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Effective Class 11 ISC Maths OP Malhotra Solutions Chapter 31 Moving Average Ex 31 can help bridge the gap between theory and application.

S Chand Class 11 ICSE Maths Solutions Chapter 31 Moving Average Ex 31

Question 1.
The table shows the number of students in a school getting at least a grade C in mathematics for the years 1994 to 2001.
(i) Represent this data as a time series.

(ii) Calculate the 3-point moving average and plot it on the same graph.
(iii) Are the school’s maths results improving?
(iv) Explain why this is not a good way to work out whether the school’s results are improving.
Solution:
(i) The table of values is given as under :


(iii) The set of data is smoothed out and is increasing.
(iv) The number is increasing but might be because the school is getting bigger.

Question 2.
The profits of a soft drink firm in thousands of litres during each month of a year were :

Calculate 3-monthly moving averages and illustrate graphically. (ISC 1992 type)
Solution:
Calculation of 3-monthly moving arrange is given as :

Question 3.
The number of traffic offences committed in a certain city over a period of 3 years is given in the following table :

Calculate 4-quarterly moving averages and illustrate these and original figures on one graph using the same axis for both. Comment briefly on a local politician’s claim that traffic offences were on the increase.
Solution:

Thus the local politician’s claim that traffic offences were on the increase be true, it is cleared from 4-quarterly moving average centred column.

Question 4.
Find the 4-quarterly moving averages in the following table which gives the quarterly index numbers of coal production (for the years 1936-1938). Also plot on the same graph the quarterly index numbers as well as the 4-quarterly moving average. Comment on the nature of the general trend.

Solution:

Question 5.
The annual incomes of a firm were recorded every quarter for 4 years. The results are shown in this table.

(i) Work out the 4-point moving average for the data.
(ii) Plot the original data and the moving average on the same graph.
(iii) Comment on how the firm’s incomes have changed over the 4-years.
Solution:


The data has been smoothed out so there is a steady increase in Income.

Question 6.
The following table shows the daily sales of milk at a local corner shop for a month.

Make a table showing the moving average using a 7-day span, and draw a graph to show the trend of milk sales over the month.
Solution:

Question 7.
The following table gives the monthly expenditure on a motor car for a period of two years.

Calculate 12-month!y moving average for the two years and display them and the original table on the same graph.
Solution:


Question 8.
A new film was shown at a theatre and ran for six w eeks. The attendances are shown in the table.

(i) Plot a line graph from the above data.
(ii) Calculate the 6-day moving average for the data and plot this on the same graph.
(iii) Comment on the weekly attendance.
Solution:

Question 9.
The table below given details of the electricity generated in million kilowatt hoars for public supply in each quarter of the years 1952 to 1955.

Draw a graph illustrating these figures.
Calculating a set of moving averages using the most suitable number of observations; give reasons of your choice. On the same diagram as before draw a graph showing the moving averages.
Solution:
We shall use 4-monthly moving average as it null eliminate the 12 monthly cycle and leave the general trend of the data.

Question 10.
The number of letters, in hundreds, posted in a certain city on each day of a fornight was as follows:

Calculate the 7-day moving averages and display these and the original figures graphically on the same diagram, using the same scale and axes. What is the general trend ?
Solution:

From graph and table, we observe that the general trend is that the no. of letters posted goes on increasing every day week after week.

Question 11.
In an influenza epidemic the numbers of cases diagnosed were :

On what days do the mode and upper and lower quartiles occur ?
Calculate 3-day moving averages and display them and the original figures on the same graph.
Solution:
Calculation of 3 yearly moving average

Question 12.

Plot these figures on a graph.
Calculate the 4-quarterly moving averages and plot on the same graph.
Solution:

Question 13.
Registered unemployed (hundreds)

Plot these monthly figures on a graph. Calculate the 12-monthly moving averages and plot these on the same graph.
Solution:

Question 14.
A Ballet Company gave a 6-weeks’ season at a large hall capable of seating 4000 people and the attendances in hundreds, at the evening performances, are recorded in the following table.
Attendance, in hundreds, to nearest hundred

Plot a graph of the above time-series and include on the same diagram the graph of 6-day moving averages.
Comment on the weekly cycle on attendances and state, with reasons, if you think, an extension of the season of eight weeks, would have been justified.
Solution:

Question 15.
Production of passenger cars, U.S.A. (tens of thousands)

Calculate the 4-quarterly moving averages and then draw the graphs of the given series and the moving averages. Briefly comment on the general trend.
Solution:

From table and graph it is observed that, the demand of cars was increasing year after year.

Question 16.
The aggregate number, in millions, of working days lost in strikes during each year of the period 1950-60 was

Draw a graph to represent this information. Calculate the 3-yearly moving averages and draw the 3-yearly moving averages graph, using the same axes and scales. What is the main purpose in drawing moving average graph ? Comment on whether the purpose is achieved in this case.
Solution:

The main purpose in drawing moving average graph is to find the general trend. The purpose is achieved in this case, because the graph shows that the number of working lost during strike is goes on increasing.

Question 17.
The profits of a soft drink firm in thousands of rupees during each month of a year were

Plot these on a graph.
Calculate 4-monthly moving averages and plot these on the same graph. Comment on the general trend.

The general comment on trend is that the profits go on increasing from Jan. to August and start decreasing from Sept, to December.

Question 18.
Calculate,5-yearIy moving averages for the following data of the commercial and industrial failures in a country from 1982 to 1997.

Display the actual and tend values on the same graph using the same axes for both.
Solution:

Question 19.
The table given below shows the daily attendance in thousands at a certain exhibition over a period of two weeks :

Calculate 7-day moving averages and illustrate these and original information on the same graph using the same scales.
Solution:
Calculation of 7-yearly moving average

Question 20.
The profit of a soft-drink firm (in thousands of rupees) during each month of the year is as given below :

Calculate the 4-monthly moving averages and plot these and the original data on a graph sheet.
Solution:
Calculation of 4 yearly moving average

Access to comprehensive ISC Class 11 S Chand Maths Solutions Chapter 10 Quadratic Equations Ex 10(f) encourages independent learning.

S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(f)

Question 1.
Show that
(a) x² – 3x + 6 > 0 for all x
(b) 4x – x² – 6 < 0 for all x
(c) 2x² – 4x + 7 is always +ve
(d) – 2x² + 3x – 4 is always -ve
(e) – x² + 3x – 3 is always -ve.
Solution:
(a) Here D = b² – 4ac = (- 3)² – 4 x 1 x 6
= 9 – 24 = – 15 < 0
Thus the expression x² – 3x + 6 and coeff. of x² have same sign.
Here coeff. of x² = a = 1 > 0
∴ x² – 3x + 6 > 0 ∀ x

(b) Here a = – 1, b = + 4; c = – 6
and Discriminant D = b² – 4ac = 4² – 4 x ( – 1) x (- 6)
= 16 – 24 = – 8 < 0
Thus the given expression 4x – x² – 6 has same sign as that of a.
Here a = – 1 < 0
∴ 4x – x² – 6 < 0 for all x.

(c) Here a = 2 ; b = – 4 ; c = 1
and discriminant D = b² – 4 ac = (- 4)² – 4 x 2 × 7 = 16 – 56 = – 40 < 0
∴ roots are imaginary. Thus the given expression 2x² – 4x + 7 and a has same sign. Here a = 2 > 0
∴ 2x² – 4x + 7 > 0 for all x

(d) Here a = – 2 ; b = – 3 and c = – 4
and discriminant D = b² – 4ac = (- 3)² – 4 x (- 2) x (- 4) = 9 – 32 = – 23 < 0
∴ roots are imaginary
Hence the given expression – 2x² + 3x – 4 and a has same sign. Here a = – 2 < 0
∴ – 2x² + 3x – 4 < 0 ∀ x

(e) Here a = – 1 ; b = 3 ; c = -3
Here discriminant D = b² – 4ac = 32 – 4 x (- 1) x (- 3) = 9 – 12 = – 3 < 0
∴ roots are imaginary. Hence given expression and ‘a’ has same sign.
Here, a = – 1 < 0
∴ – x² + 3x – 3 < 0 ∀ x.

Question 2.
Explain why 3x² + kx – 1 is never always positive for any value of k.
Solution:
Here, a = 3 ; b = k; c = – 1
Here discriminant D = b² – 4ac = k² – 4 x 3 x (- 1) = k² + 12 > 0
∴ roots are real and unequal.
Let a and P be its two roots s.t a > P
Then when x > α or x < ß, the given expression 3x² + kx – 1 has same sign as that of ‘a’
Here a = 3 > 0
∴ 3x² + kx – 1 > 0
But when ß < x < α, then 3x² + kx – 1 and a has opposite sign.
∴ 3x² + kx – 1 < 0
Hence 3x² + kx – 1 is never always positive for any value of k.

Question 3.
Under what conditions is 2x² + kx + 2 always positive ?
Solution:
Comparing 2x² + kx + 2 with ax² + bx + c, we have a = 2 ; b = k; c = 2
Here discriminant D = b² – 4ac = k² – 4 x 2 x 2 = k² – 16
Here a = 2 > 0 if D < 0 then 2x² + kx + 2 > 0 ∀ x i.e. if – 16 < 0 if k² < 16
if | k |< 4 if – 4 < k < 4
Hence, 2x² + kx + 2 > 0 ∀x, where – 4 < k < 4

Question 4.
Find the values of a so that the expression x? – (a + 2) x + 4 is always positive.
Solution:
On comparing x² – (a + 2) x + 4 with Ax² + Bx + C
we have A = 1 ; B = – (a + 2) and C = 4
Here Discriminant D = b² – 4AC = (a + 2)² – 16
Also, A = 1 > 0 if D < 0 then given expression is always be positive.
i.e. if (a + 2)² – 16 < 0 if (a + 2)² < 16 if |a + 2| < 4
if – 4 < a + 2 < 4 [∵ |x| < l ⇒ – l < x < l]
if – 6 < a < 2

Question 5.
Find the range of values of x for which the expression 12x² + 7x – 10 is negative.
Solution:
For range of x for which 12x² + 7x – 10 < 0
⇒ 12\(\left[x^2+\frac{7}{12} x-\frac{5}{6}\right]\) < 0
⇒ x² + \(\frac{7}{12} x+\frac{49}{576}-\frac{49}{576}-\frac{5}{6}\) < 0
⇒ \(\left(x+\frac{7}{24}\right)^2-\left(\frac{49+480}{576}\right)\) < 0
⇒ \(\left(x+\frac{7}{24}\right)^2<\left(\frac{23}{24}\right)^2\) ⇒ \(\left|x+\frac{7}{24}\right|<\frac{23}{24}\)
⇒ \(\frac{-23}{24}<x+\frac{7}{24}<\frac{23}{24}\)
⇒ \(\frac{-23}{24}-\frac{7}{24}<x<\frac{23}{24}-\frac{7}{24}\)
⇒ \(\frac{-30}{24}<x<\frac{16}{24} \text { i.e } \frac{-5}{4}<x<\frac{2}{3}\)

Question 6.
(i) Find the values of ‘a’ for which the expression x² – (3a – 1) x + 2a² + 2a – 11 is always positive.
(ii) If x² + 4ox + 2 > 0 for all values of x, then a lies in the interval
(a) ( – 2,4)
(b) (1, 2)
(c) (- \(\sqrt{2}\), \(\sqrt{2}\))
(d) \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
(e) (- 4, 2)
Solution:
(i) On comparing x² – (3a – 1) x + 2a² + 2a – 11 with Ax² + Bx + C, we have
A = 1 ; B = – (3a – 1) and C = 2a² + 2a – 11
Here D = B² – 4AC = (3a – 1)² – 4 (2a²+ 2a-ll) = (9a² – 6a + 1) – (8a² + 8a – 44)
= a²- 14a + 45
Here A = 1 > 0 and if D < 0 then given expression will be always positive

if a² – 14a + 45 < 0 if (a – 9) (a – 5) < 0
The critical points are a = 5, 9
Then by method of intervals, 5 < a < 9.

(ii) On comparing x² + 4ax + 2 with Ax² + Bx + C.
Here A = 1 ; B = 4a and C = 2 and discriminant
D = b² – 4AC = (4a)² – 4 x 1 x 2 = 16a² – 8
Here A = 1 > 0.
Then x² + 4ax + 2 > 0
if D < 0 if b² – 4AC < 0 if 16a² – 8 < 0
if a² < \(\frac { 1 }{ 2 }\) if | a | < \(\frac { 1 }{ 2 }\) if – \(\frac{1}{\sqrt{2}}<a<\frac{1}{\sqrt{2}}\) if a ∈ \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)

Question 7.
Find the greatest value of 3 + 5x – 2x² for all real values of x.
Solution:
Let y = 3 + 5x – 2x² ⇒ – 2x² + 5x + 3 – y = 0
Since x is real
∴ D ≥ 0
⇒ b² – 4ac ≥ 0
⇒ 5² – 4 x (- 2) (3 – y) ≥ 0
⇒ 25 + 24 – 8y ≥ 0
⇒ 49 – 8y ≥ 0
⇒ 8y ≤ 49
⇒ y ≤ \(\frac { 49 }{ 8 }\) = 6\(\frac { 1 }{ 8 }\)
∴ greatest value of y = 6\(\frac { 1 }{ 8 }\)

Question 8.
Find the least value of \(\frac{6 x^2-22 x+21}{5 x^2-18 x+17}\) for real values of x.
Solution:
Let y = \(\frac{6 x^2-22 x+21}{5 x^2-18 x+17}\)
⇒ y (5x² – 18x + 17) = 6x² – 22x + 21
⇒ x² (6 – 5y) + x (18y – 22) + 21 – 17y = 0
Since x is given to be real

∴ D ≥ 0
⇒ (18y – 22)² – 4 x (6 – 5y) (21 – 17y) ≥ 0
⇒ 4 (9y- 11 )² – 4 (6 – 5y) (21 – 17y) ≥ 0
⇒ (81y² + 121 – 198y) – (126 – 102y – 105y + 85y²) ≥ 0
⇒ – 4y² + 9y – 5 ≥ 0
⇒ 4y² – 9y + 5 ≤ 0
⇒ (y – 1) (4y – 5) ≤ 0
The critical points are y – 1 = 0 and 4y – 5 = 0 i.e. y = 1, \(\frac { 5 }{ 4 }\)
Then by method of intervals, we have 1 ≤ y ≤ \(\frac { 5 }{ 4 }\) ⇒ y ∈ [1, \(\frac { 5 }{ 4 }\) ]
∴ least value of y = 1

Question 9.
If x be real, prove that the value of \(\frac{11 x^2+12 x+6}{x^2+4 x+2}\) cannot lie between – 5 and 3.
Solution:
Let y = \(\frac{11 x^2+12 x+6}{x^2+4 x+2}\)

⇒ y (x² + 4x + 2) = 11x² + 12x + 6
⇒ x²(y – 11) + x (4y – 12) + 2y – 6 = 0
For x is to be real ∴ Disc ≥ 0 ⇒ (4y – 12)² – 4 (y – 11) (2y – 6) ≥ 0
⇒ 16 (y – 3)² – 8(y – 11) (y – 3) ≥ 0
⇒ 8[2(y² – 6y + 9) – (y² – 14y + 33)] ≥ 0
⇒ y² + 2y – 15 ≥ 0
⇒ (y – 3) (y + 5) ≥ 0
The critical points are given by y = 3, – 5.
Then by method of intervals, we have y ≤ – 5 or y ≥ 3
Hence y cannot lie between – 5 and 3.

Students appreciate clear and concise OP Malhotra Maths Class 11 Solutions Chapter 10 Quadratic Equations Ex 10(e) that guide them through exercises.

S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(e)

Draw the graph of each of the following quadratic functions.

Question 1.
y = x² – 5x + 6 ; 0 ≤ x ≤ 5.
Solution:
Given y = x² – 5x + 6 ; 0 ≤ x ≤ 5
The table of values is given as under :

Question 2.
y = – x² + 2x + 3 ; – 3 ≤ x ≤ 5.
Solution:
Given y = – x² + 2x + 3 where – 3 ≤ x ≤ 5
The table of values is given as under;

Question 3.
y = x² – 4x + 4 ; – 1 ≤ x ≤ 5.
Solution:
Given y = x² – 4x + 4 ; – 1 ≤ x ≤ 5
The table of values is given as under

Question 4.
Solve graphically and compare your answer with algebraic solution either by factorization or formula method :
(i) y = x² – 5x + 6
(ii) y = – x² + 2x + 3
(iii) y = x² – 4x + 4
(iv) y = x² – x – 6
(v) y = x² – 6x + 9
(vii) y = x² – 4x + 5 = 0
(viii) y = x² + 2x + 2 = 0
Solution:
(i) y = x² – 5x + 6
The table of values is given as under:

Clearly the graph of curve intersects x-axis at two points at
x = 2 and x = 3
∴ x = 2, 3

Verification : y = x² – 2x – 3x + 6 = x (x – 2) – 3 (x – 3)
⇒ y = (x – 2) (x – 3)
Clearly x = 2 and x = 3 are the two solutions of given curve.

(ii) Given y = – x² + 2x + 3
The table of values is given as under :

Clearly the graph of curve intersects x-axis at two points at x = – 1 and x = 3
Verification : y = – x³ + 2x + 3 = – x² + 3x – x + 3
= – x (x – 3) – 1(x – 3)
= – (x – 3) (x + 1)
Clearly x = – 1 and x = 3 are two solutions of given curve.

(iii) Given y = x² – 4x + 4
The table of values is given as under :

Clearly the graph of curve intersects x-axis at one point i.e. x = 2
Verification : y = x² – 4x + 4 = (x – 2)²
Clearly x = 2 be the solution of given curve.

(iv) Given y = x² – x – 6
The table of values is given as under

Clearly the graph meets x-axis at (- 2, 0) and (3, 0).
∴ x = – 2, 3 are the real solutions of given curve

Verification : y = x² – x – 6 = x² – 3x + 2x – 6
= x (x – 3) + 2 (x – 3) = (x + 2) (x – 3)
Clearly – 2 and 3 are the solutions of y = x² – x – 6 = 0

(v) Given eqn. of curve be y = x² – 6x + 9
The tale of values is given as under :

Clearly the graph of curve intersects x-axis at (3, 0).
∴ x = 3, 3 be the solution of given curve.

Verification : y = x² – 6x + 9 = x² – 3x – 3x + 9
= x (x – 3) – 3 (x – 3) = (x – 3)²
Clearly x = 3 be the solution of given curve.

(vi) Given eqn. of curve bey = – x² + x + 12 ⇒ y = – (x² + x – 12)
The table of values is given as under :

Clearly the graph of the curve intersects x-axis at x = – 4 and x = 3.
Hence x = – 4, 3 are two real solutions of given curve.

Verification : y = – (x² – 4x + 3x – 12) = – [x (x – 4) + 3 (x – 4)]
= – (x + 3)(x – 4)
Clearly x = – 3 and 4 are the solutions of y = – x² + x + 12 = 0

(vii) Given curve be y = x² – 4x + 5 = 0
The table of values is given as under

Clearly the curve does not meets x-axis at any point so it has no real solution.
Verification : x² – 4x + 5 = 0 ⇒ x = \(\frac{4 \pm \sqrt{16-4 \times 1 \times 5}}{2}=\frac{4 \pm 2 i}{2}=2 \pm i\)

(viii) Given curve be y = x² + 2x + 2 = 0
The table of values is given as under :

Clearly the graph of curve does not meet x-axis
∴ given curve has no real solution.
Verification : y = x² + 2x + 2 = 0
⇒ x = \(\frac{-2 \pm \sqrt{4-8}}{2}\)
⇒ x = \(\frac{-2 \pm 2 i}{2}\)
⇒ x = – 1 ± i

Interactive OP Malhotra Maths Class 11 Solutions Chapter 10 Quadratic Equations Ex 10(d) engage students in active learning and exploration.

S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(d)

Question 1.
Find the condition that one root of ax² + bx + c = 0 may be
(i) three times the other, (ii) it times the other, (iii) more than the other by h.
Solution:
(i) Let the roots of given eqn. be ax² + bx + c = 0 are α and 3α.

which gives the required condition.

(ii) Let the roots of given eqn. are α and nα

which gives the required condition.

(iii) Let the roots of given eqn. ax² + bx + c = 0 are α and α + h.

which gives the required condition.

Question 2.
Find the condition that the ratio between the roots of the equation ax² + bx + c = 0 may be m : n.
Solution:
Given quadratic eqn.be ax² + bx + c = 0 …(1)
Let the roots of eqn. (1) are mα and nα.

which gives the required condition

Question 3.
If the ratio of the roots of the equation x² + px + q = 0 is equal to the ratio of the roots of x² + lx + m = 0, prove that mp² = ql².
Solution:
Let α and ß are the roots of eqn. x² + px + q = 0
∴ α + ß = – p ; αß = q
and γ, δ are the roots of the eqn.
x² + lx + m = 0
γ + δ = – l and γδ = m
according to given condition, we have

which is the required condition.

Question 4.
For what values of a and b, the equation x² + (2a – 3) x = 3b + 4 should have both the roots zero ?
Solution:
Given quadratic eqn. be, x² + (2a – 3) x – (3b + 4) = 0 …(1)
Now both the roots of eqn. (1) are zero.
∴ coeff. of x = 0 and constant term = 0
⇒ 2a – 3 = 0 and 3b + 4 = 0 ⇒ a = \(\frac { 3 }{ 2 }\) and b = – \(\frac { 4 }{ 3 }\)

Question 5.
Find the values of X and p if both the roots of the equation (3λ + 1)x² = (2λ + 3λ) x – 3 are infinite.
Solution:
We know that if a and p are the roots of eqn. ax² + bx + c = 0.
Then \(\frac { 1 }{ α }\) and \(\frac { 1 }{ ß }\) are the roots of eqn. cx² + bx + a = 0
Hence both roots of ax² + bx + c = 0 are infinite if roots of cx² + bx + a = 0 are zero.
∴ a = 0 and b = 0
Thus both the roots of eqn. (3λ + 1) x² – (2λ + 3μ) x + 3 = 0 are infinite.
if coeff. of x² = 0 and coeff. of x = 0
if 3λ + 1 = 0 and 2λ + 3μ = 0
if λ = – \(\frac { 1 }{ 3 }\) and – \(\frac { 2 }{ 3 }\) + 3μ = 0 ⇒ μ = \(\frac { 2 }{ 9 }\)
Thus λ = – \(\frac { 1 }{ 3 }\) and μ = \(\frac { 2 }{ 9 }\)

Question 6.
Find m so that the roots of the equation \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) may be equal in magnitude and opposite in sign.
Solution:
The given quadratic equation be

Question 7.
(i) The roots of the quadratic equation 4x² – (5a + 1) x + 5a = 0, are p and q. If q = 1 + p, calculate the possible values of a, p and q.
(ii) Find the values of p for which the quadratic equation x² – px + p + 3 = 0 has
(a) coincident roots, (b) real distinct roots, (c) one positive and one negative root.
Solution:
(i) Given the roots of eqn. 4x² – (5a + 1) x + 5a = 0 are p and q.

(ii) Given quadratic eqn. be, x² – px + p + 3 = 0 …(1)
On comparing eqn. (1) with ax² + bx + c = 0
We have, a = 1 ; b = – p ; c = p + 3
(a) Roots of eqn. (1) are coincident roots
∴ discriminant = 0
⇒ b² – 4 ac = 0
⇒ (- p)² – 4 x 1 x (p + 3) = 0
⇒ p² – 4p – 12 > 0
⇒ (p + 2) (p – 6) = 0
⇒ p = – 2, 6

(b) roots of eqn. (1) are real and distinct.

∴ D > 0
⇒ b² – 4ac > 0
⇒ p² – 4p – 12 > 0
⇒ (p + 2) (p – 6) > 0
⇒ p < – 2 or p > 6

(c) given one root of eqn. (1) be positive and other be negative i.e. roots of eqn. (1) are of opposite sign.
∴ product of roots < 0
⇒ \(\frac { c }{ a }\) < 0
∴ c and a are of opposite sign
Here a = 1 > 0
∴ c < 0 ⇒ p + 3 < 0 ⇒ p < – 3

Question 8.
Find the values of m for which the quadratic equation x² – m (2x – 8) – 15 = 0 has
(i) equal roots, (ii) both roots positive.
Solution:
The given quadratic eqn. be x² – m (2x – 8) – 15 = 0
⇒ x² – 2mx + 8m – 15 = 0 …(1)
On comparing eqn. (1) with ax² + bx + c = 0 ; we have,
a = 1 ; b = – 2m and c = 8m – 15

(i) Roots of eqn. (1) are equal.
∴ discriminant = 0 ⇒ b² – 4ac = 0 (- 2m)² – 4 x 1 x (8m – 15) = 0
⇒ 4m² – 32m + 60 = 0
⇒ m² – 8m + 15 = 0
⇒ (m – 3) (m – 5) = 0
⇒ m = 3, 5

(ii) both roots of eqn. (1) are positive
∴ Sum of roots > 0 ⇒ – \(\frac { b }{ a }\) > 0 ⇒ \(\frac { b }{ a }\) < 0
i.e. b and a are opposite sign.
Here a = 1 > 0
∴ b < 0 ⇒ – 2m < 0 ⇒ m > 0
and product of roots > 0 ⇒ \(\frac { c }{ a }\) > 0
∴ c and a are of same sign.
Here a = 1 > 0 ∴ c > 0
⇒ 8m – 15 > 0
⇒ m > \(\frac { 15 }{ 8 }\)
Therefore m > 0 and m > \(\frac { 15 }{ 8 }\).

Question 9.
If a + b + c = 0, prove that the roots of ax² + bx + c = 0 are rational. Hence, show that the roots of (p + q) x² – 2px + (p – q) = 0 are rational.
Solution:
Given quadratic equation be ax² + bx + c = 0 …(1)
Clearly x = 1 satisfy eqn. (1)
∵ a + b + c = 0 (given)
Thus x = 1 be the roots of eqn. (1).
Since irrational or imaginary/non-real roots occurs in conjugate pair. Other root of eqn. (1) must be rational.
Hence the general condition for the roots of eqn. (1) are rational be a + b + c = 0
Also, the given quadratic eqn. be, (p + q) x² – 2px + p – q = 0 …(2)
Here a + b + c = p + q – 2p + p – q = 0
∴ roots of eqn. (2) are also rational.

Question 10.
Show that the roots of (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are real, and that they cannot be equal unless a = b = c.
Solution:
Given quadratic eqn. be (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0
⇒ 3x² – 2 (a + b + c) x + be + ca + ab = 0 …(1)
On comparing with Ax² + Bx + C = 0
We have, A = 3; B = – 2 (a + b + c) and c = ab + bc + ca
Now roots of eqn. (1) are real if D ≥ 0
Here D = b² – 4AC = [- 2 (a + b + c)]² – 4 x 3 (ab + bc + ca)
= 4 [(a + b + c)² – 3 (ab + bc + ca)]
= 4 [a² + b² + c² – ab – bc – ca]
= 2 [2a² + 2b² + 2c² – 2ab – 2bc – 2ca]
= 2 [(a – b)² + (b – c)² + (c – a)²] > 0 [∵ (a – b)² > 0 ; (b – c)² > 0 and (c – a)² > 0]
Hence the roots of given eqn. (1) are real.
Now the roots of given eqn. are equal
if Discriminant = 0 if (a – b)² + (b – c)² + (c – a)² = 0
i.e. if a – b = 0 and b – c = 0 and c – a = 0 i.e. if a = b = c
Hence the roots of eqn. (1) cannot be equal unless a = b = c

Question 11.
Determine the values of m for which the equations 3x² + 4mx + 2 = 0 and 2x² + 3x – 2 = 0 may have a common root.
Solution:
Let α be the common root of both given quadratic equations
3x² + 4mx + 2 = 0 and
2x² + 3x – 2 = 0
i.e. 3α² + 4mα + 2 = 0
2α² + 3α – 2 = 0
By cross multiplication method, we have

Question 12.
Find the value of k, so that the equation 2x² + kx – 5 = 0 and x² – 3x – 4 = 0 may have one root common.
Solution:
Let a be the common root of both given equations
2x² + kx – 5 = 0 and x² – 3x – 4 = 0
∴ 2a² + kα – 5 = 0 …(1)
α² – 3α – 4 = 0 …(2)
By cross multiplication method, we have

Question 13.
If ax² + bx + c = 0 and bx² + cx + a = 0 have a common root, prove that a + b + c = 0 or a = b = c.
Solution:
Let a be the common root of both equations ax² + bx + c = 0 and bx² + cx + a = 0
i.e. aα² + bα + c = 0 …(1)
bα² + cα + a = 0 …(2)
By cross multiplication method, we have

Question 14.
The equations x² + x + a = 0 and x² + ax + 1 = 0 have a common real root
(a) for no value of a
(b) for exactly one value of a
(c) for exactly two values of a
(d) for exactly three values of a
Solution:
Let α be the common real root of equations
x² + x + a = 0 and x² + ax + 1 = 0
∴ α² + α + a = 0 …(1)
and α² + aα + 1 = 0 …(2)
eqn. (1) – eqn. (2) gives ;
(1 – a)α + a – 1 = 0 ⇒ α = 1
∴ from (1); 1 + 1 + a = 0 ⇒ a = – 2

Utilizing OP Malhotra Maths Class 11 Solutions Chapter 10 Quadratic Equations Ex 10(c) as a study aid can enhance exam preparation.

S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(c)

Question 1.
Without solving, find the nature of the roots of the following equations :
(i) 3x² – 7x + 5 = 0.
(ii) 4x² + 4x + 1 = 0.
(iii) 3x² + 7x + 2 = 0.
(iv) x² + px – q² = 0.
Solution:
(i) Given quadratic eqn. be 3x² – 7x + 5 = 0
On comparing with ax² + bx + c = 0 ; we have
a = 3; b = – 7; c = 5
Here discriminant D = b² – 4ac
= (- 7)² – 4 x 3 x 5
= 49 – 60 = – 11 < 0
Hence the roots of given quadratic eqn. are imaginary.

(ii) Given quadratic eqn. be 4x² + 4x + 1 = 0
On comparing with ax² + bx + c = 0, we have a = 4; b = 4; c = 1
Here discriminant D = b² – 4ac
= 16 – 4 x 4 x 1 = 0
∴ roots are real and equal.

(iii) Given quadratic eqn. be 3x² + 7x + 2 = 0
On comparing with ax² + bx + c = 0, we have a = 3 ; b = 7 ; c = 2
Here discriminant D = b² – 4ac
= 7² – 4 x 3 x 2
= 49 – 24
= 25 > 0
∴ roots are real, distinct and rational

(iv) Given quadratic eqn. be
x² + px – q² = 0 …(1)
On comparing eqn. (1) with
ax² + bx + c = 0,
we have a = 1 ; b = p ; c = – q²
Here discriminant D = b² – 4ac
= p² – 4 x 1 x (- q²)
= p² + 4q² > 0
Hence the roots are real and unequal if p and q both not equal to 0 and roots are real and equal if p = 0 and q = 0

Question 2.
If the equation
(1 + m²) x² + 2mcx + c² – a² = 0 has equal roots, show that c² = a² (1 + m²).
Solution:
Given quadratic eqn. be,
(1 + m²)x² + 2mcx + c² – a² = 0 …(1)
On comparing with Ax² + Bx + C = 0
we have, A = 1 + m²; B = 2mc
and C = c² – a²
given eqn. (1) has equal roots.
∴ b² – 4AC = 0
⇒ 4m²c² – 4 (1 + m²) (c² – a²) = 0
⇒ 4m²c² – 4 {c² – a² + m²c² – a²m²} = 0
⇒ – 4 (c² – a² – a²m²) = 0
⇒ c² = a²(1 + m²)

Question 3.
Find the value of m so that the roots of the equation (4 – m)x² + (2m + 4) x + (8m + 1) = 0 may be equal.
Solution:
Given quadratic equation be,
(4 – m)x² + (2m + 4)x + 8m + 1 = 0
On comparing with ax² + bx + c = 0 ; we have
a = 4 – m ; b = 2m + 4 and c = 8m + 1
Since it is given that roots of eqn. (1) are equal
∴ discriminant = 0 ⇒ b² – 4ac = 0
⇒ (2m + 4)² – 4 x (4 – m) (8m + 1) = 0
⇒ 4m² + 16m + 16 – 4 (32m + 4 – 8m² – m) = 0
⇒ 36m² – 108m = 0
⇒ 36m (m – 3) = 0
⇒ m = 0, 3

Question 4.
If the roots of ax² + x + b = 0be real and unequal, show that the roots of \(\frac{x^2+1}{x}=4 \sqrt{a b}\) are imaginary.
Solution:
Given roots of ax² + x + b = 0 be real and unequal.
∴ discriminant > 0
⇒ 1 – 4ab > 0 …(1)
Also, given quadratic eqn. can be written as,
x² – 4\(\sqrt{ab}\) x + 1 = 0 …(2)
On comparing eqn. (2) with Ax² + Bx + C = 0
we have, A = 1 ; B = – 4\(\sqrt{ab}\) ; C = 1
Here discriminant = B² – 4AC
= (- 4\(\sqrt{ab}\))² – 4 x 1 x 1
= 16ab – 4 = 4 (4ab – 1)
= – 4(1 – 4ab) < 0 [using eqn. (1)]
Hence the roots of eqn. (2) are imaginary.

Question 5.
Find a so that the sum of the roots of the equation ax² + 2x + 3a = 0 may be equal to their product.
Solution:
Given quadratic eqn. be
ax² + 2x + 3a = 0 … (1)
On comparing eqn. (1) with
Ax² + Bx + C = 0
we have, A = a; B = 2; C = 3a
∴ Sum of roots = – \(\frac { B }{ A }\) = – \(\frac { 2 }{ a }\)
product of roots = \(\frac { C }{ A }\) = \(\frac { 3a }{ a }\) = 3
It is given that, sum of roots = product of roots
⇒ \(\frac { – 2 }{ a }\) = 3 ⇒ a = \(\frac { -2 }{ 3 }\)

Question 6.
If α, ß are the roots of the equation x² + x + 1 = 0, find the value of α³ + ß³.
Solution:
Since α, ß are the roots of eqn. x² + x + 1 = 0
∴ S = α + ß = \(\frac { – 1 }{ 1 }\) = – 1
and P = αß = \(\frac { 1 }{ 1 }\) = 1
∴ α³ + ß³ = (α + ß)³ – 3αß (α + ß)
= (- 1)³ – 3 (1)(- 1) = – 1 + 3 = 2

Question 7.
If α, ß are the roots of the equation x² + px + q = 0, find the value of
(a) α³ß + αß³
(b) α⁴ + α²ß² + ß⁴.
Solution:
Given α and ß are the roots of equation x² + px + q = 0
∴ α + ß = – p ; ap = q
(a) α³ß + αß³ = αß (α² + ß²)
= αß [(α + ß)² – 2αß]
= q (- P)² – 2q] = q(p² – 2q)

(b) b⁴ + α²ß² + ß⁴ = (α² + ß²)² – α²ß²
= [(α + ß)² – 2αß]² – (αß )²
= [(- P)² – 2q]² – q²
= (p² – 2q)² – q²
= p⁴ – 4p²q + 3q²

Question 8.
If the roots of the equation x² + px + 7 = 0 are denoted by α and ß, and α² + ß² = 22, find the possible values of p.
Solution:
Since it is given that a and p are the roots of equation x² + px + 7 = 0
∴ α + ß = – p ; αß = 7
Also given α² + ß² = 22
⇒ (α + ß)² – 2αß = 22
⇒ (- p)² – 2 x 7 = 22
⇒ p² = 22+ 14 = 36
⇒ p = ± 6

Question 9.
If α, ß are the roots of the equation 3x² – 6x + 4 = 0, find the value of
\(\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3 \alpha \beta\)
Solution:
Given a and p are the roots of equation 3x² – 6x + 4 = 0

Question 10.
If α, ß are the roots of ax² + bx + c = 0, find the values of
(i) \(\left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2\)
(ii) \(\frac{\alpha^3}{\beta}+\frac{\beta^3}{\alpha}\)
Solution:
Given a and p are the roots of eqn. ax² + bx + c = 0

Question 11.
If the sum of the roots of the equation x² – px + q = 0 be m times their difference, prove that p² (m² – 1) = 4m²p.
Solution:
Let α, ß are the roots of equation
x² – px + q = 0 …(1)
∴ α + ß = p ; αß = q
It is given that sum of roots of eqn. (1)
= m x difference of their roots
⇒ α + ß = m|α – ß|
On squaring both sides ; we have
(α + ß)² = m² (α – ß)²
⇒ (α + ß)² = m² [(α + ß)² – 4αß]
⇒ p² = m² [p² – 4q]
⇒ p² (m² – 1) = 4m²q

Question 12.
If one root of the equation x² + ax + 8 = 0 is 4 while the equation x² + ax + b = 0 has equal roots, find b.
Solution:
Since 4 be the root of equation
x² + ax + 8 = 0 ⇒ 4² + 4a + 8 = 0
⇒ 4a + 24 = 0 ⇒ a = – 6
Since the equation x² + ax + b = 0 has equal roots
∴ Discriminant = 0
⇒ α² – 4b = 0 ⇒ (- 6)² – 4b = 0
⇒ 4b = 36 ⇒ b = 9

Question 13.
Find the value of a for which one root of the quadratic equation (a² – 5a + 3) x² + (3a – 1) x + 2 = 0 is twice as large as the other.
Solution:
Let α and 2α are the two roots of given equation (a² – 5a + 3)x² + (3a – 1)x + 2 = 0

Question 14.
If α, ß are the roots of the equation ax² – bx + b = 0, prove that
\(\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}-\sqrt{\frac{b}{a}}\) = 0.
Solution:
Given α and ß are the roots of the equation ax² – bx + b = 0
∴ α + ß = \(\frac { b }{ a }\) and αß = \(\frac { b }{ a }\)
Now, \(\frac { 1 }{ 2 }\)
= \(\frac { 1 }{ 2 }\)

Question 15.
If α and ß are the roots of the equation x² + x – 7 = 0, form the equation whose roots are α² and ß².
Solution:
Given a and P are the roots of the eqn. x² + x – 7 = 0
∴ α + ß = – 1; αß = – 7
Sum of roots of required eqn. = α² + ß² = S
= (α + ß)² – 2αß = (- 1)² – 2 (- 7)
= 15
and product of roots of required equation
= α² p² = (αß)² = (- 7)² = 49
Thus the required quadratic eqn. by given by x² – Sx + P = 0 ⇒ x² – 15x + 49 = 0

Question 16.
If α and ß are the roots of the equation 2x² + 3x + 2 = 0, find the equation whose roots are α + 1 and ß + 1.
Solution:
Given α and ß are the roots of the eqn.
2x² + 3x + 2 = 0
∴ α + ß = – \(\frac { 3 }{ 2 }\) ; αß = \(\frac { 2 }{ 2 }\) = 1
∴ S = sum of roots of quadratic eqn.
= α + 1 + ß + 1
= – \(\frac { 3 }{ 2 }\) + 2 = \(\frac { 1 }{ 2 }\)
P = product of roots of quadratic eqn.
= (α + 1)(α + 1)
= αß + (α + ß) + 1
= 1 – \(\frac { 3 }{ 2 }\) + 1 = \(\frac { 1 }{ 2 }\)
Hence the required quadratic eqn. be given by x² – Sx + P = 0 ⇒ x² – \(\frac { 1 }{ 2 }\)x + \(\frac { 1 }{ 2 }\) = 0
⇒ 2x² – x + 1 = 0

Question 17.
Find the equation whose roots are \(\frac { α }{ ß }\) and \(\frac { ß }{ α }\), where α and ß are the roots of the equation x² + 2x + 3 = 0.
Solution:
Given a and p are the roots of equation x² + 2x + 3 = 0
∴ α + ß = – 2 ; αß = 3
S = sum of roots of required eqn.
= \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}\)
= \(\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}=\frac{(-2)^2-2 \times 3}{3}=-\frac{2}{3}\)
P = product of roots of required eqn.
= \(\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha}\) = 1
Hence the required quadratic eqn. having roots \(\frac { α }{ ß }\) and \(\frac { ß }{ α }\) is given by
x² – Sx + P = 0 ⇒ x² – (- \(\frac { 2 }{ 3 }\)) x + 1 = 0
⇒ 3x² + 2x + 3 = 0

Question 18.
If a and p are the roots of the equation 2x² – 3x + 1 = 0, form the equation whose roots are \(\frac{\alpha}{2 \beta+3} \text { and } \frac{\beta}{2 \alpha+3}\)
Solution:
Given a and p are the roots of the eqn. 2x² – 3x + 1 = 0

Question 19.
If a ≠ b and a² = 5a – 3, b² = 5b – 3, then form that equation whose roots are \(\frac { a }{ b }\) and \(\frac { b }{ a }\).
Solution:
Given a ≠ b and a² = 5a – 3 and b² = 5b – 3
∴ a and b are roots of the quadratic equation x² – 5x + 3 = 0
∴ a + b = 5 ; ab = 3
S = sum of roots of required eqn.
= \(\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{a b}\)
= \(\frac{(a+b)^2-2 a b}{a b}=\frac{5^2-2 \times 3}{3}=\frac{19}{3}\)
and P = product of roots of required eqn.
= \(\frac{a}{b} \times \frac{b}{a}\) = 1
Hence the required quadratic eqn. having roots \(\frac { a }{ b }\) and \(\frac { b }{ a }\) be given by
x² – Sx + P = 0
⇒ x² – x + 1 = 0 ⇒ 3x² – 19x + 3 = 0

Question 20.
Given that a and P are the roots of the equation x² = x + 7.
(i) Prove that
(a) \(\frac { 1 }{ α }\) = \(\frac{\alpha-1}{7}\) and (b) α³ = 8α + 7.
(ii) Find the numerical value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\).
Solution:
(i) Since α be the root of eqn.
x² = x + 7 …(1)
∴ it must satisfies eqn. (1)
∴ α² = α + 7 ⇒ α² – α = 7
⇒ α (α – 1) = 7 ⇒ \(\frac { 1 }{ α }\) = \(\frac{\alpha-1}{7}\)

(b) Further α³ = α² . α = (α + 7) α
= α² + 7α = α + 7 + 7α
⇒ α³ = 8α + 7

(ii) Since α, ß are the roots of eqn. (1)
∴ α + ß = 1 ; αß = – 7
∴ \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}\)
= \(\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}\)
= \(\frac{1^2-2 \times(-7)}{-7}=\frac{-15}{7}\)

Question 21.
Given that α and ß are the roots of the equation x² – x + 7 = 0, find
(i) the numerical value of \(\frac{\alpha}{\beta+3}+\frac{\beta}{\alpha+3}\) ;
(ii) an equation whose roots are \(\frac { α }{ ß+3 }\) and \(\frac { ß }{ α +3 }\).
Solution:
Given α and ß are the roots of equation x² – x + 7 = 0
∴ α + ß = + 1 ; αß = 7
(i) S = sum of roots of required eqn.

(ii) Hence the required quadratic eqn. having roots \(\frac { α }{ ß+3 }\) and \(\frac { ß }{ α + 3 }\)
x² – Sx + P = 0
⇒ x² + \(\frac { 10 }{ 19 }\)x + \(\frac { 7 }{ 19 }\) = 0
⇒ 19x² + 10x + 7 = 0

Question 22.
Given that α and ß are the roots of the equation 2x² – 3x + 4 = 0, find an equation whose roots are α + \(\frac { 1 }{ α }\) and ß + \(\frac { 1 }{ ß }\).
Solution:
Given α and ß are the roots of the eqn.

Thus the required quadratic eqn. having roots α + \(\frac { 1 }{ α }\) and ß + \(\frac { 1 }{ ß }\) be given by
x² – Sx + p = 0 ⇒ x² – \(\frac { 9 }{ 4 }\)x + \(\frac { 13 }{ 8 }\) = 0
⇒ 8x² – 18x + 13 = 0

Question 23.
The roots of the quadratic equation x² + px + 8 = 0 are α and ß. Obtain the values of p, if
(i) α = ß²
(ii) α – ß = 2.
Solution:
Given α and ß are the roots of the eqn.
x² + px + 8 = 0
∴ α + ß = – p ; αß = 8 … (1)
(i) When α = ß² ∴ from (1); we have
ß³ = 8 ⇒ ß = 2 ∴ α = 2² = 4
∴ from (1) ; – p = 4 + 2 ⇒ p = – 6

(ii) Given α – ß = 2 ;
On squaring both sides ; (α – ß)² = 4
⇒ (α + ß)² – 4αß = 4
⇒ (- p)² – 4 x 8 = 4
⇒ p² = 36 ⇒ p = ± 6

Question 24.
If the roots of x² – bx + c = 0 be two consecutive integers, then find the value of b² – 4c.
Solution:
Let the roots of eqn.
x² – 6x + c = 0 are α, α + 1
∴ α + α + 1 = b ⇒ 2α + 1 = b …(1)
and α (α + 1) = c …(2)
Eliminating a from eqn. (1) and eqn. (2)
From (1); α = \(\frac { b – 1 }{ 2 }\)
∴ from (2); \(\left(\frac{b-1}{2}\right)\left(\frac{b-1}{2}+1\right)\) = c
⇒ \(\left(\frac{b-1}{2}\right)\left(\frac{b+1}{2}\right)\) = c
⇒ b² – 1 = 4c
⇒ b² – 4c = 1

Question 25.
The roots of the equation
px² – 2(p + 2)x + 3p = 0 are α and ß. If α – ß = 2, calculate the value of α, ß and p.
Solution:
Given a and p are the roots of the equation
px² – 2(p + 2)x + 3p = 0 = 0
∴ α + ß = \(\frac{2(p+2)}{p}\) ; αß = \(\frac { 3p }{ p }\) = 3
Also given α – ß = 2 ; on squaring both sides; we have

On solving eqn. (2) and (3); we have α = – 1 and ß = – 3

Question 26.
The roots of the equation ax² + bx + c = 0 are α and ß. Form the quadratic equation whose roots are α + \(\frac { 1 }{ ß }\) and ß + \(\frac { 1 }{ α }\).
Solution:
Given α and ß are the roots of the eqn. ax² + bx + c = 0

Thus the required quadratic eqn. having roots α + \(\frac { 1 }{ ß }\) and ß + \(\frac { 1 }{ α }\) is given by
x² – Sx + P = 0
⇒ x² + \(\frac{b(c+a) x}{a c}+\frac{(a+c)^2}{a c}\) = 0
⇒ acx² + b (c + a) x + (a + c)² = 0

Question 27.
Two candidates attempt to solve a quadratic equation of the form x² + px + q = 0. One starts with a wrong value of p and finds the roots to be 2 and 6. The other starts with a wrong value of q and finds the roots to be 2 and – 9. Find the correct roots and the equation.
Solution:
One candidate find the roots of given eqn.
x² + px + q = 0 …(1)
are 2 and 6 ∴ it forms the quadratic eqn. as
(x – 2) (x – 6) = 0
⇒ x² – 8x + 12 = 0 …(2)
∴ Correct value of q = 12 [on comparing eqn. (1) and eqn. (2)]
[since value of p is given to be wrong]
Other candidate find the roots of eqn. (1) are 2 and – 9 ∴ he forms the quadratic eqn. as : (x – 2) (x + 9) = 0
⇒ x² + 7x – 18 = 0 …(3)
So on comparing eqn. (1) and eqn. (3) ;
we have Correct value of p = 7
Hence the correct values of p and q are 7 and 12
∴ required eqn. becomes; x² + 7x + 12 = 0
⇒ (x + 3) (x + 4) = 0 ⇒ x = – 3, – 4
Hence the correct roots of given eqn. are – 3 and – 4.

Question 28.
Given that a and P are the roots of the equation x² = 7x + 4.
(i) Show that α³ = 53α + 28
(ii) find the value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\).
Solution:
Since α and ß are the roots of eqn. x² = 7x + 4
i. e. x² – 7x – 4 = 0 …(1)
∴ α + ß = 7 and αß = – 4
Since a be the root of eqn. (1) ∴ it satisfies eqn. (1)
∴ α² = 7α + 4
⇒ α³ = 7α² + 4α = 7 (7α + 4) + 4α = 53α + 28

(ii) \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}\)
= \(\frac{7^2-2 \times(-4)}{-4}=\frac{57}{-4}\)

Question 29.
The ratio of the roots of the equation x² + αx + α + 2 = 0 is 2. Find the values of the parameter a.
Solution:
Since the ratio of the roots of the equation x² + αx + α + 2 = 0 is 2.
Let 2ß and ß are the roots of given equation.
∴ 2ß + ß = – a ⇒ 3ß = – α …(1)
and (2ß) ß = α + 2 ⇒ 2ß² = α + 2 …(2)
From (1) and (2); we have
\(2\left(-\frac{\alpha}{3}\right)^2=\alpha+2 \Rightarrow 2 \times \frac{\alpha^2}{9}=\alpha+2\)
⇒ 2α² – 9α – 18 = 0
∴ α = \(\frac{9 \pm \sqrt{81+4 \times 2 \times 18}}{2 \times 2}=\frac{9 \pm 15}{4}\)
∴ α = 6, – \(\frac { 3 }{ 2 }\)

Question 30.
If (1 – p) is a root of the quadratic equation x² + px + (1 – p) = 0, then its roots are
(a) 0, – 1
(b) – 1, 1
(c) 0, 1
(d) – 1, 2
Solution:
Since (1 – p) be the root of quadratic equation
x² + px + (1 – p) = 0 …(1)
⇒ (1 – p)² + p(1 – p) + (1 – p) = 0
⇒ (1 – P) [1 – p + p + 1] = 0
⇒ 2(1 – p) = 0
⇒ p = 1
∴ eqn. (1) becomes ; x² + x = 0
⇒ x (x + 1) = 0 ⇒ x = 0, – 1
Hence the required roots of given equation are 0 and – 1.

The availability of step-by-step OP Malhotra Maths Class 11 Solutions Chapter 10 Quadratic Equations Ex 10(b) can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(b)

Solve the following equations :

Question 1.
x⁴ – 5x² + 6 = 0.
Solution:
Given eqn. be, x⁴ – 5x² + 6 = 0 … (1)
put x² = t in eqn. (1) ; we have
t² – 5t + 6 = 0
⇒ (t – 2) (t – 3) = 0
⇒ t – 2 = 0 or t – 3 = 0
⇒ x² = 2 or x² = 3
⇒ x = ± \(\sqrt{2}\) or x = ± \(\sqrt{3}\)
∴ x = ± \(\sqrt{2}\), ± \(\sqrt{3}\)

Question 2.
x⁵ + 242 = \(\frac{243}{x^5}\).
Solution:
Given eqn. be,
x⁵ + 242 = \(\frac{243}{x^5}\) … (1)
putting x⁵ = y in eqn. (1) ; we have
y + 242 = \(\frac { 243 }{ y }\) ⇒ y² + 242y – 243 = 0
⇒ y² – y + 243y – 243 = 0
⇒ y(y – 1) + 243 (y – 1) = 0
⇒ (y – 1) (y + 243) = 0
⇒ y = 1 = 0 or y + 243 = 0
⇒ y = 1 or y = – 243
⇒ x⁵ = 1 or x⁵ = (- 3)⁵
⇒ x = 1 or x = – 3
Thus, s = – 1, – 3

Question 3.
10x^-2 – 9 – x^-4 = 0.
Solution:
Given eqn. be, 10x^-2 – 9 – x^-4 = 0 … (1)
putting x^-2 = t in eqn. (1); we have
10t – 9 – t² = 0
⇒ t² – 10t + 9 = 0
⇒ t² – t – 9t + 9 = 0
⇒ t (t – 1) – 9 (t – 1) = 0
⇒ (t – 9)(t – 1) = 0
either t = 9 = 0 or t – 1 = 0
⇒ x^-2 = 9 or x^-2 – 1 = 0
⇒ \(\frac{1}{x^2}\) = 9 or \(\frac{1}{x^2}\) = 1
⇒ x = ± \(\frac { 1 }{ 3 }\) or x = ± 1
Thus, x = ± \(\frac { 1 }{ 3 }\), ± 1

Question 4.
3^2x – 10 x 3^x + 9 = 0.
Solution:
Given eqn. be,
3^2x – 10 x 3^x + 9 = 0 …(1)
putting 3^x = t in eqn. (1) ; we have
t² – 10t + 9 = 0
⇒ t² – 9t – t + 9 = 0
⇒ t(t – 9) – 1 (t – 9) = 0
⇒ (t – 9) (t – 1) = 0
either t – 9 = 0 or t – 1 = 0
⇒ t = 9 or t = 1
⇒ 3^x = 3^x or 3^x = 3°
⇒ x = 2 or x = 0
Thus, x = 0, 2

Question 5.
2^2x-1 – 9 x 2^x-2 + 1 = 0.
Solution:
Given equation be,
2^2x-1 – 9 x 2^x-2 + 1 = 0
⇒ (2^x)² x 2^-1 – 9 x 2^x x 2^-2 + 1 = 0 …(1)
putting 2^x = t in eqn. (1); we have
t² x \(\frac { 1 }{ 2 }\) – 9 x t x \(\frac { 1 }{ 4 }\) + 1 = 0
⇒ 2t² – 9t + 4 = 0
⇒ 2t² – 8t – t + 4 = 0
⇒ 2t (t – 4) – 1 (t – 4) = 0
⇒ (2t – 1) (t – 4) = 0
either 2t – 1 = 0 or t – 4 = 0
⇒ t = \(\frac { 1 }{ 2 }\) or t = 4
⇒ 2^x = \(\frac { 1 }{ 2 }\) = 2^-1 or 2^x = 2²
⇒ x = – 1 or x = 2
∴ x = – 1 or 2

Question 6.
3^2x+1 + 3² = 3^x+3 + 3^x.
Solution:
Given eqn. be,
3^2x+1 + 3² = 3^x+3 + 3^x
⇒ (3^x)² x 3 + 9 = 3^x (3³ + 1) …(1)
putting 3^x = t in eqn. (1) ; we have
3t² + 9 = 28t ⇒ 3t² – 28t + 9 = 0
⇒ 3t² – 27t – t + 9 = 0
⇒ 3t(t – 9) – 1 (t – 9) = 0
⇒ (t – 9) (3t – 1) = 0
either t – 9 = 0 or 3t – 1 = 0
⇒ t = 9 or t = \(\frac { 1 }{ 3 }\)
⇒ 3^x = 3² or 3^x = 3^-1
⇒ x = 2 or x = – 1
Thus, x = 2, – 1

Question 7.
\(\sqrt{x^2-3 x}=4 x^2-12 x-3\)
Solution:
Given equation be,
\(\sqrt{x^2-3 x}=4 x^2-12 x-3\) – 3 …(1)
putting x² – 3x = t in eqn. (1); we have
\(\sqrt{t}\) = 4t – 3 ;
on squaring both sides ; we have
⇒ t = (4t – 3)² ⇒ t = 16t² – 24t + 9 ⇒ 16t² – 25t + 9 = 0

Question 8.
\(\sqrt{\frac{x^2+2}{x^2-2}}+6 \sqrt{\frac{x^2-2}{x^2+2}}\) = 5.
Solution:
Given equation be
\(\sqrt{\frac{x^2+2}{x^2-2}}+6 \sqrt{\frac{x^2-2}{x^2+2}}\) = 5
putting \(\sqrt{\frac{x^2+2}{x^2-2}}\) = y in eqn. (1) ; we have
y + \(\frac { 6 }{ y }\) = 5
⇒ y² – 5y + 6 = 0
⇒ (y – 2)(y – 3) = 0
⇒ y – 2 = 0 or y – 3 = 0
⇒ y = 2 or y = 3
Case-I: When y = 2 ⇒ \(\sqrt{\frac{x^2+2}{x^2-2}}\) = 2
On squaring both sides ; we have
\(\frac{x^2+2}{x^2-2}\) = 4
⇒ x² + 2 = 4x² – 8
⇒ 3x² = 10
⇒ x² = \(\frac { 10 }{ 3 }\)
⇒ x = ± \(\sqrt{\frac{10}{3}}\)

Case-II: When y = 3 ⇒ \(\sqrt{\frac{x^2+2}{x^2-2}}\) = 3
On squaring both sides ; we have
\(\frac{x^2+2}{x^2-2}\) = 9 ⇒ x² + 2 = 9x² – 18
⇒ 8x² = 20
⇒ x² = \(\frac { 5 }{ 2 }\)
⇒ x = ± \(\sqrt{\frac{5}{2}}\)
Hence x = ± \(\sqrt{\frac{10}{3}}, \pm \sqrt{\frac{5}{2}}\)

Question 9.
\(\sqrt{\frac{2 x^2+1}{x^2-1}}+6 \sqrt{\frac{x^2-1}{2 x^2+1}}\) = 5.
Solution:
Given equation be
\(\sqrt{\frac{2 x^2+1}{x^2-1}}+6 \sqrt{\frac{x^2-1}{2 x^2+1}}\) = 5 … (1)
putting \(\sqrt{\frac{2 x^2+1}{x^2-1}}\) = t in eqn. (1); we have
t + \(\frac { 6 }{ t }\) = 5
⇒ t² – 5t + 6 = 0
⇒ (t – 2)(t – 3) = 0
either t – 2 = 0 or t – 3 = 0
⇒ t = 2 or t = 3
Case-I: When t = 2
⇒ \(\sqrt{\frac{2 x^2+1}{x^2-1}}\) = 2
On squaring both sides ; we have 2*2 + 1
\(\frac{2 x^2+1}{x^2-1}\) = 4
⇒ 2x² + 1 = 4x² – 4
⇒ 2x² = 5
⇒ x = \(\pm \sqrt{\frac{5}{2}}\)

Case-II: When t = 3

Question 10.
x (x – 1) (x + 2) (x- 3) + 8 = 0.
Solution:
Given equation be
x (x – 1) (x + 2) (x – 3) + 8 = 0
⇒ (x² – x) [x² – x – 6] + 8 = 0 … (1)
putting x² – x = t in eqn. (1) ; we have
t(t – 6) + 8 = 0 ⇒ t² – 6t + 8 = 0
⇒ t² – 4t – 2t + 8 = 0
⇒ t(t – 4) – 2(t – 4) = 0
⇒ (t – 4)(t – 2) = 0
either t – 4 = 0 or t – 2 = 0
⇒ t = 4 or t = 2
Case-I: When t – 4
⇒ x² – x – 4 = 0
⇒ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
⇒ x = \(\frac{-(-1) \pm \sqrt{(-1)^2-4 \times 1 \times(-4)}}{2}\)
⇒ x = \(\frac{1 \pm \sqrt{17}}{2}\)

Case-II: When t = 2 ⇒ x² – x – 2 = 0
⇒ x² – 2x + x – 2 = 0
⇒ x (x – 2) + 1 (x – 2) = 0
⇒ (x + 1) (x – 2) = 0
⇒ x = – 1, 2
Hence x = – 1, 2, \(\frac{1 \pm \sqrt{17}}{2}\)

Question 11.
(x – 7) (x – 3) (x + 1) (x + 5) = 1680.
Solution:
Given eqn. be
(x – 7) (x – 3) (x + 1)(x + 5) = 1680
⇒ {(x – 7) (x + 5)} {(x-3) (x+ 1)} = 1680
⇒ {x² – 2x – 35} {x² – 2x – 3} = 1680 … (1)
putting x² – 2x = t in eqn. (1); we get
(t – 35) (t – 3) = 1680
⇒ t² – 38t + 105 – 1680 = 0
⇒ t² – 38t – 1575 = 0
∴ t = \(\frac{38 \pm \sqrt{1444+6300}}{2}=\frac{38 \pm 88}{2}\)
⇒ t = 63, – 25

Case-I: When t = 63 ⇒ x² – 2x – 63 = 0

Question 12.
(2x – 7) (x² – 9) (2x + 5) = 91.
Solution:
Given equation be,
(2x – 7) (x² – 9) (2x + 5) = 91
⇒ {(2x – 7) (x + 3)} {(x – 3)(2x + 5)} = 91
⇒ (2x² – x – 21)(2x² – x – 15) = 91 …(1)
putting 2x² – x = t in eqn. (1) ; we have
(t – 21) (t – 15) = 91
⇒ t² – 36t + 224 = 0
⇒ t² – 8t – 28t + 224 = 0
⇒ t (t – 8) – 28 (t – 8) = 0
⇒ (t – 8) (t – 28) = 0
either t – 8 = 0 or t – 28 = 0
⇒ t = 8 or t = 28
Case-I: When t = 8 ⇒ 2x² – x – 8 = 0
⇒ x = \(\frac{-(-1) \pm \sqrt{(-1)^2-4 \times 2 \times(-8)}}{2 \times 2}\)
⇒ x = \(\frac{1 \pm \sqrt{65}}{4}\)

Case-II: When t = 28
⇒ 2x² – x – 28 = 0
⇒ 2x² – 8x + 7x – 28 = 0
⇒ 2x (x – 4) + 7 (x – 4) = 0
⇒ (x – 4) (2x + 7) = 0
either x – 4 = 0 or 2x + 7 = 0
⇒ x = 4 or x = – \(\frac { 7 }{ 2 }\)
Hence, x = + 4, – \(\frac { 7 }{ 2 }\), \(\frac{1 \pm \sqrt{65}}{4}\)

Question 13.
By substituting y = 2^x, or otherwise, solve the equation
2^2x + 2^x+2 – 4 x 2³ = 0.
Solution:
Given eqn. be,
2^2x + 2^x+2 – 4 x 2³ = 0 …(1)
putting 2^2x = y in eqn. (1) ; we have
y² + y . 2² – 32 = 0
⇒ y² + 4y – 32 = 0
⇒ y² + 8y – 4y – 32 = 0
⇒ y(y + 8) – 4(y + 8) = 0
⇒ (y + 8)(y – 4) = 0
either y + 8 = 0 or y – 4 = 0
⇒ y = – 8 or y = 4
⇒ 2^x = – 8
it has no real solution (∵ 2^x > 0)
or 2^x = 4 = 2² ∴ x = 2
Thus x = 2

Question 14.
2^x² : 2^x = 8 : 1.
Solution:

Question 15.
2^2x+3 + 2^x+3 = 1 + 2^x
Solution:
Given eqn. be 2^2x+3 + 2^x+3 = 1 + 2^x
⇒ (2^x)² x 2³ + 2^x x 2³ = 1 + 2^x …(1)
putting 2^x = t in eqn. (1) ;
we have 8t² + 8t = 1 + t
⇒ 8t² + 7t – 1 = 0
⇒ 8t² + 8t – t – 1 = 0
⇒ 8t (t + 1) – 1 (t + 1) = 0
⇒ (t + 1) (8t – 1) = 0
either t + 1 = 0 or 8t – 1 = 0
⇒ t = – 1 or t = \(\frac { 1 }{ 8 }\)
⇒ 2^x = – 1, which is not possible as 2^x > 0
when t = \(\frac { 1 }{ 8 }\) ⇒ 2^x = 2^-3 ⇒ x = – 3

Question 16.
\(4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2 x-1}\)
Solution:
Given equation be

Students often turn to OP Malhotra Maths Class 11 Solutions Chapter 10 Quadratic Equations Ex 10(a) to clarify doubts and improve problem-solving skills.

S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(a)

Find the roots of the following equations.

Question 1.
2x² + x – 3 = 0
Solution:
Given 2x² + x – 3 = 0
On comparing with ax² + bx + c = 0, we have
a = 2 ; 4 = 1 and c = – 3
Then by quadratic formula, we have
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{-1 \pm \sqrt{1-4 \times 2 \times(-3)}}{2 \times 2}=\frac{-1 \pm 5}{4}\)
⇒ x = \(\frac { – 6 }{ 4 }\), 1
⇒ x = \(\frac { – 3 }{ 2 }\), 1

Question 2.
6x² + 7x – 20 = 0
Solution:
Given 6x² + 7x – 20 = 0
On comparing with ax² + bx + c = 0, we have
a = 6; b = 7 and c = – 20
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{-7 \pm \sqrt{7^2-4 \times 6 \times(-20)}}{2 \times 6}\)
= \(\frac{-7 \pm \sqrt{49+480}}{12}\)
= \(\frac{-7 \pm 23}{12}=\frac{-30}{12}, \frac{16}{12}\)
⇒ x = \(\frac { – 15 }{ 6 }\), \(\frac { 4 }{ 3 }\) i.e. x = – \(\frac { 5 }{ 2 }\), \(\frac { 4 }{ 3 }\)

Question 3.
36x² + 23 = 60x
Solution:
Given 36x² – 60x + 23 = 0
On comparing with ax² + bx + c = 0 ; we have
a = 36 ; b = – 60 ; c = 23
Then using quadratic formula,

Question 4.
x² – 2x + 5 = 0
Solution:
Given x² – 2x + 5 = 0
On comparing with ax² + bx + c = 0, we have
a = 1 ; b = – 2 and c = 5
Then by quadratic formula, we have
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{2 \pm \sqrt{4-4 \times 1 \times 5}}{2}=\frac{2 \pm \sqrt{-16}}{2}\)
= \(\frac{2 \pm 4 i}{2}\) = 1 ± 2i

Question 5.
3x² – 17x + 25 = 0
Solution:
Given quadratic equation be,
3x² – 17x + 25 = 0
On comparing with ax² + bx + c = 0 ; we have
a = 3 ; b = – 17 ; c = 25
Then by quadratic formula, we have
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{-(-17) \pm \sqrt{(-17)^2-4 \times 3 \times 25}}{2 \times 3}\)
= \(\frac{17 \pm \sqrt{289-300}}{6}\)
⇒ x = \(\frac{17 \pm \sqrt{-11}}{6}=\frac{17 \pm \sqrt{11} i}{6}\)

Question 6.
x² + 3x – 3 = 0, giving your answer correct to two decimal places.
Solution:
Given quadratic eqn. be, x² + 3x – 3 = 0
On comparing with ax² + bx + c = 0 ; we have
a = 1 ; b = 3 ; c = – 3
Then by using quadratic formula,
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{-3 \pm \sqrt{3^2-4 \times 1 \times(-3)}}{2 \times 1}\)
= \(\frac{-3 \pm \sqrt{21}}{2}=\frac{-3 \pm 4.5826}{2}\)
= 0.7913 – 3.7913
⇒ x = 0.79 – 3.79 (correct to two decimal places)

Question 7.
5x² – x + 4 = 0
Solution:
Given quadratic eqn. be 5x² – x + 4 = 0
On comparing with ax² + bx + c = 0 ; we have
a = 5; b = – 1; c = 4
Then by quadratic formula, we have
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{-(-1) \pm \sqrt{(-1)^2-4 \times 5 \times 4}}{2 \times 5}\)
⇒ x = \(\frac{1 \pm \sqrt{-79}}{10}=\frac{1 \pm \sqrt{79} i}{10}\)

Question 8.
\(\sqrt{3}\)x² – \(\sqrt{2}\)x + 3\(\sqrt{3}\) = 0
Solution:
Given quadratic eqn. be,
\(\sqrt{3}\)x² – \(\sqrt{2}\)x + 3\(\sqrt{3}\) = 0
On comparing with ax² + bx + c = 0, we have
a = \(\sqrt{3}\) ; b = – \(\sqrt{2}\) ; c = 3\(\sqrt{3}\)
Then by quadratic formula, we have
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{\sqrt{2} \pm \sqrt{(-\sqrt{2})^2-4 \times \sqrt{3} \times 3 \sqrt{3}}}{2 \times \sqrt{3}}\)
x = \(\frac{\sqrt{2} \pm \sqrt{2-36}}{2 \sqrt{3}}=\frac{\sqrt{2} \pm \sqrt{34} i}{2 \sqrt{3}}\)

Question 9.
\(\frac{x^2+8}{11}\) = 5x – x² – 5
Solution:
Given quadratic eqn. can be written as
x² + 8 = 11 (5x – x² – 5)
⇒ 12x² + 55x + 63 = 0
On comparing with ax² + bx + c = 0 ; we
have
a = 12; b = – 55 ; c = 63
By quadratic formula, we have
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{55 \pm \sqrt{(-55)^2-4 \times 12 \times 63}}{2 \times 12}\)
= \(\frac{55 \pm 1}{24}=\frac{56}{24}, \frac{54}{24} \text { i.e. } \frac{7}{3}, \frac{9}{4}\)

Question 10.
\(\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=8 \frac{1}{3}\)
Solution:
Given quadratic equation be,

Question 11.
The number of real solutions of the equation x² – 3|x| + 2 = o is
(a) 3
(b) 4
(c) 1
(d) 3
Solution:
Given eqn. be x² – 3|x| + 2 = o = 0
⇒ |x|² – 3|x| + 2 = 0 [∵ |x|² = x²]
⇒ (|x| – 1)(|x| – 2) = 0
either | x | – 1 = 0 or | x | – 2 = 0
⇒ | x | = 1 or |x| = 2
⇒ x = ± 1 or x = ± 2
∴ x = ± 1, ± 2

Accessing OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Chapter Test can be a valuable tool for students seeking extra practice.

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Chapter Test

Question 1.
Find the square root of 5 – 12i.
Solution:
Let \(\sqrt{5-12 i}\) = x – iy ; on squaring both sides; we have
5 – 12i = (x – iy)² = x² – y² – 2ixy
On equating real and imaginary parts on both sides ; we have
x² – y² = 5 …(1)
and 2xy = 12 …(2)
Now x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{5^2+12^2}\)
= \(\sqrt{25+144}\) = 13 … (1)
On adding eqn. (1) and (3) ; we have
2x² = 18 ⇒ x² = 9 ⇒ x = ± 3
eqn. (3) – eqn. (1) gives ; 2y² = 8 ⇒ y = ± 2
Since xy be of positive sign ∴ x and y are of same sign.
∴ When x = 3, y – 2 or when x = – 3, y = – 2
Thus, \(\sqrt{5-12 i}\) = (3 – 2i) or – (3 – 2i)
= ± (3 – 2i)

Question 2.
Find the locus of a complex number Z = x + iy, satisfying the relation |z + i| = |z + 2|.
Illustrate the locus of z in the Argand plane.
Solution:
Given |z + i| = |z + 2|; where z = x + iy
⇒ | x + iy + i| = | x + iy + 2 |
⇒ |x + i(y + 1)| = |x + 2 + iy|
⇒ \(\sqrt{x^2+(y+1)^2}=\sqrt{(x+2)^2+y^2}\)

On squaring both sides ; we have x
x² + (y + 1)² = (x + 2)² + y²
⇒ x² + y² + 2y + 1 = x² + y² + 4x + 4
⇒ 2y + 1 = 4x + 4
⇒ 4x – 2y + 3 = 0
Thus locus of z is a straight line intersecting coordinate axes at \(\left(-\frac{3}{4}, 0\right) \text { and }\left(0, \frac{3}{2}\right)\)

Question 3.
Express \(\frac{13 i}{2-3 i}\) in the form A + Bi.
Solution:
Let z = \(\frac{13 i}{2-3 i} \times \frac{2+3 i}{2+3 i}=\frac{13 i(2+3 i)}{2^2+3^2}=\frac{26 i-39}{13}\) = 2i – 3 = – 3 + 2i

Question 4.
If z = x + yi and \(\frac{|z-1-i|+4}{3|z-1-i|-2}\) = 1, show that x² +y² – 2x- 2y – 7 = 0.
Solution:
Given z = x + iy and \(\frac{|z-1-i|+4}{3|z-1-i|-2}\) = 1
⇒ |x + iy – 1 – i| + 4 = 3|x + iy – 1 – i | – 2
⇒ | (x – 1) + i(y – 1) | = 3 | (x – 1) + i(y – 1) | – 6
\(\sqrt{(x-1)^2+(y-1)^2}=3 \sqrt{(x-1)^2+(y-1)^2}-6\)
⇒ 2\(\sqrt{(x-1)^2+(y-1)^2}\) = 6
⇒ \(\sqrt{(x-1)^2+(y-1)^2}\) = 3
On squaring both sides ; we have
(x – 1)² + (y – 1)² = 9 ⇒ x² + y² – 2x – 2y – 7 = 0

Question 5.
If ω and ω² are cube roots of unity, prove that (2 – ω + 2ω²) (2 + 2ω – ω²) = 9.
Solution:
(2 – ω + 2ω²) (2 + 2ω – ω²) = [2(1 + ω²) – ω] [2 (1 + ω) – ω²] [∵ 1 + ω + ω² = 0]
= [2 (- ω) – ω] [- 2ω² – ω²]
= (- 3ω) (- 3ω²) = 9ω³ = 9 [∵ ω³ = 1]

Question 6.
If z₁, z₂ ∈ C (set of complex numbers), prove that | z₁ + z₂ | ≤ | z₁ | + | z₂ |.
Solution:

Question 7.
If z = x + yi, ω = \(\frac{2-i z}{2 z-i}\) and | ω | = 1, find the locus of z and illustrate it in the complex plane.
Solution:
Given z = x + iy and w = \(\frac{2-i z}{2 z-i}\)
and | ω | = 1
⇒ \(\frac{2-i z}{2 z-i}\) = 1 ⇒ \(\frac{2-i z}{2 z-i}\) = 1 [∵ \(\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\) ]
⇒ | 2 – i(x + iy) | = |2(x + iy) – i |
⇒ | (2 + y) – ix | = | 2x + i (2y – 1) |
⇒ \(\sqrt{(2+y)^2+x^2}=\sqrt{(2 x)^2+(2 y-1)^2}\)
On squaring both sides ; we have
(2 + y)² + x² = 4x² + (2y – 1)²
⇒ y² + 4y + 4 + x² = 4x² + 4y² – 4y + 1
⇒ 3x² + 3y² – 8y – 3 = 0
⇒ x² + y² – \(\frac { 8 }{ 3 }\)y – 1 = o
⇒ x² + y² – \(\frac { 8 }{ 3 }\)y + \(\frac { 16 }{ 9 }\) – 1 – \(\frac { 16 }{ 9 }\) = o
⇒ x² + (y – \(\frac { 4 }{ 3 }\))² = \(\frac { 25 }{ 9 }\) = (\(\frac { 5 }{ 3 }\))²
Thus locus of z represents a circle with centre (0, \(\frac { 4 }{ 3 }\)) and radius \(\frac { 5 }{ 3 }\).

Question 8.
Simplify : (1 – 3ω + ω²) (1 + ω – 3ω²).
Solution:
(1 – 3ω + ω²) (1 + ω – 3ω²) = (1 + ω² – 3ω) (1 + ω – 3ω²)
= (- ω – 3ω) (- ω² – 3ω²) [∵ 1 + ω + ω² = 0]
= (- 4ω) (- 4ω²)
= 16ω³
= 16 x 1 = 16

Question 9.
Sketch in the complex plane the set of points z satisfying \(\left|\frac{z-3}{z+1}\right|\) = 3.
Solution:
Given \(\left|\frac{z-3}{z+1}\right|\) = 3 ; where z = x + iy

On squaring both sides ; we have
(x – 3)² + y² = 9[(x + 1)² + y²]
⇒ 9[x² + 2x + 1 + y²] = x ² + y² – 6x + 9
⇒ 8x² + 8y² + 24x = 0
⇒ x² + y² + 3x = 0
which represents the set of points in the circle whose centre (\(\frac { -3 }{ 2 }\), 0) and radius \(\frac { 3 }{ 2 }\) units.

Question 10.
Given that
\(\frac{2 \sqrt{3} \cos 30^{\circ}-2 i \sin 30^{\circ}}{\sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right)}\) = A + Bi, find the values of A and B.
Solution:
Given \(\frac{2 \sqrt{3} \cos 30^{\circ}-2 i \sin 30^{\circ}}{\sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right)}\) = A + Bi

On comparing real and imaginary parts on both sides ; we have A = 1; B = – 2

Question 11.
Simplify :
(1 – ω) (1 – ω²) (1 – ω⁴) (1 – ω⁸).
Solution:
(1 – ω) (1 – ω²) (1 – ω⁴) (1 – ω⁸)
= (1 – ω) (1 – ω²) (1 – ω) (1 – ω²) [∵ ω³ = ω⁶ = 1]
= [(1 – ω)(1 – ω²)]²
= [1 – ω – ω² + ω³]²
= [1 – (ω + ω²) + 1]²
= [1 – (- 1) + 1]² = 9 [∵ 1 + ω + ω² = 0]

Question 12.
Find the locus of a complex number z = x + yi, satisfying the relation | 2z + 3i | ≥ | 2z + 5 |.
Illustrate the locus in the Argand plane.
Solution:
Given | 2z + 3i | ∈ | 2z + 5 |;
where z = x + iy
⇒ | 2 (x + iy) + 3i | ≥ | 2 (x + iy) + 5 |
⇒ | 2x + i (2y + 3) | ≥ (2x + 5) + 2iy |
\(\sqrt{4 x^2+(2 y+3)^2} \geq \sqrt{(2 x+5)^2+4 y^2}\)
On squaring both sides ; we have
4x² + 4y² + 12y + 9 ≥ 4x² + 20x + 4y² + 25
⇒ 12y – 20x ≥ 16
⇒ 3y – 5x ≥ 4
⇒ 3y ≥ 5x + 4
Hence locus be the set of points in the region not containing origin lying on the line intersecting coordinate axes at (-\(\frac { 4 }{ 5 }\), 0) and (0, \(\frac { 4 }{ 3 }\).

Question 13.
Find the real values of x and y satisfying the equality \(\frac{x-2+(y-3) i}{1+i}\) = 1 – 3i.
Solution:
Given \(\frac{(x-2)+(y-3) i}{1+i}\) = 1 – 3i
⇒ (x – 2) + (y – 3)i = (1 – 3i)(1 + i)
⇒ (x – 2) + (y – 3) i = 1 + i – 3i + 3 = 4 – 2i
On comparing real and imaginary parts on both sides; we have
x – 2 = 4 ⇒ x = 6
and y – 3 = – 2 ⇒ y = 1

Question 14.
If i = \(\sqrt{-1}\), prove the following :
(x + 1 + i) (x + 1 – i) (x – 1 – i) (x – 1 + i) = x⁴ + 4.
Solution:
L.H.S = (x + 1 + i) (x + 1 – i) (x – 1 – i) (x – 1 + i)
= [(x+ 1 )² – i²] [(x – 1 )² – i²]
= [x² + 2x + 2] [x² – 2x + 2]
= (x² + 2 + 2x) (x² + 2 – 2x)
= (x² + 2)² – (2x)²
= x⁴ + 4x² + 4 – 4x²
= x⁴ + 4 = R.H.S

Question 15.
If z = x + yi and | 2z + 1 | = | z – 2i |, show that 3 (x² + y²) + 4 (x – y) = 3.
Solution:
Given z = x + iy
Also, | 2z + 1 | = | z – 2i |
⇒ | 2 (x + iy) + 1 | = | x + iy – 2i |
⇒ | 2x + 1 + 2iy | = | x + i (y – 2) |
⇒ \(\sqrt{(2 x+1)^2+(2 y)^2}=\sqrt{x^2+(y-2)^2}\)
On squaring both sides ; we have
4x² + 4x + 1 + 4y² = x² + y² – 4y + 4
⇒ 3x² + 3y² + 4x + 4y – 3 = 0

Question 16.
Find the amplitude of the complex number \(\sin \frac{6 \pi}{5}+i\left(1-\cos \frac{6 \pi}{5}\right)\).
Solution:

Question 17.
Express \(\frac{1-2 i}{2+i}+\frac{3+i}{2-i}\) in the form a + bi.
Solution:
Let z = \(\frac{(1-2 i)}{2+i}+\frac{3+i}{2-i}=\frac{(1-2 i)(2-i)+(3+i)(2+i)}{(2+i)(2-i)}=\frac{2-5 i-2+6+5 i-1}{2^2-i^2}\)
= \(\frac{5}{2^2+1}=\frac{5}{5}\) = 1 = 1 + 0i

Question 18.
Find the value of x and y given that (x + yi) (2 – 3i) = 4 + i.
Solution:
Given (x + iy) (2 – 3i) = 4 + i
⇒ x + iy = \(\frac{4+i}{2-3 i} \times \frac{2+3 i}{2+3 i}=\frac{8+12 i+2 i-3}{2^2-(3 i)^2}=\frac{5+14 i}{4+9}\)
⇒ x + iy = \(\frac{5}{13}+\frac{14}{13} i\)
On equating real and imaginary parts on both sides ; we have
x = \(\frac { 5 }{ 13 }\) and y = \(\frac { 14 }{ 13 }\)

Question 19.
If the ratio \(\frac{z-i}{z-1}\) is purely imaginary, prove that the point z lies on the circle whose centre is the point \(\frac { 1 }{ 2 }\)(1 + i) and radius is \(\frac{1}{\sqrt{2}}\).
Solution:

Question 20.
If (- 2 + \(\sqrt{-3}\))(- 3 + 2\(\sqrt{-3}\)) = a + bi, find the real numbers a and b. With these values of a and b, also find the modulus of a + bi.
Solution:
(- 2 + \(\sqrt{3}\))(- 3 + 2\(\sqrt{-3}\)) = a + ib
⇒ (- 2 + \(\sqrt{3}\)i)(- 3 + 2\(\sqrt{3}\) i) = a + ib
⇒ (6 – 4\(\sqrt{3}\) i – 3\(\sqrt{3}\) i – 6) = a + ib
⇒ – 7\(\sqrt{3}\) i = a + ib
∴ a = 0; b = – 7\(\sqrt{3}\)
Thus |a + ib| = | – 7\(\sqrt{3}\)i| = 7\(\sqrt{3}\)

Question 21.
if 1, ω, ω² are the three cube roots of unity, then simplify : (3 + 5ω + 3ω²)² (1 + 2ω + ω²)
Solution:

Question 22.
Find the locus of a complex number z = x + iy, satisfying the relation | 3z – 4i | < | 3z + 2 |. Illustrate the locus in the Argand plane.
Solution:
Given z = x + iy

Also, | 3z – 4i | ≤ | 3z + 2 |
⇒ | 3 (x + iy) | ≤ | 3 (x + yi) + 2|
⇒ | 3x + iy (3y – 4) | ≤ | (3x + 2) + 3iy|
⇒ \(\sqrt{(3 x)^2+(3 y-4)^2} \leq \sqrt{(3 x+2)^2+(3 y)^2}\)
On squaring both sides ; we have
⇒ 9x² + 9y² – 24y + 16 ≤ 9x² + 12x + 9y² + 4
⇒ 12x + 24y – 12 ≥ 0
⇒ x + 2
Clearly the line x + 2y – 1 = 0 intersecting coordinate axes at (1, 0) and (0, \(\frac { 1 }{ 2 }\))
Hence locus of z consisting of set of points lying in region not containing (0, 0) satisfying x + 2y – 1 ≥ 0.

Question 23.
Find the modulus and argument of the complex number \(\frac{2+i}{4 i+(1+i)^2}\).
Solution:

Question 24.
If | z – 3 + i | = 4, then the locus of z is
(i) x² + y² – 6 = 0
(ii) x² + y² – 3x + y = 0
(iii) x² + y² – 6x – 2 = 0
(iv) x² + y² – 6x + 2y – 6 = 0
Solution:
Given | z – 3 + i | = 4 ; where z = x + iy
⇒ | x + iy – 3 + i | = 34
⇒ \(|(x-3)+i(y+1)|\) = 4
⇒ \(\sqrt{(x-3)^2+(y+1)^2}\) = 4
⇒ (x – 3)² + (y + 1)² = 16
⇒ x² + y² – 6x + 2y – 6 = 0

Question 25.
The locus of the point z is the Argand plane for which | z + 1 |² + | z – 1|² = 4 is a
(a) Straight line
(b) Pair of straight lines
(c) Parabola
(d) Circle
Solution:
Given | z + 1 |² + | z – 1 |² = 4, where z = x + iy ⇒
⇒ | x + 1 + iy |² + | x – 1 + iy |² = 4
⇒ (x + 1)² + y² + (x – 1)² + y² = 4
⇒ 2x² + 2y² = 2
⇒ x² + y² = 1
which clearly represents a circle with centre (0, 0) and radius unity.