# Class 12 RD Sharma Solutions – Chapter 18 Maxima and Minima – Exercise 18.2

### Question 1. f(x) = (x – 5)4

Solution:

Given function

f(x) = (x – 5)4

Now, differentiate the given function w.r.t. x

f ‘(x) = 4(x-5)3

Now, for local maxima and minima

Put f ‘(x) = 0

â‡’ 4(x – 5)3 = 0

â‡’ x – 5 = 0

â‡’ x = 5

So, at x = 5, f'(x) changes from negative to positive. Hence, x = 5 is the point of local minima

So, the minimum value is f(5) = (5 – 5)4 = 0

### Question 2. f(x) = x3– 3x

Solution:

Given function

f(x) = x3– 3x

Now, differentiate the given function w.r.t. x

f ‘(x) =  3x2– 3

Now, for local maxima and minima

Put f ‘(x) = 0

â‡’ 3x2– 3 = 0

â‡’ x = Â±1

Now, again differentiating f'(x) function w.r.t. x

f “(x) = 6x

Put x = 1 in f”(x)

f “(1)= 6 > 0

So, x = 1 is point of local minima

Put x = -1 in f”(x)

f “(-1)= -6 < 0

So, x = -1 is point of local maxima

So, the minimum value is f(1) = x3– 3x = 13 – 3 = -2

and the maximum value is f(-1) = x3– 3x = (-1)3 – 3(-1) = 2

### Question 3. f(x) = x3 (x – 1)2

Solution:

Given function

f(x) = x3(x – 1)2

Now, differentiate the given function w.r.t. x

f ‘(x) = 3x2 (x- 1)2 + 2x3(x- 1)

= (x – 1) (3x2(x – 1) + 2x3)

= (x – 1) (3x3 – 3x2 + 2x3)

= (x – 1) (5x3 – 3x2)

= x2(x – 1) (5x – 3)

Now, for all maxima and minima,

Put f ‘(x) = 0

= x2(x – 1) (5x- 3) = 0

x = 0, 1, 3/5

So, at x = 3/5, f'(x) changes from negative to positive. Hence, x = 3/5 is a point of minima

So, the minimum value is f(3/5) = (3/5)3(3/5 – 1)2 = 108/3125

At x = 1, f'(x) changes from positive to negative. Hence, x = 1 is point of maxima.

So, the maximum value is f(1) = (1)3(1 – 1)2 = 0

### Question 4. f (x) = (x – 1) (x + 2)2

Solution:

Given function

f(x) = (x – 1)(x + 2)2

Now, differentiate the given function w.r.t. x

f ‘(x) = (x + 2)2 + 2(x – 1)(x + 2)

= (x+ 2) (x+ 2 + 2x – 2)

=(x + 2) (3x)

Now, for all maxima and minima,

Put f ‘(x) = 0

â‡’ (x + 2) (3x) = 0

x = 0,-2

So, at x = -2, f(x) changes from positive to negative. Hence, x = -2 is a point of Maxima

So, the maximum value is f(-2) = (-2 – 1)(-2 + 2)2 = 0

At x = 0, f ‘(x) changes from negative to positive. Hence, x = 0 is point of Minima.

So, the minimum value is f(0) = (0 – 1)(0 + 2)2 = -4

### Question 5. f(x) = (x – 1)3 (x + 1)2

Solution:

Given function

f(x) = (x – 1)3 (x + 1)2

Now, differentiate the given function w.r.t. x

f ‘(x) = 3(x – 1)2 (x + 1)2 + 2(x – 1)3 (x + 1)

= (x – 1)2 (x + 1) {3(x + 1) + 2(x – 1)}

= (x – 1)2 (x + 1) (5x + 1)

Now, for local maxima and minima,

Put f ‘(x) = 0

â‡’ (x – 1)2 (x + 1) (5x + 1) = 0

â‡’ x = 1, -1, -1/5

So, at x = -1, f ‘(x) changes from positive to negative. Hence, x = -1 is point of maxima

So, the maximum value is f(-1) = (-1 – 1)3 (-1 + 1)2 = 0

At x = -1/5, f ‘(x) changes from negative to positive so x= -1/5 is point of minima

So, the minimum value is f(-1/5) = (-1/5 – 1)3 (-1/5 + 1)2 = -3456/3125

### Question 6. f(x) = x3 – 6x2 + 9x +15

Solution:

Given function

f(x) = x3 – 6x2 + 9x + 15

Now, differentiate the given function w.r.t. x

f ‘(x) = 3x2 – 12x + 9

= 3 (x2 – 4x + 3)

= 3 (x – 3) (x – 1)

Now, for local maxima and minima,

Put f ‘(x) = 0

â‡’ 3 (x – 3) (x – 1) – 0

â‡’ x = 3, 1

At x = 1, f'(x) changes from positive to negative. Hence, x = 1 is point of local maxima

So, the maximum value is f(1) = (1)3 – 6(1)2 + 9(1) + 15 = 19

At x = 3, f'(x) changes from negative to positive. Hence, x = 3 is point of local minima

So, the minimum value is f(x) = (3)3 – 6(3)2 + 9(3) + 15 = 15

### Question 7. f(x) = sin2x, 0 < x < Ï€

Solution:

Given function

f(x) = sin2x, 0 < x, Ï€

Now, differentiate the given function w.r.t. x

f'(x) = 2 cos 2x

Now, for local maxima and minima,

Put f'(x) = 0

â‡’ 2x =

â‡’ x =

At x = Ï€â€‹/4, f'(x) changes from positive to negative. Hence, x = Ï€â€‹/4, Is point of local maxima

So, the maximum value is f(Ï€â€‹/4) = sin2(Ï€â€‹/4) = 1

At x = 3Ï€â€‹/4, f'(x) changes from negative to positive. Hence, x = 3Ï€â€‹/4 is point of local minima,

So, the minimum value is f(3Ï€â€‹/4) = sin2(3Ï€â€‹/4) = -1

### Question 8. f(x) = sin x – cos x, 0 < x < 2Ï€

Solution:

Given function

f(x) = sin x – cos x, 0 < x < 2Ï€

Now, differentiate the given function w.r.t. x

f'(x)= cos x + sin x

Now, for local maxima and minima,

Put f'(x) =0

cos x = -sin x

tan x = -1

x = âˆˆ (0, 2Ï€)

Now again differentiate the given function w.r.t. x

f”(x) = -sin x + cos x

<0

>0

Therefore, by second derivative test, is a point of local maxima

Hence, the maximum value is

However, is a point of local minima

Hence, the minimum value is

### Question 9. f(x) = cos x, 0< x < Ï€

Solution:

Given function

f(x) = cos x, 0< x < Ï€

Now, differentiate the given function w.r.t. x

f'(x) = – sin x

Now, for local maxima and minima,

Put f ‘(x) – 0

â‡’ – sin x = 0

â‡’ x = 0, and Ï€

But, these two points lies outside the interval (0, Ï€)

So, no local maxima and minima will exist in the interval (0, Ï€).

### Question 10. f(x) = sin 2x – x,  -Ï€/2 â‰¤ x â‰¤  Ï€/2

Solution:

Given function

f(x) = sin2x – x

Now, differentiate the given function w.r.t. x

f ‘(x) = 2 cos 2x – 1

Now, for local maxima and minima,

Put f'(x) = 0

â‡’ 2cos 2x – 1 = 0

â‡’ cos 2x = 1/2 = cos Ï€/3

â‡’ 2x = Ï€/3, -Ï€/3

â‡’ x =

At x = -Ï€/6, f'(x) changes from negative to positive. Hence, x = Ï€/6 is point of local minima.

So, the minimum value is

At x = Ï€/6, f'(x) changes from positive to negative. Hence, x = Ï€/6 is point of local maxima

The maximum value is

### Question 11. f(x) = 2sin x – x,  -Ï€/2 â‰¤ x â‰¤ Ï€/2

Solution:

Given function

f(x) = 2sin x – x, -Ï€/2â‰¤ x â‰¤ Ï€/2

Now, differentiate the given function w.r.t. x

f ‘(x) = 2cos x – 1 = 0

Now, for local maxima and minima,

Put f'(x) = 0

â‡’ cos x = 1/2 = cos Ï€/3

â‡’ x = -Ï€/3, Ï€/3

So, at x = -Ï€/3, f'(x) changes from negative to positive. Hence, x = -Ï€/3 is point of local minima

So, the minimum value is f(-Ï€/3) = 2sin(-Ï€/3) – (-Ï€/3) = -âˆš3 – Ï€/3

At x = Ï€/3, f'(x) changes from positive to negative. Hence, x = Ï€/3 is point of local minima

The maximum value is f(Ï€/3) = 2sin(Ï€/3) – (Ï€/3) = âˆš3 – Ï€/3

### Question 12. f(x) = x, x > 0

Solution:

Given function

f(x) = x, x > 0

Now, differentiate the given function w.r.t. x

f'(x) =

Now, for local maxima and minima,

Put f'(x) = 0

â‡’

â‡’ 2 – 3x = 0

â‡’ x = 2/3

f “(x) =

=

= (3x – 4)/4(1 – x)2

f “(2/3) =

Therefore, x = 2/3 is a point of local maxima and the local maximum value of f at x = 2/3 is

f(2/3) = 2/3(âˆš1/3) = (2âˆš3)/9

### Question 13. f(x) = x3(2x – 1)3

Solution:

Given function

f(x) = x3(2x – 1)3

Now, differentiate the given function w.r.t. x

f'(x) = 3x2(2x – 1)2 + 6x3(2x – 1)2

= 3x2(2x – 1)2(2x – 1 + 2x)

= 3x2(4x – 1)

Now, for local maxima and minima,

Put f'(x) = 0

â‡’ 3x2(4x – 1) = 0

â‡’ x = 0, 1/4

At x = 1/4, f'(x) changes from negative to positive. Hence, x = 1/4 is the point of local minima,

So, the minimum value is f(1/4)= (1/4)3(2(1/4) – 1)3= -1/512

### Question 14. f(x) = x/2 + 2/x,  x > 0

Solution:

Given function

f(x) = x/2 + 2/x, x > 0

Now, differentiate the given function w.r.t. x

f'(x) = 1/2 – 2/x2, x > 0

Now, for local maxima and minima,

Put f'(x) = 0

â‡’ 1/2 – 2/x2 = 0

â‡’ x2 – 4 = 0

â‡’ x = 2, -2

At x = 2, f'(x) changes from negative to positive. Hence, x = 2 is point of local minima

So, the local minimum value is f(2) = 2/2 + 2/2 = 2

### Question 15. f(x) = 1/(x2 + 2)

Solution:

Given function

f(x) = 1/(x2 + 2)

Now, differentiate the given function w.r.t. x

f'(x) = -(2x)/(x2 + 2)2

Now, for local maxima and minima,

Put f'(x) = 0  f'(x) = 0

f'(x) = -(2x)/(x2 + 2)2 = 0

â‡’ x = 0

At x = 0, f'(x) > 0

At x = 0+, f'(x) < 0

Therefore, local minimum and maximum value of f(0) = 1/2

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