RD Sharma Class 12 Ex 18.2 Solutions Chapter 18 Maxima and Minima

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Here we provide RD Sharma Class 12 Ex 18.2 Solutions Chapter 18 Maxima and Minima for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 18.2 Solutions Chapter 18 Maxima and Minima book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter18
Exercise18.2
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 18.2 Solutions Chapter 18 Maxima and

Find the points of local maxima or local minima, if any, of the following functions, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be:

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

⇒ 4(x – 5)3 = 0

⇒ x – 5 = 0

⇒ 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

⇒ 3x2– 3 = 0

⇒ x = ±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

⇒ (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

⇒ (x – 1)2 (x + 1) (5x + 1) = 0

⇒ 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

⇒ 3 (x – 3) (x – 1) – 0

⇒ 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 < π

Solution:

Given function 

f(x) = sin2x, 0 < x, π

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

f'(x) = 2 cos 2x

Now, for local maxima and minima,

Put f'(x) = 0

⇒ 2x = \frac{π}{2},\frac{3π}{2}

⇒ x = \frac{π}{4},\frac{3π}{4}

At x = π​/4, f'(x) changes from positive to negative. Hence, x = π​/4, Is point of local maxima

So, the maximum value is f(π​/4) = sin2(π​/4) = 1

At x = 3π​/4, f'(x) changes from negative to positive. Hence, x = 3π​/4 is point of local minima,

So, the minimum value is f(3π​/4) = sin2(3π​/4) = -1

Question 8. f(x) = sin x – cos x, 0 < x < 2π

Solution:

Given function 

f(x) = sin x – cos x, 0 < x < 2π

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 = \frac{3π}{4},\frac{7π}{4} ∈ (0, 2π)

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

f”(x) = -sin x + cos x

f"(\frac{3π}{4})=-sin\frac{3π}{4}+cos\frac{3π}{4}=-\frac{1}{√2}-\frac{1}{√2}=-√2 <0

f"(\frac{7π}{4})=-sin\frac{7π}{4}+cos\frac{7π}{4}=\frac{1}{√2}+\frac{1}{√2}=√2 >0

Therefore, by second derivative test, x=\frac{3π}{4} is a point of local maxima 

Hence, the maximum value is f(\frac{3π}{4})=sin\frac{3π}{4}-cos\frac{3π}{4}=\frac{1}{√2}+\frac{1}{√2}=√2.

However, x=\frac{7π}{4} is a point of local minima 

Hence, the minimum value is f(\frac{7π}{4})=sin\frac{7π}{4}-cos\frac{7π}{4}=-\frac{1}{√2}-\frac{1}{√2}=-√2

Question 9. f(x) = cos x, 0< x < π

Solution:

Given function 

f(x) = cos x, 0< x < π

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

f'(x) = – sin x

Now, for local maxima and minima,

 Put f ‘(x) – 0

⇒ – sin x = 0

⇒ x = 0, and π

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

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

Question 10. f(x) = sin 2x – x,  -π/2 ≤ x ≤  π/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

⇒ 2cos 2x – 1 = 0

⇒ cos 2x = 1/2 = cos π/3

⇒ 2x = π/3, -π/3

⇒ x = \frac{π}{6},-\frac{π}{6}

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

So, the minimum value is f(-\frac{π}{6})=\frac{-√3}{2}+\frac{π}{6}

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

The maximum value is f(\frac{π}{6})=\frac{√3}{2}-\frac{π}{6}

Question 11. f(x) = 2sin x – x,  -π/2 ≤ x ≤ π/2

Solution:

Given function 

f(x) = 2sin x – x, -π/2≤ x ≤ π/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

⇒ cos x = 1/2 = cos π/3

⇒ x = -π/3, π/3

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

So, the minimum value is f(-π/3) = 2sin(-π/3) – (-π/3) = -√3 – π/3

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

The maximum value is f(π/3) = 2sin(π/3) – (π/3) = √3 – π/3

Question 12. f(x) = x\sqrt{1- x} , x > 0

Solution:

Given function 

f(x) = x\sqrt{1- x} , x > 0

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

f'(x) = \sqrt{1- x}+ x.\frac{1}{\sqrt{1- x}}(-1)=\sqrt{1- x}-\frac{x}{2\sqrt{1- x}}

= \frac{2(1-x)-x}{2\sqrt{1- x}}=\frac{2-3x}{2\sqrt{1- x}}

Now, for local maxima and minima,

Put f'(x) = 0

⇒ \frac{2-3x}{2\sqrt{1-x}}=0

⇒ 2 – 3x = 0

⇒ x = 2/3

f “(x) = \frac{1}{2}[\frac{\sqrt{1-x}(-3)-(2-3x)(\frac{-1}{2\sqrt{1-x}})}{1-x}]

=\frac{\sqrt{1-x}(-3)+(2-3x)(\frac{1}{2\sqrt{1-x}})}{2(1-x)}

= \frac{-6(1-x)+(2-3x)}{4(1-x)^{3/2}}

= (3x – 4)/4(1 – x)2

f “(2/3) = \frac{3(\frac{2}{3})-4}{4(1-\frac{2}{3})^{3/2}}

= \frac{2-4}{4(\frac{1}{3})^{3/2}}

= \frac{-1}{2(\frac{1}{3})^{3/2}}<0

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(√1/3) = (2√3)/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

⇒ 3x2(4x – 1) = 0

⇒ 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

⇒ 1/2 – 2/x2 = 0

⇒ x2 – 4 = 0

⇒ 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

⇒ 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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