒ (5t – 1) (t – 25) = 0
either 5t – 1 = 0 or t – 25 = 0
⇒ t = \(\frac { 1 }{ 5 }\) or t = 25
⇒ 5^x = 5^-1 or 5^x = 5²
⇒ x = – 1 or x = 2
Thus, x = – 1, 2

Question 2.
\(\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}\)
Solution:
Given equation be
\(\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}\) … (1)
Putting \(\sqrt{\frac{x}{1-x}}\) = t in eqn. (1); we have
t + \(\frac { 1 }{ t }\) = \(\frac { 13 }{ 6 }\)
⇒ 6t² – 13t + 6 = 0
⇒ 6t² – 9t- 4t + 6 = 0
⇒ 3t (2t – 3) – 2 (2t – 3) = 0
⇒ (3t – 3) (2t – 3) = 0
either 3t – 2 = 0 ⇒ 2t – 3 = 0
⇒ t = \(\frac { 2 }{ 3 }\) or t = \(\frac { 3 }{ 2 }\)

Case-I: When t = \(\frac { 2 }{ 3 }\) ⇒ \(\sqrt{\frac{x}{1-x}}=\frac{2}{3}\)
On squaring both sides, we have
\(\frac{x}{1-x}=\frac{4}{9}\) ⇒ 9x = 4 – 4x
⇒ 13x = 4
⇒ x = \(\frac { 4 }{ 13 }\)

Case-II: When t = \(\frac { 3 }{ 2 }\) ⇒ \(\sqrt{\frac{x}{1-x}}=\frac{3}{2}\)
On squaring both sides ; we have
\(\frac { 4 }{ 13 }\) ⇒ 4x = 9 – 9x
⇒ 13x = 9
⇒ x = \(\frac { 9 }{ 13 }\)
Thus, x = \(\frac { 4 }{ 13 }\) or \(\frac { 9 }{ 13 }\)

Question 3.
(x + 1) (x + 2) (x + 3) (x + 4) = 120
Solution:
Given equation be,
(x + 1) (x + 2) (x + 3) (x + 4) = 120
{(x + 1) (x + 4)} {(x + 2) (x + 3)} = 120
⇒ (x² + 5x + 4) (x² + 5x + 6) = 120 … (1)
putting x² + 5x = t in eqn. (1) ; we have
(t + 4)(t + 6) – 120 = 0
⇒ t² + 10t – 96 = 0
⇒ t² + 16t – 6t – 96 = 0
⇒ t(t + 16) – 6(t + 16) = 0
⇒ (t – 6)(t + 16) = 0
either t – 6 = 0 or t + 16 = 0
⇒ t = 6 or t = – 16

Case-I: When t = 6
⇒ x² + 5x = 6
⇒ x² + 5x – 6 = 0
⇒ x² + 6x – x – 6 = 0
⇒ x (x + 6) – 1 (x + 6) = 0
⇒ (x – 1)(x + 6) = 0
⇒ x = 1, – 6

Case-II: When t = – 16
⇒ x² + 5x + 16 = 0
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
⇒ x = \(\frac{-5 \pm \sqrt{25-64}}{2}=\frac{-5 \pm \sqrt{39} i}{2}\)
Hence, x = 1, – 6, \(\frac{-5 \pm \sqrt{39} i}{2}\)

Question 4.
Prove that both the roots of the equation x² – x – 3 = 0 are irrational.
Solution:
Given quadratic eqn. be
x² – x – 3 = 0 …(1)
On comparing eqn. (1) with
ax² + bx + c = 0
we have, a = 1; b = – 1 ; c = – 3
Here discriminant D = b² – 4ac
= (- 1 )² – 4 x 1 x (- 3)
= 1 + 12 = 13 >0
∴ roots are real and distinct
Further D is not a perfect square.
Thus both roots are irrational. Since irrational roots are always occur in conjugate pairs.

Question 5.
For what values of m will the equation x² – 2mx + 1m – 12 = 0 have
(i) equal roots, (it) reciprocal roots ?
Solution:
Given quadratic eqn. be
x² – 2mx + 7m – 12 = 0 …(1)
On comparing eqn. (1) with
ax² + bx + c = 0 ; we have
a = 1 ; b = – 2m ; c = 7m – 12
(i) Since eqn. (1) have equal roots ∴ D = 0
⇒ b² – 4ac = 0
⇒ (- 2m)² – 4 x 1 (7m – 12) = 0
⇒ 7m² – 28m + 48 = 0
⇒ m² – 7m+12 = 0
⇒ (m – 3) (m – 4) = 0
⇒ m = 3, 4

(ii) Since the roots of eqn. (1) are reciprocal to each other.
∴ product of roots = 1
⇒ \(\frac { c }{ a }\) = 1
⇒ \(\frac{7 m-12}{1}\) = 1
⇒ m = \(\frac { 13 }{ 7 }\)

Question 6.
If one root of 2x² – 5x + k = 0 be double the other, find the value of k.
Solution:
Let α, 2α be the roots of eqn.

Question 7.
If α, ß be the roots of the equation x² – x – 1 = 0, determine the value of
(i) α² + ß² and (ii) α³ + ß³.
Solution:
Given a and P are the roots of the eqn. x² – x – 1 =0
∴ α + ß = – (- 1) = 1 ; αß = – 1
(i) ∴ α² + ß² = (α + ß)² – 2αß
= (1)² – 2(- 1) = 3

(ii) α³ + ß³ = (α + ß)³ – 3αß (α + ß)
= 1³ – 3 (- 1) x 1 = 4

Question 8.
If the roots of the equation ax² + bx + c = 0 be in the ratio 3 : 4, show that 12b² = 49ac.
Solution:
Since the roots of the equation ax² + bx + c = 0 be in the ratio 3 : 4.
Let the roots of given equation are 3α, 4α.
∴ 3α + 4α = – \(\frac { b }{ a }\)
⇒ 7α = – \(\frac { b }{ a }\)
⇒ α = – \(\frac { b }{ 7a }\)
Also (3α) (4α) = \(\frac { c }{ a }\)
⇒ 12α² = \(\frac { c }{ a }\)
⇒ 12 \(\left(-\frac{b}{7 a}\right)^2=\frac{c}{a}\)
⇒ \(\frac{12 b^2}{49 a^2}=\frac{c}{a}\)
⇒ 12b² = 49ac

Question 9.
If x is real, prove that the quadratic expression
(i) (x – 2) (x + 3) + 7 is always positive.
(ii) 4x – 3x² – 2 is always negative.
Solution:
(i) The given expression can be written as (x – 2) (x + 3) + 7 = x² + x + 1
On comparing with ax² + bx + c, we have a = 1 ; b = 1 ; c = 1
Here discriminant D = b² – 4ac
= 1² – 4 x 1 x 1 = – 3 < 0 Here a = 1 > 0
∴ x² + x + 1 > 0 for all x.

(ii) Comparing – 3x² + 4x – 2 with ax² + bx + c
We have, a = – 2, b = 4 ; c = – 2
Here D = b² – 4ac = 4² – (- 3) x (- 1)
= 16 – 24 = – 8 < 0
and Here a = – 3 < 0.
Thus – 3x² + 4x – 2 < 0 ∀ x.

Question 10.
Draw the graph of the quadratic function x² – 4x + 3 and hence find the roots of the equation x² – 4x + 4 = 0. What is the minimum value of the function ?
Solution:
Let y = x² – 4x + 3
The table of values is given as under :

x	0	1	2	3	4	5
y	3	0	-1	0	3	8

it is clear from graph that, it intersects x- axis at two points (+ 1,0) and (3, 0)
Thus x = 1, 3 are the real solutions of given graph.
Further from graph, Minimum value of y = – 1.

Question 11.
For what real values of a, will be expression x² – ax + 1 – 2a², for the real x, be always positive ?
Solution:
On comparing x² – ax + 1 – 2a² with Ax² + Bx + C,
we have A = 1; B = – a; C = 1 – 2a²
Here discriminant D = B² – 4AC = (- a)² – 4(1 – 2a²)
= a² – 4 + 8 a²
= 9a² – 4
Also A = 1 > 0
∴ given expression be always be positive
if D < 0 if 9a² – 4 < 0 if a² < \(\frac { 4 }{ 9 }\)
if |a| < \(\frac { 2 }{ 3 }\) i.e. if – \(\frac { 2 }{ 3 }\) < a < \(\frac { 2 }{ 3 }\)

Question 12.
If x be real, prove that the value of \(\frac{2 x^2-2 x+4}{x^2-4 x+3}\) cannot lie between – 7 and 1.
Solution:
Let y = \(\frac{2 x^2-2 x+4}{x^2-4 x+3}\)
⇒ y (x² – 4x + 3) = 2x² – 2x + 4
⇒ x² (y – 2) + x (2 – 4y) + 3y – 4 = 0
Since x is real ∴ D > 0
⇒ b² – 4ac ≥ 0
⇒ (2 – 4y)² – 4 (y – 2) (3y – 4) ≥ 0
⇒ 4 [(1 – 2y)² – (y – 2) (3y – 4)] ≥ 0
⇒ [4y² – 4y + 1 – (3t – 10y + 8)] ≥ 0
⇒ y² + 6y – 7 ≥ 0
⇒ (y – 1)(y + 7) ≥ 0
The critical points are given by 1 and – 7 Then by method of intervals, we have y ≤ – 7 or y ≥ 1

Hence y can’t lies between – 7 and 1.
i.e. value of \(\frac{2 x^2-2 x+4}{x^2-4 x+3}\) cannot lie between – 7 and 1.

Question 13.
If the roots of the equation qx² + 2px + 2q = 0 are real and unequal, prove that the roots of the equation (p + q) x² + 2qx +(p – q) = 0 are imaginary.
Solution:
Given quadratic equation be,
qx² + 2px + 2q = 0 …(1)
On comparing eqn. (1) with
ax² + bx + c = 0, we have
a = q; b = 2p ; c = 2q
Since the roots of eqn. (1) are real and unequal.
∴ D > 0
⇒ b² – 4ac > 0
⇒ (2p)² – 4 x q x 2q > 0
⇒ 4p² – 8q² > 0
⇒ p² – 2q² > 0 …(2)
Also given quadratic eqn. be
(p + q)x² + 2qx + (p – q) = 0 …(3)
Here discriminant
D = (2q)² – 4(p + q)(p – q)
= 4 q² – 4(p² – q²)
= 4 [q² – p² + q²]
= 4(2q² – p²)
= – 4 (p² – 2q²) < 0 [using (2)]
Hence the roots of eqn. (3) are imaginary.

Question 14.
If α, ß be the roots of x² – px + q = 0, find the value of α⁵ß⁷ + α⁷ß⁵ in terms of p and q.
Solution:
Since α and ß are the roots of
x² – px + q = 0
∴ α + ß = p; αß = q
Thus, α⁵ß⁷ + α⁷ß⁵ = α⁵ß⁵ (ß² + α²)
= (αß)⁵ [(α + ß)² – 2αß]
= q⁵[P² – 2q]

Question 15.
If the difference between the roots of the equation x² + ax + 1 = 0 is less than \(\sqrt{5}\), then the set of possible values of a is (a) (3, ∞)
(b) (- ∞, – 3)
(c)(- 3, 3)
(d) (- 3, ∞)
Solution:
Let a and P are the roots of eqn. x² + ax + 1 = 0
∴ α + ß = – a ; αß = 1 …(1)
Also, it is given that | α – ß | < \(\sqrt{5}\)
⇒ \(\sqrt{(\alpha+\beta)^2-4 \alpha \beta}<\sqrt{5}\)
⇒ \(\sqrt{(-a)^2-4}<\sqrt{5}\)
⇒ \(\sqrt{a^2-4}<\sqrt{5}\)
On squaring both sides ; we have
a² – 4 < 5
⇒ a² < 9
⇒ | a | < 3
⇒ – 3 < a < 3
⇒ a ∈ (- 3, 3)

Question 16.
Let α, ß be the roots of the equation x² – px + r = 0 and α/2, 2ß be the roots of the equation x² – qx + r = 0, then the value of r is
(a) \(\frac { 2 }{ 9 }\) (p – q)(2q – p)
(b) \(\frac { 2 }{ 9 }\) (q – p)(2p – q)
(c) \(\frac { 2 }{ 9 }\) (q – 2p)(2q – p)
(d) \(\frac { 2 }{ 9 }\) (2p – q) (2q – p)
Solution:
Given a and p are the roots of equation
x² – px + r = 0
∴ α + ß = p ; αß = r …(1)
Also, given \(\frac { α }{ 2 }\), 2ß are the roots of eqn.

Question 17.
α, ß are the roots of ax² + 2bx + c = 0 and α + δ, ß + δ are the roots of Ax² + 2Bx + C = 0, then what is (b² – ac)/(B² – AC) equal to ?
(a) (8/B)²
(b) (a/A)²
(c) (a² b²)/(A²B²)
(d) ab/AB
Solution:
Given α, ß are the roots of ax² + 2bx + c = 0

Question 18.
If α, ß are the roots of the equation x² – 2x – 1 = 0, then what is the value of α²ß^-2 + α^-2ß² ?
(a) – 2
(8) 0
(c) 30
(d) 34
Solution:
Given α and ß are the roots of eqn. x² – 2x – 1 =0

Question 19.
If the roots of the quadratic equation x² + px + q = 0 are tan 30° and tan 15°, then value of 2 + q – p is
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
Given tan 30° and tan 15° are the roots of quadratic equation x² + px + q = 0
∴ tan 30° +tan 15° = – p …(1)
and tan 30° tan 15° = q …(2)
Now tan 45° = tan (30° + 15°)
tan 45° = \(\frac{\tan 30^{\circ}+\tan 15^{\circ}}{1-\tan 30^{\circ} \tan 15^{\circ}}\)
⇒ 1 = \(\frac{-p}{1-q}\) [using (1) and (2)]
⇒ 1 – q = – p
⇒ p – q + 1 = 0
⇒ q – p – 1 = 0
⇒ q – p + 2 = 3

Question 20.
If both the roots of the quadratic equation x² – 2kx + k² + k – 5 = 0 are less than 5, then k lies in the interval
(a) (5, 6]
(b) (6, ∞)
(c) (- ∞, 4)
(d) [4, 5]
Solution:
Given quadratic eqn. be
x² – 2kx + k² + k – 5 = 0 …(1)
since each root of eqn. (1) be less than 5
∴ sum of roots of eqn. (1) be less than 10
⇒ \(\frac { – b }{ a }\) < 10 ⇒ 2k < 10 ⇒ k < 5 …(2)
On comparing eqn. (1) with
ax² + bx + c = 0, we have
a = 1 ; b = – 2k and c = k² + k – 5
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{2 k \pm \sqrt{4 k^2-4\left(k^2+k-5\right)}}{2}\)
⇒ x = k ± \(\sqrt{k^2-k^2+5-k}=k \pm \sqrt{5-k}\)
Further, each root be less than 5.
⇒ k + \(\sqrt{5-k}\) < 5
⇒ \(\sqrt{5-k}\) < 5 – k
On squaring both sides ; we have
5 – k < (5 – k)² ⇒ k² – 10k + 25 + k – 5 > 0
⇒ k² – 9k + 20 > 0
⇒ (k – 4) (k – 5) > 0
Critical points are given by k = 4, 5
Then by method of intervals, we have
k < + 4 or k > 5
But k < 5 ∴ k < 4

Question 21.
If α and ß are the roots of ax² + bx + c = 0 and if px² + qx + r = 0 has roots \(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\) then r =
(a) a + 2b
(b) a + b + c
(c) ab + bc + ca
(d) abc
Solution:
Given α and ß are the roots of ax² + bx + c = 0

Hence the quadratic equation having roots
\(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\) be given by x² – Sx + P = 0
⇒ x² + (\(\left(\frac{b+2 c}{c}\right)\))x + \(\frac{a+b+c}{c}\) = 0
⇒ cx² + (b + 2c)x + a + b + c = 0 …(4)
On comparing eqn. (3) and eqn. (4) ; we have
p – c ; q = b +2c and r = a + b + c

Question 22.
The quadratic equations x² – 6x + a = 0 and x² – cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is
(a) 1
(b) 4
(c) 3
(d) 2
Solution:
Let α and 4ß are the roots of
x² – 6x + a = 0 …(*)
∴ α + 4ß = 6 …(1)
4αß = a …(2)
Also, let α and 3ß are the roots of
x² – cx + 6 = 0 …(**)
∴ α + 3ß = c …(3)
3αß = 6 …(4)
where α be the common root of both given eqns.
From (2) and (4); we have
\(\frac{4 \alpha \beta}{3 \alpha \beta}=\frac{a}{6}\) ⇒ a = \(\frac { 24 }{ 3 }\) = 8
∴ eqn. (*) becomes : x² – 6x + 8 = 0
⇒ (x – 2) (x – 4) = 0 ⇒ x = 2, 4
Taking α = 2, αß = 4 ⇒ ß = 1
∴ c = 2 + 3 = 5
∴ eqn. (**) becomes ;
x² – 5x + 6 = 0 ⇒ x = 2, 3
Clearly 2 be the common root.
When α = 4 ; 4ß = 2 ⇒ ß = \(\frac { 1 }{ 2 }\)
∴ c = 4 + \(\frac { 3 }{ 2 }\) = \(\frac { 11 }{ 2 }\)
∴ eqn. (**) becomes ;
x² – \(\frac { 11 }{ 2 }\) x + 6 = 0
⇒ 2x² – 11x + 12 = 0
⇒ x = \(\frac{11 \pm 5}{4}\) = 4, \(\frac { 3 }{ 2 }\)
But it is given that other roots of both equations are integers.

Question 23.
If α, ß are the roots of the equation λ(x² – x) + x + 5 = 0 and if λ₁ and λ₂ are two values of λ obtained from \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{4}{5}\), then \(\frac{\lambda_1}{\lambda_2^2}+\frac{\lambda_2}{\lambda_1^2}\) equals
(a) 4192
(b) 4144
(c) 4096
(d) 4048
Solution:
Given α and ß are the roots of eqn.

Question 24.
If α, ß be the roots of x² – a (x – 1) + b = 0, then the value of
\(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-\alpha \beta}+\frac{2}{a+b}\) is
(a) \(\frac{4}{a+b}\)
(b) \(\frac{1}{a+b}\)
(c) 0
(d) – 1
Solution:
Given α and ß are the roots of quadratic eqn.
x² – a (x – 1) + b = 0
⇒ x² – ax + a + b = 0 …(1)
∴ α + ß = a ; αß = a + b
Also, a and P both satisfies eqn. (1)
∴ α² – aα + a + b = 0
⇒ α² – aα = – (a + b)
and ß² – αß + a + b = 0
⇒ ß² – aß = – (a + b)
∴ \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}\)
= \(\frac{1}{-(a+b)}+\frac{1}{-(a+b)}+\frac{2}{a+b}\)
= \(\frac{-2}{a+b}+\frac{2}{a+b}\) = 0

Practicing ISC Mathematics Class 12 Solutions is the ultimate need for students who intend to score good marks in examinations.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.9

Evaluate the following (1 to 4) integrals:

Question 1.
(i) ∫ \(\frac{d x}{2+\cos x}\)
(ii) ∫ \(\frac{d x}{5+4 \cos x}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{2+\cos x}\)
put tan \(\frac{x}{2}\) = t
⇒ sec² \(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
and cos x = \(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\)
= \(\frac{1-t^2}{1+t^2}\)

(ii) Let I = ∫ \(\frac{d x}{5+4 \cos x}\)

Question 2.
(i) ∫ \(\frac{d x}{1+2 \cos x}\)
(ii) ∫ \(\frac{d x}{4 \cos x-1}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{1+2 \cos x}\)
put tan \(\frac{x}{2}\) = t
⇒ sec² \(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
and cos x = \(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\)
= \(\frac{1-t^2}{1+t^2}\)

(ii) Let I = ∫ \(\frac{d x}{4 \cos x-1}\)
put tan \(\frac{x}{2}\) = t
⇒ sec² \(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt

Question 3.
(i) ∫ \(\frac{d x}{4+5 \sin x}\)
(ii) ∫ \(\frac{d x}{1-\sin x+\cos x}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{4+5 \sin x}\)
put tan \(\frac{x}{2}\) = t
⇒ sec² \(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt

(ii) Let I = ∫ \(\frac{d x}{1-\sin x+\cos x}\)
put tan \(\frac{x}{2}\) = t
⇒ sec² \(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt

= – log |1 – tan x/2| + c

Question 4.
(i) ∫ \(\frac{d x}{1-\tan x}\) (NCERT)
(ii) ∫ \(\frac{d x}{1+\cot x}\) (NCERT)
Solution:

(i) Let I = ∫ \(\frac{d x}{1-\tan x}\)
= ∫ \(\frac{\cos x d x}{\cos x-\sin x}\)
Let cos x = l (cos x – sin x) + m \(\frac{d}{d x}\) (cos x – sin x)
cos x = l (cos x – sin x) + m (- sin x – cos x)
Comparing the coefficients of sin x and cos x on both sides,
we have 0 = – l – m
and 1 = l – m
⇒ m = \(-\frac{1}{2}\) ; l = \(\frac{1}{2}\)
∴ I = ∫ \(\frac{l(\cos x-\sin x)+m(-\sin x-\cos x)}{\cos x-\sin x}\)
= ∫ l dx + m \(\frac{-\sin x-\cos x}{\cos x-\sin x}\) dx
= lx + m ∫ \(\frac{d t}{t}\)
where t = cos x – sinx
⇒ dt = (- sin x – cos x) dx
= \(\frac{x}{2}\) – \(\frac{1}{2}\) log |cos x – sin x| + c

(ii) Let I = ∫ \(\frac{d x}{1+\cot x}\)
= ∫ \(\frac{\sin x d x}{\sin x+\cos x}\)
Let Numerator = l (denominator + m \(\frac{d}{d x}\) (derivative of denominator)
⇒ 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)
Equating the coefficients of sin x and cos x on both sides,
we have 1 = l – m
and 0 = l + m
On solving the eqns ; we have
⇒ l = \(\frac{1}{2}\) ; m = \(-\frac{1}{2}\)
∴ I = ∫ \(\frac{l(\sin x+\cos x)+m(\cos x-\sin x)}{\sin x+\cos x}\)
= ∫ l dx + m ∫ \(\frac{(\cos x-\sin x) d x}{\sin x+\cos x}\) dx
= \(\frac{1}{2}\) x – \(\frac{1}{2}\) log |sin x + cos x| + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]

Question 5.
If ∫ \(\frac{5 \tan x}{\tan x-2}\) dx = ax + b log |sin x – 2 cos x| + C, then find the values of a and b.
Solution:
Let I = ∫ \(\frac{5 \tan x}{\tan x-2}\) dx
= ∫ \(\frac{5 \frac{\sin x}{\cos x} d x}{\frac{\sin x}{\cos x}-2}\)
= ∫ \(\frac{5 \sin x d x}{\sin x-2 \cos x}\)
Let 5 sin x = l (sin x – 2 cos x) + m \(\frac{d}{d x}\) (sin x – 2 cos x)
⇒ 5 sin x = l (sin x – 2 cos x) + m (cos x + 2 sin x) ………… (1)
comparing the coefficients of sin x and cos x on both sides ; we have
5 = l + 2m …………(2)
0 = – 2l + m ……………(3)
On solving eqn. (2) and (3) ; we have
m = 2 and l = 1
∴ I = ∫ \(\frac{5 \sin x d x}{\sin x-2 \cos x}\)
= ∫ \(\frac{l(\sin x-2 \cos x)+m(\cos x+2 \sin x) d x}{\sin x-2 \cos x}\)
= l ∫ dx + m ∫ \(\frac{(\cos x+2 \sin x) d x}{\sin x-2 \cos x}\) dx
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]
= lx + m log |sin x – 2 cos x| + C
= x + 2 log |sin x – 2 cos x| + C ……………. (1)
Also, Given I = ∫ \(\frac{5 \tan x d x}{\tan x-2}\)
= ax + b log |sin x – 2 cos x| + C …………….(2)
∴ From (1) and (2) ; we have
a = 1 ; b = 2.

Students can cross-reference their work with ML Aggarwal Class 12 ISC Solutions Chapter 7 Applications of Derivatives Ex 7.9 to ensure accuracy.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.9

Typical Problems:

Question 1.
The radius of the base of right circular cone is increasing at the rate of 3 cm per minute and the altitude is decreasing at the rate of 4 cm per minute. Find the rate of change of total surface area of the cone when its radius is 7cm and altitude is 24 cm.
Solution:

Let r be the radius and h be the height of right circular cone.
Let S = Total surface area of right circular cone = πr² + πrl
= πr² + πr \(\sqrt{r^2+h^2}\) …..(1)
[where l = slant height of cone = \(\sqrt{r^2+h^2}\)]
Diff. (1) w.r.t. t ; we get
\(\frac{d \mathrm{~S}}{d t}=2 \pi r \frac{d r}{d t}+\pi\left[\sqrt{r^2+h^2} \frac{d r}{d t}+\frac{r}{2 \sqrt{r^2+h^2}}\left[2 r \frac{d r}{d t}+2 h \frac{d h}{d t}\right]\right]\)
= \(2 \pi r \frac{d r}{d t}+\pi\left[\sqrt{r^2+h^2} \frac{d r}{d t}+\frac{r}{\sqrt{r^2+h^2}}\left\{r \frac{d r}{d t}+h \frac{d h}{d t}\right\}\right]\)
Given \(\frac{d r}{d t}\) = 3 cm / min.
\(\frac{d h}{d t}\) = – 4 cm/min ;
r = 7 cm and h = 24 cm
∴ from (1) ; we have
\(\frac{d S}{d t}\) = 2π × 7 × 3 + π \(\left[\sqrt{7^2+24^2} \times 3+\frac{7}{\sqrt{7^2+24^2}}\{7 \times 3+24(-4)\}\right]\)
= 42π + π [25 × 3 + \(\frac{7}{25}\) (21 – 96)]
= 42π + π [75 – 21]
= 96π cm² /min.

Question 2.
Find the equation of normal to the curve y = x + \(\frac{2}{x}\) at the point where abscissa is 2. If this normal meets the axes in points A and B find the length of AB.
Solution:
Given eqn. of curve y = x + \(\frac{2}{x}\) …………..(1)
given point on the curve (1) whose abscissa is 2.
Thus, x = 2
∴ from (1) ;
y = 2 + \(\frac{2}{2}\) = 3
∴ point on given curve becomes (2, 3).
Differentiating eqn. (1) w.r.t. x; we have
\(\frac{d y}{d x}\) = 1 – \(\frac{2}{x^2}\)
∴ Slope of normal to given curve at point (2, 3) = \(-\frac{1}{\left(\frac{d y}{d x}\right)_{(2,3)}}\)
= \( \frac{-1}{\left(1-\frac{2}{4}\right)}=\frac{-1}{\frac{1}{2}}\) = – 2
;. required eqn. of normal to given curve at point (2, 3) is given by
y – 3 = – 2(x – 2)
⇒ 2x + y = 7 ………..(2)
Now let eqn. (2) meets x-axis at A ¡e. y = 0
∴ from (2);
x = \(\frac{7}{2}\)
∴ coordinates of A are (\(\frac{7}{2}\), 0)
Let eqn. (2) meets y-axis at B i.e. x = O
∴ from (2) ;
y = 7
Thus coordinates of point B are (0, 7)
∴ |AB| = length of AB
= \(\sqrt{\left(0-\frac{7}{2}\right)^2+(7-0)^2}\)
= \(\sqrt{\frac{49}{4}+49}\)
= \(\frac{7}{2}\) √5 units.

Question 3.
Show that for all a ≥ 1, the function f defined by f(x) = √3 sin x – cos x – 2ax + b is decreasing on R.
Solution:
Given f(x) = √3 sin x – cos x – 2ax + b
putting √3 = r cos α ; – 1 = r sin α
On squaring and adding; we have
r = 2 and on dividing; we have
tan α = – \(\frac{1}{\sqrt{3}}\)
= tan (- \(\frac{\pi}{6}\))
⇒ α = – \(\frac{\pi}{6}\)
∴f(x) = 2 sin (x – \(\frac{\pi}{6}\)) – 2ax + b
Diff. both sides w.r.t. x, we have
f’ (x) = 2 cos (x – \(\frac{\pi}{6}\)) – 2ax + b
Diff. both sides w.r.t. x, we have
f'(x) = 2 cos (x – \(\frac{\pi}{6}\)) – 2a
Now – 1 ≤ cos (x – \(\frac{\pi}{6}\)) ≤ 1
⇒ – 2 ≤ 2 cos (x – \(\frac{\pi}{6}\)) ≤ 2
⇒ – 2 – 2a ≤ 2 cos (x – \(\frac{\pi}{6}\)) – 2a ≤ 2 – 2a ≤ 0
[if a ≥ 1
⇒ 2a ≥ 2
⇒ 2 – 2a ≤ 0]
⇒ f'(x) ≤ 0
Thus f(x) be a decreasing function if a ≥ 1.

Question 4.
Find the greatest and the least value of f(x) = x² log x in [1, e].
Solution:
Given f(x) = x² log x in [ 1, e]
On differentiating both sides, w.r.t. x, we have
f’ (x) = x² × \(\frac{1}{x}\) + (log x) 2x
= x + 2x log x
= x [ 1 + 2 log x]
For critical points ;
f’ (x) = 0
⇒ x (1 + 2 log x) = 0
⇒ x = 0, e^{– 1/2}
but x ∈ e [1, e]
Since 0, e^{– 1/2} ∉ [1, e]
Thus only critical points are 1 and e.
f(1) = 1 log 1 = 0
f(e) = e² log e = e²
∴ greatest value of f(x) = e²
and it attains at x = e
and least value of f(x) = 0
and it attains at x = 0.

Question 5.
A lighthouse is located at A, 2 km off share from the nearest point O on a straight beach and a shop is located B on the beach at a distance of 4 km from O. If the housekeeper can row at the speed of 4 km/h and can walk at the rate of 6 km/h. Where should be plane to reach the shore so as to cover the distance to the shop in the least possible time?
Solution:
Let C be the point between O and B and the man should reach the shore at the point C.
Let OC = x km.
The distance rowed = AC = \(\sqrt{2^2+x^2}\) km.
Let t₁ = rowing time
= distance/speed
⇒ t₁ = \(\frac{\sqrt{2^2+x^2}}{4}\) hr

The distance walked = CB = (4 – x) km
∴ t₂ = walking time = \(\frac{4-x}{6}\) hr
Thus total time required T = t₁ + t₂
∴ T = \(\frac{\sqrt{4+x^2}}{4}+\frac{4-x}{6}\)
Diff. both sides w.r.t. x, we have

On squaring; we have
36x² = 16 (4 + x)
⇒ 9x² = 4 (4 + x²)
⇒ 5x² = 16
⇒ x = ± \(\frac{4}{\sqrt{5}}\) [but x > 0]
∴ x = \(\frac{4}{\sqrt{5}}\)
at x = \(\frac{4}{\sqrt{5}}\) ;
\(\frac{d^2 \mathrm{~T}}{d x^2}=\frac{1}{\left(4+\frac{16}{5}\right)^{3 / 2}}\)
= \(\frac{5 \sqrt{5}}{216}\) > 0
Thus T is minimise for x = \(\frac{4}{\sqrt{5}}\) km
Hence the man should reach the shore at the point whose distance from the point O = \(\frac{4}{\sqrt{5}}\) km
= \(\sqrt{\frac{32}{10}}\) km
= \(\sqrt{3.2}\) km.

Question 6.
A cone is circumscribed about a sphere of radius R. Show that the volume of the cone is minimum if its height is 4R.
Solution:
Let x cm be the radius of the circumscribed right circular cone
and Let h cm be the height of cone.
Then AM = \(\sqrt{\mathrm{OA}^2-\mathrm{OM}^2}\)
= \(\sqrt{(h-\mathrm{R})^2-\mathrm{R}^2}\)
= \(\sqrt{h^2-2 h \mathrm{R}}\)

Let ∠DAC = α,
In right angled ∆ADC, we have
tan α = \(\frac{\mathrm{DC}}{\mathrm{AD}}=\frac{x}{h}\) ………..(1)
Also in right angled ∆OMA, we have
tan α = \(\frac{\mathrm{OM}}{\mathrm{AM}}=\frac{\mathrm{R}}{\sqrt{h^2-2 h \mathrm{R}}}\)
From eqn. (1) and (2) ; we have
\(\frac{x}{h}=\frac{\mathrm{R}}{\sqrt{h^2-2 h \mathrm{R}}}\)
⇒ x = \(\frac{\mathrm{R} h}{\sqrt{h^2-2 h \mathrm{R}}}\)
Let V be the volume of cone which is to be minimised
Then V = latex]\frac{\pi}{3}[/latex] r²h
= \(\frac{\pi}{3} \frac{\mathrm{R}^2 h^2}{\left(h^2-2 h \mathrm{R}\right)}\) × h
= \(\frac{\pi}{3} \frac{\mathrm{R}^2 h^2}{h-2 \mathrm{R}}\)
Diff. both sides w.r.t. h, we have
\(\frac{d \mathrm{~V}}{d h}=\frac{\pi \mathrm{R}^2}{3}\left[\frac{(h-2 \mathrm{R}) 2 h-h^2}{(h-2 \mathrm{R})^2}\right]\)
= \(\frac{\pi \mathrm{R}^2}{3} \frac{\left(h^2-4 \mathrm{R} h\right)}{(h-2 \mathrm{R})^2}\)
⇒ \(\frac{d \mathrm{~V}}{d h}=\frac{\pi \mathrm{R}^2 h(h-4 \mathrm{R})}{3(h-2 \mathrm{R})^2}\)
For maxima/minima, \(\frac{d V}{d h}\) = 0
⇒ h (4 – 4R) = 0
⇒ h = 4R (∵ h > 0)
When h slightly < 4R
⇒ h – 4R < 0
∴ \(\frac{d V}{dh}\) < 0 When h > 4R slightly
⇒ h – 4R > 0
∴ \(\frac{d V}{dh}\) > 0
Thus \(\frac{d V}{dh}\) changes its sign from -ve to positive (+ve) as we move h <4R slightly to h > 4R slightly.
∴ h = 4R be a point of minima.
Hence volume of cone is minimum when height of cone is 4R.

Interactive ISC Mathematics Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.9 engage students in active learning and exploration.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.9

Differentiate the following functions (1 to 7) w.r.t. x :

Question 1.
(i) (x + 3)² (x + 4)³ (x + 5)⁴ (NCERT)
(ii) cos x . cos 2x . cos 3x (NCERT)
Solution:
(i) Let y = (x + 3)² (x + 4)³ (x + 5)⁴
Taking logarithm on both sides of eqn. (1) ; we get
log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)
[∵ log a^b = b log a
and log ab = log a + log b]
Diff. bothsides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\)
∴ \(\frac{d y}{d x}\) = (x + 3)² (x + 4)³ (x + 5)⁴ \(\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]\) [using (1)]

(ii) Let y = cos x . cos 2x . cos 3x …………….(1)
Taking logarithm on both sides of eqn. (1) ; we get
log y = log cos x + log cos 2x + log cos 3x ;
Diff. bothsides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=\frac{1}{\cos x}(-\sin x)+\frac{1}{\cos 2 x}(-2 \sin 2 x)+\frac{1}{\cos 3 x}(-3 \sin 3 x)\)
∴ \(\frac{d y}{d x}\) = cos x cos 2x cos 3x . [- tan x – 2 tan 2x – 3 tan 3x] [using (1)]

Question 2.
(i) e^x cos³ x sin² x
(ii) \(\frac{e^{x^2} \tan ^{-1} x}{\sqrt{1+x^2}}\).
Solution:
(i) Let y = e^x cos³ x sin² x
Taking logarithm on both sides, we have ;
log y = x log e + 3 log cos x + 2 log sin x
Diff. bothsides w.r.t. x, we get
∴ \(\frac{1}{y} \frac{d y}{d x}=1+\frac{3}{\cos x}(-\sin x)+\frac{2}{\sin x}(\cos x)\)
Thus \(\frac{d y}{d x}\) = e^x cos³ x sin² x [1 – 3 tan x + 2 cot x] [using (1)]

(ii) Let y = \(\frac{e^{x^2} \tan ^{-1} x}{\sqrt{1+x^2}}\)
Taking logarithm on both sides, we get ;
log y = log e^x2 + log tan^-1 x – \(\frac{1}{2}\) log (1 + x²)
⇒ log y = x² + log tan^{-1</sup x – \(\frac{1}{2}\) log (1 + x2)
Diff. bothsides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=2 x+\frac{1}{\tan ^{-1} x} \frac{1}{1+x^2}-\frac{1}{2} \times \frac{1}{1+x^2} \times 2 x\)
⇒ \(\frac{d y}{d x}=\frac{e^{x^2} \tan ^{-1} x}{\sqrt{1+x^2}}\left[2 x+\frac{1}{\left(1+x^2\right) \tan ^{-1} x}-\frac{x}{1+x^2}\right]\)}

Question 2 (old).
(i) \(\frac{x \sqrt{x^2+1}}{(x+1)^{2 / 3}}\), x > 0
(ii) \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\) (NCERT)
Solution:
(i) Let y = \(\frac{x \sqrt{x^2+1}}{(x+1)^{2 / 3}}\), x > 0
Taking natural logarithm on both sides ; we get
log y = log x + \(\frac{1}{2}\) log (x² + 1) – \(\frac{2}{3}\) log (x + 1)
[[∵ log a^b = b log a ;
log \(\frac{a}{b}\) = log a – log b
and log ab = log a + log b]
Diff. bothsides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=\frac{1}{x}+\frac{1}{2} \frac{1}{x^2+1} \times 2 x-\frac{2}{3(x+1)}\)
∴ \(\frac{d y}{d x}=y\left[\frac{1}{x}+\frac{x}{x^2+1}-\frac{2}{3(x+1)}\right]\) ; x ≠ 0
= \(\frac{x \sqrt{x^2+1}}{(x+1)^{2 / 3}}\left[\frac{1}{x}+\frac{x}{x^2+1}-\frac{2}{3(x+1)}\right]\) ; x ≠ 0

(ii) Let y = \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\)
Taking natural logarithm on both sides ; we get
log y = \(\frac{1}{2}\) [log (x – 1) + log (x – 2) – log (x – 3) – log (x – 4) – log (x – 5)]
[using properties of logarithm]
Diff. both sides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=\frac{1}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]\)
⇒ \(\frac{d y}{d x}=\frac{y}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]\)
= \(\frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]\).

Question 3.
(i) (sin x)^{cos x}, 0 < x < π (NCERT Exampler)
(ii) (sin x)^{sin x}, 0 < x< π (NCERT)
Solution:
(i) Let y = (sin x)^{cos x} …………..(1)
Taking natural logarithm on both sides ; we have
log y = cos x. log (sin x)
Differentiate both sides w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}\) = cos x \(\frac{1}{\sin x}\) cos x + log (sin x) (- sin x)
⇒ \(\frac{d y}{d x}=\frac{(\sin x)^{\cos x}}{\sin x}\) [cos² x – sin² x log (sin x)] [using (1)\

(ii) Let y = (sin x)^{sin x} ; 0 < x< π
Taking natural logarithm on both sides ; we have
log y = sin x log sin x ;
Diff. both sides w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}\) = sin x \(\frac{1}{\sin x}\) cos x + (log sin x) cos x
⇒ \(\frac{d y}{d x}\) = y cos x [1 + log (sin x)]
⇒ \(\frac{d y}{d x}\) = (sin x)^{sin x} cos x [1 + log sin x].

Question 4.
(i) x^{sin x}, x > 0 (NCERT)
(ii) (2x + 3)^{x – 5}, x > – \(\frac{3}{2}\)
Solution:
(i) Let y = x^{sin x}, x > 0
Taking logarithm on both sides we get ;
log y = sin x . log x ;
Differentiate w.r.t. x
∴ \(\frac{1}{y} \frac{d y}{d x}\) = sin x . \(\frac{1}{x}\) + log x cos x
⇒ \(\frac{d y}{d x}\) = x^{sin x} [\(\frac{sin x}{x}\) + log x . cos x] [using (1)]

(ii) Let y = (2x + 3)^{x – 5}, x > – \(\frac{3}{2}\)
Taking logarithm on both sides we get ;
log y = (x – 5) log (2x + 3)
Diff. both sides w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}\) = \(\frac{x-5}{2 x+3}\) . 2 + log (2x + 3) . 1
[using product rule]
⇒ \(\frac{d y}{d x}\) = (2x + 3)^{x – 5} [\(\frac{2(x-5)}{2 x+3}\) + log (2x + 3)]

Question 5.
(i) (log x)^{cos x}, x > 1 (NCERT)
(ii) (log x)^{log x}, x > 1 (NCERT)
Solution:
(i) Let y = (log x)^{cos x}
Taking logarithm on both sides ; we have
log y = cos x . log (log x)
Differentiate both sides w.r.t. x, we have
∴ \(\frac{1}{y} \frac{d y}{d x}\) = cos x . \(\frac{1}{\log x} \cdot \frac{1}{x}\) + log (log x) (- sin x)
⇒ \(\frac{d y}{d x}\) = (log x)^{cos x} [\(\frac{\cos x}{x \log x}\) – sin x . log (log x)] [using (1)]

(ii) Let y = (log x)^{log x}
Taking logarithm on both sides ; we have
log y = log x . log (log x)
Differentiate both sides w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}\) = log x . \(\frac{1}{\log x} \cdot \frac{1}{x}\) + log (log x) . \(\frac{1}{x}\)
⇒ \(\frac{1}{y} \frac{d y}{d x}\) = \(\frac{1}{x}\) [1 + log (log x)]
∴ \(\frac{d y}{d x}\) = \(\frac{(\log x)^{\log x}}{x}\) [1 + log (log x)]

Question 6.
(i) If y = x^y, prove that x \(\frac{d y}{d x}\) = \(\frac{y^2}{1-y \log x}\).
(ii) If x = e^x/y, prove that \(\frac{d y}{d x}=\frac{x-y}{x \log x}\). (NCERT Exampler)
(iii) If x^y = e^x-y, prove that \(\frac{d y}{d x}=\frac{\log x}{(\log x e)^2}\).
(iv) If x¹⁶ y⁹ = (x² + y)¹⁷, prove that \(\frac{d y}{d x}=\frac{2 y}{x}\).
Solution:
(i) Given y = x^y ………(1)
Taking logarithm on both sides ; we get ;
log y = y log x ;
Differentiate both sides w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}=\frac{y}{x}+\log x \frac{d y}{d x}\)
⇒ \(\left(\frac{1}{y}-\log x\right) \frac{d y}{d x}=\frac{y}{x}\)
⇒ \(\left(\frac{1-y \log x}{y}\right) \frac{d y}{d x}=\frac{y}{x}\)
⇒ \(\frac{d y}{d x}=\frac{y^2}{x(1-y \log x)}\)

(ii) Given x = e^x/y ;
Taking logarithm on both sides ; we get ;
log x = log e^x/y
= \(\frac{x}{y}\) . 1
⇒ y log x = x
On differentiating both sides w.r.t. x ; we get
\(\frac{d}{d x}\) (y log x) = \(\frac{d}{d x}\) (x)
⇒ y \(\frac{d}{d x}\) log x + log x \(\frac{d y}{d x}\) = 1
⇒ \(\frac{y}{x}\) + log x \(\frac{d y}{d x}\) = 1
⇒ log x \(\frac{d y}{d x}\) = 1 – \(\frac{y}{x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{x-y}{x \log x}\)

(iii) Given, x^y = e^x-y ;
Taking logarithm on both sides ; we get ;
y log x = (x – y) log e = x – y …………(1)
⇒ y (1 + log x) = x
⇒ y = \(\frac{x}{1+\log x}\) ……….(2)
Diff. eqn. (2) both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{(1+\log x) \cdot 1-x\left(0+\frac{1}{x}\right)}{(1+\log x)^2}\)
= \(\frac{\log x}{(1+\log x)^2}\)
= \(\frac{(x-y)}{y[\log e+\log x]^2}\) [using eqn. (1)]
= \(\frac{x-y}{y(\log e x)^2}\)
[∵ log a + log b = log ab]

(iv) Given, x¹⁶ y⁹ = (x² + y)¹⁷
Taking logarithm on both sides ; we get ;
log (x¹⁶ y⁹) = log (x² + 17)¹⁷
⇒ log x¹⁶ + log y⁹ = 17 log (x² + y)
⇒ 16 log x + 9 log y = 17 log (x² + y)
Diff. both sides w.r.t. x ; we get

Question 7.
Find the derivative of x^x + a^x + x^a + a^a for some fixed a > 0, x > 0. (NCERT)
Solution:
Let y = x^x + a^x + x^a + a^a, a > 0, x > 0
Diff. both sides w.r.t. x ; we get
\(\frac{d y}{d x}=\frac{d}{d x} x^x+\frac{d}{d x} a^x+\frac{d}{d x} x^a+\frac{d}{d x} a^a\)
= \(\frac{d}{d x}\) e log x^x + a^x log a + ax^a-1 + 0
[∵ a^a be a constant
∴ \(\frac{d}{d x}\) (a^a) = 0]
= e^{x log x} [x × \(\frac{1}{x}\) + log x . 1] + a^x log a + ax^a-1
= x^x [1 + log x] + a^x log a + ax^a-1

Question 7 (old).
(i) x^{sin x + cos x}
(ii) (x² sin x)^1/x
Solution:
(i) Let y = x^{sin x + cos x}
Taking logarithm on both sides ; we have
log y = (sin x + cos x) log x
Differentiate both sides w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}\) = (sin x + cos x) \(\frac{1}{x}\) + log x (cos x – sin x)
\(\frac{d y}{d x}\) = x^{sin x + cos x} [\(\frac{\sin x+\cos x}{x}\) + (cos x – sin x) log x]

(ii) Let y = (x² sin x)^1/x
Taking logarithm on both sides ; we have
log y = \(\frac{1}{x}\) log (x² sin x)
Differentiate both sides w.r.t. x, we have

Question 8.
Differentiate the following functions w.r.t. x:
(i) x^{log x} + (log x)^x
(ii) (sin x)^{cos x} + x^{sin x}
(iii) x^{cos x} + (cos x)^x
(iv) (x)^{cos x} + (sin x)^{tan x}
(v) (sin 2x)^x + sin^-1 \(\sqrt{3x}\)
(vi) e^{sin x} + (tan x)^x
(vii) x^x – 2^{sin x} (NCERT)
(viii) x^{x cos x} + \(\frac{x^2+1}{x^2-1}\)
Solution:
(i) Let y = x^{log x} + (log x)^x
= u + v
where u = x^{log x} ; v = (log x)^x
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………….(1)
Now u = x^{log x}
so that log u = (log x)²
Diff. both sides w.r.t. x ; we get
\(\frac{1}{u} \frac{d u}{d x}=\frac{2 \log x}{x}\)
⇒ \(\frac{d u}{d x}=x^{\log x}\left(\frac{2 \log x}{x}\right)\) …………(2)
Also v = (log x)^x
so that log v = x log (log x)
Differentiate both sides w.r.t. x, we have
\(\frac{1}{v} \frac{d v}{d x}=x \cdot \frac{1}{\log x} \cdot \frac{1}{x}\) + log (log x) . 1
⇒ \(\frac{d v}{d x}\) = (log x)^x [\(\frac{1}{\log x}\) + log (log x)] ………….(3)
Putting eqn. (2) and (3) in eqn. (1) ; we get
∴ \(\frac{d y}{d x}=x^{\log x}\left(\frac{2 \log x}{x}\right)+(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right]\)

(ii) Let y = (sin x)^{cos x} + (x)^{sin x} = u + v ;
where u = sin x^{cos x} ;
v = (x)^{sin x}
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
Now u = (sin x)^{cos x} so that
log u = cos x log (sin x) ;
Diff. both sides w.r.t. x ; we get
\(\frac{1}{u} \frac{d u}{d x}\) = cos x . cot x + log (sin x) (- sin x)
∴ \(\frac{d u}{d x}\) = (sin x)^{cos x} [cos x cot x – sin x log (sin x)] ………..(2)
Also v = (x)^{sin x}
so that log v = sin x . log x
Diff. both sides w.r.t. x ; we get
\(\frac{1}{v} \frac{d v}{d x}\) = \(\frac{sin x}{x}\) + cos x log x
∴ \(\frac{d v}{d x}\) = x^{sin x} [\(\frac{sin x}{x}\) + cos x log x]………….(3)
putting the values of eqn. (2) and (3) in eqn. (1), we have
\(\frac{d y}{d x}\) = (sin x)^{cos x} [cos x cot x – sin x log (sin x)] + x^{sin x} [\(\frac{sin x}{x}\) + cos x log x]

(iii) Let y = x^{cos x} + (cos x)^x
= u + v ;
where u = x^{cos x} ;
v = (cos x)^x
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
Now u = x^{cos x}
so that log u = cos x log x ;
Diff. both sides w.r.t. x ; we have
\(\frac{1}{u} \frac{d u}{d x}\) = cos x . \(\frac{1}{x}\) + log x (- sin x)
⇒ \(\frac{d u}{d x}\) = x^{cos x} [\(\frac{cos x}{x}\) – sin x log x] ……..(2)
also, v = (cos x)^x so that
log v = x log cos x ;
Diff. both sides w.r.t. x ;
\(\frac{1}{v} \frac{d u}{d x}\) = – x tan x + log (cos x)
⇒ \(\frac{d v}{d x}\) = (cos x)^x [- x tan x + log (cos x)] …………..(3)
putting eqn. (2) and eqn. (3) in eqn. (1) ; we get
\(\frac{d y}{d x}\) = x^{cos x} [\(\frac{cos x}{x}\) – sin x log x] + (cos x)^x [- x tan x + log (cos x)]

(iv) Let y = x^{cos x} + (sin x)^{tan x}
= u + v
where u = x^{cos x} ; v = (sin x)^{tan x}
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
Now u = x^{cos x} so that log x = cos x log x
Differentiate both sides w.r.t. x ; we have
∴ \(\frac{1}{u} \frac{d u}{d x}\) = \(\frac{cos x}{x}\) – sin x log x
∴ \(\frac{d u}{d x}\) = x^{cos x} [\(\frac{cos x}{x}\) – sin x log x] ……….(2)
Also v = (sin x)^{tan x} so that
log v = tan x log sin x ;
Differentiate w.r.t. x
\(\frac{1}{v} \frac{d v}{d x}\) = tan x . cot x + log (sin x) sec² x
\(\frac{d v}{d x}\) = (sin x)^{tan x} [1 + sec² log sin x]
putting the values of eqn’s (2) and (3) in eqn. (1) ; we have
∴ \(\frac{d y}{d x}\) = x^{cos x} [\(\frac{cos x}{x}\) – sin x log x] + (sin x)^{tan x} [1 + sec² log sin x]

(v) Let y = (sin 2x)^x + sin^-1 \(\sqrt{3x}\)
= u + v ………….(1)
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(2)
Given u = (sin 2x)^x ;
Taking logarithm on both sides ; we have
log u = x log (sin 2x)^x = x log (sin 2x) ;
Diff. both sides w.r.t. x
∴ \(\frac{1}{u} \frac{d u}{d x}\) = log (sin 2x) + \(\frac{x}{sin 2x}\) . 2 cos 2x
⇒ \(\frac{d u}{d x}\) = (sin 2x)^x [log sin 2x + 2x cot 2x]
and v = sin^-1 \(\sqrt{3x}\) ………..(3)
∴ \(\frac{d v}{d x}\) = \(\frac{1}{\sqrt{1-(\sqrt{3} x)^2}} \frac{d}{d x}\) √3 √x
= \(\frac{1}{\sqrt{1-3 x}} \frac{\sqrt{3}}{2} \frac{1}{\sqrt{x}}\) …………..(4)
putting eqn. (3) and eqn. (4) in eqn. (2) ; we have
∴ \(\frac{d y}{d x}\) = (sin 2x)^x [log sin 2x + 2x cot 2x] + \(\frac{\sqrt{3}}{2 \sqrt{x-3 x^2}}\)

(vi) Let y = e^{sin x} + (tan x)^x
= u + v
where u = e^{sin x}
and v = (tan x)^x
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
Now u = e^{sin x}
⇒ \(\frac{d u}{d x}\) = e^{sin x} cos x …………..(2)
Also v = (tan x)^x so that log v = x log (tan x)
Differentiate both sides w.r.t. x ; we have
\(\frac{1}{v} \frac{d v}{d x}\) = [log (tan x) + x \(\frac{\sec ^2 x}{\tan x}\)]
⇒ \(\frac{d v}{d x}\) = (tan x)^x [log (tan x) + x sec² x cot x] ……………(3)
Putting (2) and (3) in eqn. (1) ; we get
∴ \(\frac{d y}{d x}\) = e^{sin x} cos x + (tan x)^x [log (tan x) + x sec x cosec x]

(vii) Let y = x^x – 2^{sin x} ;
Diff. both sides w.r.t. x, we get
∴ \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x^x) – 2^{sin x} log 2 . cos x ………..(1)
Let u = x^x so that
log u = x log x
∴ \(\frac{d u}{d x}\) = x^x (1 + log x) ………..(2)
Putting eqn. (2) in (1) ; we get
∴ \(\frac{d y}{d x}\) = x^x (1 + log x) – cos x . 2^{sin x} log 2.

(viii) Let y = x^{x cos x} + \(\frac{x^2+1}{x^2-1}\)
⇒ y = e^{log xx cos x} + \(\frac{x^2+1}{x^2-1}\)
⇒ y = e^{(x cos x) log x} + \(\frac{x^2+1}{x^2-1}\)

Question 8 (old).
(v) (sin x)^x + sin^-1 √x
Solution:
Let y = (sin x)^x + sin^-1 √x
= u + v ;
where u = (sin x)^x
and v = sin^-1 √x
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
Now u = (sin x)^x so that log u = x . log (sin x)
Differentiate both sides w.r.t. x, we get
\(\frac{1}{u} \frac{d u}{d x}\) = log (sin x) . 1 + x cot x
⇒ \(\frac{d u}{d x}\) = (sin x)^x [log (sin x) + x cot x] ………….(2)
Now v = sin^-1 √x
⇒ \(\frac{d v}{d x}\) = \(\frac{1}{\sqrt{1-(\sqrt{x})^2}} \frac{1}{2 \sqrt{x}}\)
= \(\frac{1}{2 \sqrt{x-x^2}}\) ………..(3)
putting the values of eqn’s (2) and (3) in eqn. (1) ; we have
∴ \(\frac{d y}{d x}\) = (sin x)^x [log (sin x) + x cot x] + \(\frac{1}{2 \sqrt{x-x^2}}\)

Question 9.
(i) If y = (log x)^{cos x} + \(\frac{x^2+1}{x^2-1}\), find \(\frac{d y}{d x}\).
(ii) If y = e^{x2 cos x} + (cos x)^x, find \(\frac{d y}{d x}\).
Solution:
Let y = (log x)^{cos x} + \(\frac{x^2+1}{x^2-1}\)
= u + v,
where u = (log x)^{cos x}
and v = \(\frac{x^2+1}{x^2-1}\)
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
Now u = (log x)^{cos x}
so that log u = cos x (log log x) ;
Differentiate both sides w.r.t. x, we have
\(\frac{1}{u} \frac{d u}{d x}\) = cos x . \(\frac{1}{x \log x}\) + log (log x) (- sin x)
⇒ \(\frac{d u}{d x}\) = (log x)^{cos x} [\(\frac{\cos x}{x \log x}\) – sin x log (log x)] ………..(2)
Also v = \(\frac{x^2+1}{x^2-1}\) ;
Diff. both sides w.r.t. x, we get
\(\frac{d v}{d x}\) = \(\frac{\left(x^2-1\right) 2 x-\left(x^2+1\right) 2 x}{\left(x^2-1\right)^2}\)
= \(\frac{-4 x}{\left(x^2-1\right)^2}\) …………..(3)
Putting eqn. (2) and (3) in eqn. (1) ; we have
\(\frac{d y}{d x}\) = (log x)^{cos x} [\(\frac{\cos x}{x \log x}\) – sin x log (log x)] – \(\frac{4 x}{\left(x^2-1\right)^2}\)

(ii) Given y = e^{x2 cos x} + (cos x)^x
⇒ y = u + v
⇒ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
where u = e^{x2 cos x}
diff. both sides w.r.t. x
\(\frac{d u}{d x}\) = e^{x2 cos x} \(\frac{d}{d x}\) (x² cos x)
= e^{x2 cos x} [x² (- sin x) + cos x . 2x] ……………..(2)
and v = (cos x)^x ;
taking logarithm on both sides
log v = x log cos x
∴ \(\frac{1}{v} \frac{d v}{d x}\) = log cos x + \(\frac{x}{cos x}\) (- sin x)
= log cos x – x tan x
⇒ \(\frac{d v}{d x}\) = (cos x)^x [log cos x – x tan x] ……….(3)
putting equation (2) and (3) in equation (1) ; we have
\(\frac{d y}{d x}\) = e^{x2 cos x} [- x² sin x + 2x cos x] + (cos x)^x [log cos x – x tan x]

Question 10.
Find \(\frac{d y}{d x}\) when
(i) x^y y^x = a^b
(ii) x^y + y^x = log a^b
(iii) x^x + y^x = 1
(iv) (sin x)^y = x + y
(v) xy = e^{x – y} (NCERT)
(vi) (cos x)^y = (cos y)^x
Solution:
(i) Given, x^y y^x = a^b ;
Taking logarithm both sides ; we have
log (x^y . y^x) = log a^b
⇒ log x^y + log y^x = 0
⇒ y log x + x log y = 0
Diff. both sides w.r.t. x ; we get
y \(\frac{d}{d x}\) log x + log x \(\frac{d y}{d x}\) + \(\frac{x}{y} \frac{d y}{d x}\) + log y \(\frac{d}{d x}\) (x) = 0
⇒ \(\frac{y}{x}\) + log x \(\frac{d y}{d x}\) + \(\frac{x}{y} \frac{d y}{d x}\) + log y . 1 = 0
⇒ (log x + \(\frac{y}{x}\)) \(\frac{d y}{d x}\) = – (log y + \(\frac{y}{x}\))
⇒ \(\frac{d y}{d x}=-\frac{y(x \log y+y)}{x(y \log x+x)}\)

(ii) Given x^y + y^x = log a^b
⇒ u + v = log a^b ……………(1)
where u = x^y
and v = y^x
Diff. eqn. (1) w.r.t. x ; we have
\(\frac{d u}{d x}\) + \(\frac{d v}{d x}\) = 0 …………(2)
Since u = x^y ;
Taking logarithm on both sides, we have
log u = y log x ;
Diff. both sides w.r.t. x, we have
\(\frac{1}{u} \frac{d u}{d x}\) = \(\frac{y}{x}\) + log x \(\frac{d y}{d x}\)
⇒ \(\frac{d u}{d x}\) = x^y [\(\frac{y}{x}\) + log x \(\frac{d y}{d x}\)]
= yx^y-1 + x^y log x \(\frac{d y}{d x}\)
Also v = y^x ;
Taking logarithm on both sides ; we have
log v = log y^x
= x log y
Diff. both sides w.r.t. x ; we have
\(\frac{1}{v} \frac{d v}{d x}\) = \(\frac{x}{y} \frac{d y}{d x}\) + log y . 1
⇒ \(\frac{d v}{d x}\) = y^x [\(\frac{x}{y} \frac{d y}{d x}\) + log y]
= xy^x-1 \(\frac{d y}{d x}\) + y^x log y
Using eqn. (3) and eqn. (4) in eqn. (2) ; we have
yx^y-1 + x^y log x \(\frac{d y}{d x}\) + xy^x-1 \(\frac{d y}{d x}\) + y^x log y = 0
⇒ (x^y log x + x y^{x – 1}) \(\frac{d y}{d x}\) = – (y x^x-1 + y^x log y)
⇒ \(\frac{d y}{d x}\) = – \(\frac{y\left(x^{y-1}+y^{x-1} \log y\right)}{x\left(x^{y-1} \log x+y^{x-1}\right)}\).

(iii) Given x^y + y^x = 1
⇒ u + v = 1
where u = x^x
and v = y^x
∴ \(\frac{d u}{d x}\) + \(\frac{d v}{d x}\) = 0 ………..(1)
Now u = x^x
so that lo g u = x log x
Differentiate both sides w.r.t. x, we get
\(\frac{1}{u} \frac{d u}{d x}\) = x . \(\frac{1}{x}\) + log x
⇒ \(\frac{d u}{d x}\) = x^x (1 + log x)
Also v = y^x
so that log v = x log y
Differentiate both sides w.r.t. x, we get
\(\frac{1}{v} \frac{d v}{d x}\) = \(\frac{x}{y} \frac{d y}{d x}\) + log y
⇒ \(\frac{d v}{d x}\) = y^x [\(\frac{x}{y} \frac{d y}{d x}\) + log y] ………….(30
putting eqn. (2) and (3) in eqn. (1) ; we get
x^x (1 + log x) + xy^x-1 \(\frac{d y}{d x}\) + y^x log y = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{-\left[y^x \log y+x^x(1+\log x)\right]}{x y^{x-1}}\).

(iv) Given (sin x)^y = x + y ;
Taking logarithm on both sides ; we have
log (sin x)^y = log (x + y)
⇒ y log sin x = log (x + y)
Diff. both sides w.r.t. x ; we get
\(\frac{d}{d x}\) y log sin x = \(\frac{d}{d x}\) log (x + y)
⇒ y \(\frac{d}{d x}\) log sin x + log sin x \(\frac{d y}{d x}\) = \(\frac{1}{x+y}\) \(\frac{d}{d x}\) (x + y)
⇒ \(\frac{y}{\sin x} \frac{d}{d x}\) (sin x) + log sin x \(\frac{d y}{d x}\) = \(\frac{1}{x+y}\left[1+\frac{d y}{d x}\right]\)
⇒ [y cot x + log sin x \(\frac{d y}{d x}\)] (x + y) = 1 + \(\frac{d y}{d x}\)
⇒ [(x + y) log sin x – 1] \(\frac{d y}{d x}\) = 1 – y (x + y) cot x
∴ \(\frac{d y}{d x}\) = \(\frac{1-y(x+y) \cot x}{(x+y) \log \sin x-1}\).

(v) Given xy = e^x-y ……….(1),
Taking logarithm on both sides, we get
log x + log y = (x – y) log e ;
Differentiate boht sides w.r.t. x, we have
\(\frac{1}{x}+\frac{1}{y} \frac{d y}{d x}=1-\frac{d y}{d x}\)
⇒ \(\left(\frac{1}{y}+1\right) \frac{d y}{d x}=1-\frac{1}{x}\)
⇒ \(\frac{d y}{d x}=\frac{(x-1) y}{(y+1) x}\)

(vi) Given (cos x)^y = (cos y)^x ;
Taking logarithm on both sides, we have
y log cos x = x log cos y
Diff. both sides w.r.t. x ; we get
⇒ \(\frac{d}{d x}\) [y log cos x] = \(\frac{d}{d x}\) [x log cos y]
⇒ y \(\frac{d}{d x}\) log cos x + log cos x \(\frac{d y}{d x}\) = x \(\frac{d}{d x}\) log cos y + log cos y . 1
⇒ \(\frac{y}{cos x}\) \(\frac{d}{d x}\) (cos x) + log cos x \(\frac{d y}{d x}\) = \(\frac{x}{cos y}\) \(\frac{d}{d x}\) (cos y) + log cos y
⇒ – y tan x + log cos x \(\frac{d y}{d x}\) = – x tan y \(\frac{d y}{d x}\) + log cos y
⇒ \(\frac{d y}{d x}\) (log cos x + x tan y) = log cos y + y tan x
∴ \(\frac{\log \cos y+y \tan x}{\log \cos x+x \tan y}\)

Question 11.
If y = x^{xx ………… ∞}, prove that \(\frac{d y}{d x}\) = \(\frac{y^2}{x(1-y \log x)}\).
Solution:
Given y = x^y …………(1)
Taking logarithm on both sides, we have ;
log y = y log x ;
\(\frac{1}{y} \frac{d y}{d x}=\frac{y}{x}+\log x \frac{d y}{d x}\)
⇒ \(\left(\frac{1}{y}-\log x\right) \frac{d y}{d x}=\frac{y}{x}\)
⇒ \(\left(\frac{1-y \log x}{y}\right) \frac{d y}{d x}=\frac{y}{x}\)
⇒ \(\frac{d y}{d x}=\frac{y^2}{x(1-y \log x)}\)

Exercise 5.9 (old)

Evaluate the following (1 to 5) limits (if they exist) :

Question 1.
(i) \(\ {Lt}_{x \rightarrow \infty} \frac{1}{(x-3)^2}\)
(ii) \(\ {Lt}_{x \rightarrow \infty} \frac{1}{(1-x)^2}\)
Solution:
(i) Let f(x) = \(\frac{1}{(x-3)^2}\) ;
D_f = R – {3}
as x Lakes positive values only and successive values of x go on increasing and becomes greater than any pre-assigned real number. however large it may be.
Then we say that x tends (goes) to infinity i.e. x → ∞
as x → ∞, x – 3 → ∞
(x – 3)² → ∞
i.e. \(\frac{1}{(x-3)^2}\) → 0
∴ \(\ {Lt}_{x \rightarrow 0} \frac{1}{(x-3)^2}\) = 0

(ii) Let f(x) = \(\frac{1}{(1-x)^2}\) ;
D_f = R – {1}
as x takes positive values only and successive values of x go on increasing and becomes greater than any pre assigned real number, however large it may be. Then we say that x → ∞
⇒ (1 – x)² → + ∞
⇒ \(\frac{1}{(1-x)^2}\) → 0
∴ \(\ {Lt}_{x \rightarrow \infty} \frac{1}{(1-x)^2}\) = 0.

Question 2.
(i) \(\underset{x \rightarrow \infty}{\ {Lt}}\) cos x
(ii) \(\underset{x \rightarrow 0}{\mathbf{L t}}\) cos \(\frac{1}{x}\)
Solution:
(i) Let cos x = f(x), D_f = R
When x = (2n + 1) \(\frac{\pi}{2}\) (n ∈ N)
Then cos x = 0, even for large values of n.
When x = 2nπ (n ∈ N)
Then cos x = 1, even for large values of n
When x = (2n+ 1) π (n ∈ N)
Then cos x = – 1, even for large values of n
Thus we observe that, cos x oscillates between – 1 and 1 as x → ∞
∴ \(\underset{x \rightarrow \infty}{\ {Lt}}\) cos x does not exists.

(ii) Let f(x) = cos \(\frac{1}{x}\) ; D_f = R – {0}
When x = \(\frac{1}{(2 n+1) \frac{\pi}{2}}\), n ∈ N and n is large
then x → 0 and
cos \(\frac{1}{x}\) = cos (2n + 1) \(\frac{\pi}{2}\) = 0
When x = \(\frac{1}{2 n \pi}\), n ∈ N and n is large then x → 0
∴ cos \(\frac{1}{x}\) = cos 2nπ = 1
When x = \(\frac{1}{2 n \pi+\pi}\), n ∈ N, n is large then x → 0
∴ cos \(\frac{1}{x}\) = cos (2n + 1) π = – 1
Thus, cos \(\frac{1}{x}\) oscillates between – 1 and 1,
as x → 0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) cos \(\frac{1}{x}\) does not exist.

Question 3.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^{-x}-1}{x}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}}{x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^{-x}-1}{x}\) = – \(\ {Lt}_{x \rightarrow 0} \frac{e^{-x}-1}{-x}\)
= – log e
= – 1
[∵ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{e^{-x}-1}{x}\) = log e = 1]

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}}{x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x-1-\left(e^{-x}-1\right)}{x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x-1}{x}-\ {Lt}_{x \rightarrow 0} \frac{e^{-x}-1}{x}\)
= log e + \(\ {Lt}_{x \rightarrow 0} \frac{e^{-x}-1}{-x}\)
= log e + log even= 1 + 1 = 2

Question 4.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{7^x-1}{\tan x}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{3^x-1}{\sqrt{2+x}-\sqrt{2}}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{7^x-1}{\tan x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{7^x-1}{x} \times \frac{x}{\tan x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{7^x-1}{x} \cdot \ {Lt}_{x \rightarrow 0} \frac{x}{\tan x}\)
= log 7 . 1 = log 7
[∵ \(\underset{\theta \rightarrow 0}{\ {Lt}} \frac{\theta}{\tan \theta}\) = 1]

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{3^x-1}{\sqrt{2+x}-\sqrt{2}}\)

Question 5.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-x-1}{x}\)
(ii) \(\ {Lt}_{x \rightarrow 1} \frac{x-1}{\log x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-x-1}{x}\)
= \(\ {Lt}_{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)-\ {Lt}_{x \rightarrow 0} 1\)
[∵ \(\ {Lt}_{x \rightarrow 0} \frac{e^x-1}{x}\) = log e = 1]
= log e – 1
= 1 – 1 = 0

(ii) put x = 1 + h
as x → 1 ⇒ h → 0
∴ \(\ {Lt}_{x \rightarrow 1} \frac{x-1}{\log x}=\ {Lt}_{h \rightarrow 0} \frac{1+h-1}{\log (1+h)}\)
= \(\ {Lt}_{h \rightarrow 0} \frac{h}{\log (1+h)}\)
= \(\frac{1}{\ {Lt}_{h \rightarrow 0} \frac{1}{h} \log (1+h)}\)
= \(\frac{1}{1}\) = 1
[∵ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{1}{x}\) log (1 + x) = 1]

Question 6.
Examine the function f(x) = \(\left\{\begin{array}{cc}
e^{1 / x} & , x \neq 0 \\
1 & , x=0
\end{array}\right.\) for continuity at x = 0.
Solution:
L.H.L. = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) e^1/x = 0
[as x → 0^–
⇒ x < 0
∴ \(\frac{1}{x}\) → – ∞
Thus e^1/x → e^{– ∞} = 0]
R.H.L. = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) e^1/x → + ∞
[as x → 0⁺ ⇒ x > 0
∴ \(\frac{1}{x}\) → + ∞
∴ e^1/x → ∞]
∴ L.H.L. ≠ R.H.L.
Thus f(x) is discontinuous at x = 0.

Continuous practice using S Chand Class 10 Maths Solutions ICSE Chapter 14 Circle Ex 14(c) can lead to a stronger grasp of mathematical concepts.

S Chand Class 10 ICSE Maths Solutions Chapter 14 Circle Ex 14(c)

Question 1.
(a) In figure, APB is tangent to circle with centre O. If ∠QPB = 50°, find ∠POQ.

(b) In figure, AB and AC are tangents. If AB = 4 cm, find AC.

(c) In the figure, PQ and PR are tangents to circle, centre O. If ∠QPR = 80°, find ∠QOR.

Solution:
(a) In the figure, APB is tangent to the circle
with centre O
∠QPB = 50°
∵ OP is the radius and APB is tangent
∴ OP ⊥ APB
∴ ∠OPB = 90° ⇒ ∠OPQ + ∠QPB = 90°
⇒ ∠OPQ + 50° = 90°
⇒ ∠OPQ = 90° – 50°
⇒ OPQ = 40°
But in △OPQ, OP = OQ (radii of the circle)
∴ ∠OPQ = ∠OQP = 40°
∠POQ + ∠OPQ + ∠OQP = 180°
⇒ ∠POQ + 40° + 40° = 180°
⇒ ∠POQ + 80°= 180°
∠POQ = 180° – 80 = 100°

(b) In circle, two tangent AB and AC are drawn, a point A outside the circle
∴ AC = AB = 4 cm

(c) In the figure, a circle with centre O from a point P outside the circle two tangents PQ and PR are drawn and ∠QPR = 80°
∴ ∠QPR and ∠QOR are supplementary ∠QPR + ∠QOR = 180°
⇒ 80° + ∠QOR = 180°
⇒ ∠QOR = 180° – 80° = 100°
∠QOR = 100°

Question 2.
In figure, O is the centre of the circle. Find ∠POS.

Solution:
In the figure, a circle with centre O
From a point P outside of it, tangents PT and
PS are drawn to the circle, and ∠TPO = 30°
In △PTO, OT ⊥ PT
∴ ∠OTP = 90°
∴ ∠TOP + ∠TPO = 90°
⇒ ∠TOP + 30° = 90°
⇒ ∠TOP = 90°- 30° = 60°
∴ OP is the bisector of ∠TOS
∴ ∠TOP = ∠POS = 60°

Question 3.
In figure, PQ is tangent to the circle at A, DB is a diameter, ∠ADB = 30° and ∠CBD = 60°, calculate
(i) ∠QAB
(ii) ∠PAD
(iii) ∠CDB

Solution:
In the figure, BD is the diameter of the circle PQ is tangent to the circle at A
∠ADB = 30°, ∠DBC = 60°

(i) ∵ QAP is tangent and AB is chord of the circle
∴ ∠QAB = ∠ADB = 30°

(ii) ∠PAD + ∠DAB + ∠QAB = 180° (Angles of a line)
⇒ ∠PAD + 90° + 30° = 180°
(∵ ∠DAB = 90° angle in semicircle)
⇒ ∠PAD + 120° = 180°
⇒ ∠PAD = 180° – 120° = 60°

(iii) In ABCD,
∠CDB + ∠CBD + ∠BCD = 180° (Sum of angles of a triangle)
⇒ ∠CDB + 60° + 90°= 180°
⇒ ∠CDB + 150° = 180°
⇒ ∠CDB = 180°- 150° = 30°

Question 4.
In figure, PR and PQ are the tangents, each of them being equal to 9 cm, ∠QPR = 60°. Find the length of the chord QR which joins their points of contact.

Solution:
In the figure, PQ and PR the tangents drawn
from P outside the circle such that
PQ = PR = 9 cm and ∠QPR = 60°
QR is joined
In △PQR, ∠QPR = 60°
∵ PQ = PR
∴ ∠PQR = ∠PRQ = 60°
∴ △PQR is an equilateral triangle
PQ = PR = QR = 9 cm

Question 5.
Find the length of the tangent drawn to a circle of radius 3 cm, from a point distant 5 cm from the centre.
Solution:
In a circle of radius 3 cm, point P is 5 cm away from the centre O of the circle
PQ and PR are the tangents drawn from P to the circle

∵ OQ is radius and PQ is tangent
∴ OQ ⊥ QP or ∠OQP = 90°
Now in right angled △OPQ,
OP² = OQ² + PQ² (Pythagoras Theorem)
⇒ (5)² = (3)² + PQ²
⇒ 25 = 9 + PQ²
⇒ PQ² = 25 – 9 = 16 = (4)²
∴ PQ = 4 cm
But PQ = PR
∴ PQ = PR = 4 cm

Question 6.
A circle touches the side BC of △ABC at P and touches AB and AC produced at Q and R respectively. If AQ = 5 cm, find the perimeter of △ABC.

Solution:

∵ From A, AQ and AR are the tangents drawn to the circle
∴ AQ = AR = 5cm ….(i)
Similarly from B, tangent BQ and BP are drawn
∴ BQ = BP ….(ii)
and from C,
CR = CR ….(iii)
Now perimeter of △ABC,
= AB + AC + BC = AB + AC + BP + CP = AB + AC + BQ + CR [From (ii) and (iii)]
= AB + BQ + AC + CR
= AQ + AR = 5 cm + 5 cm [From (i)] = 10 cm

Question 7.
There are two concentric circles of radii 3 cm and 5 cm respectively. Find the length of the chord of the outer circle which touches the inner circle
Solution:
Two circles which are concentric and their centre is O, are of radii 5 cm and 3 cm
i. e. OA = 5 cm and OP = 3 cm
AB is chord of the larger circle which touches the smaller circle at P

∴ OP ⊥ AB ⇒ AP = PB
In right △OAP,
OA² = OP² + AP² (Pythagoras Theorem)
⇒ (5)² = (3)² + AP²
⇒ 25 = 9 + AP²
⇒ AP² = 25 – 9 = 16 = (4)²
∴ AP = 4 cm
Hence AB = 2AP = 2 × 4 = 8 cm

Question 8.
Three circles with centres A, B, C touch each other externally; AB = 4 cm, BC = 6 cm, CA = 7 cm; find their radii.
Solution:
ABC is a triangle and with centre A, B and C, three circles are drawn touching each other externally a t P, Q and R respectively AB = 4 cm, BC = 7 cm and AC = 6 cm

Let radii of circles with centre A, B and C respectively be x, y and z
∴ AB = x + y, BC = y + z, CA = z + x
⇒ x + y = 4 cm, y + z = 6 cm, z + x = 7 cm
∴ AB + BC + CA = x +y +y + z + z + x
⇒ 4 + 6 + 7 = 2(x + y + z)
⇒ x + y + z = \(\frac { 17 }{ 2 }\) = 8.5 cm
Subtracting from x + y + z, we get
x = 8.5 – 4 = 4.5 cm
y = 8.5 – 6 = 2.5 cm
z = 8.5 – 7 = 1.5 cm
Hence their radii are 2.5 cm, 1.5 cm and 4.5 cm

Question 9.
Equal circles, centres O and O’ touch each other at X. OO’ is produced to meet the circle O’ at A. AC is tangent to the circle whose centre is O. O’ is perpendicular to AC. Find the value of

Solution:
Two equal circles with centres O and O’ touch each other externally at X. 00′ is produce to meet the circle O’ at A. Through A, a tangent AC is drawn to the circle with
centre O. O’D ⊥ AC
Let r be the radius of each circle
In △AO’D and △AOC,
∠D = ∠C (each 90°)
∠A = ∠A (common)

Question 10.
In figure, P and Q are centres of two circles of radii 12 cm and 3 cm respectively. A and B are the points of contact of the common tangent XY. Find AB.

Solution:
Two circles with centres P and Q touch externally at R
XY is their common tangent
A and B are their points of contact
Join PA, QB and PQ
From Q, draw QS || XY

∵ PA and QB are perpendicular to XY and
QS || AB
∴ QS = AB
PA = 12 cm, QB = 3 cm, PQ = 12 + 3 = 15 cm
∴ PS = PA – SA = 12 – 3 = 9 cm
Now in right △PSQ,
PQ² = PS² + QS² (Pythagoras Theorem)
⇒ (15)² = (9)² + QS²
⇒ 225 = 81 + QS²
⇒ QS² = 225 – 81 = 144 = (12)²
∴ QS = 12 cm
But AB = QS = 12 cm