Open in App
Not now

# Class 12 RD Sharma Solutions- Chapter 18 Maxima and Minima – Exercise 18.4

• Last Updated : 25 Jan, 2021

### (i) f(x) = 4x – x2/2 in [-2, 9/2]

Solution:

Given: f(x) = 4x – x2/2 in [-2, 9/2]

On differentiation, we get,

f'(x) = 4 – x

For local maxima and local minima we have f'(x) = 0

4 – x = 0

â‡’ x = 4

Let’s evaluate the value of f at the critical point x=4 and at the interval [-2, 9/2]

f(4) = 4(4) – 42/2 = 16 – 16/2 = 16 – 8 = 8

f(-2) = 4(-2) – (-2)2/2 = -8 – 4/2 = -8 – 2 = -10

f(9/2) = 4(9/2) – (9/2)2/2 = 18 – 81/8 = 18 – 10.125 = 7.875

Hence, the absolute maximum value of f in [-2, 9/2] is 8 at x = 4 and

the absolute minimum value of f in [-2, 9/2] is -10 at x = -2.

### (ii) f(x) = (x – 1)2+3 in [-3, 1]

Solution:

Given: f(x) = (x – 1)2+3 in [-3, 1]

On differentiation, we get,

f'(x) = 2(x – 1)

For local maxima and local minima we have f'(x) = 0

2(x – 1) = 0

â‡’ x = 1

Let’s evaluate the value of f at the critical point x = 1 and at the interval [-3, 1]

f(1) = (1 – 1)2 + 3 = 3

f(âˆ’3) = (-3 – 1)2 + 3 = 16 + 3 = 19

Hence, the absolute maximum value of f in [-3, 1] is 19 at x = -3 and

the absolute minimum value of f in [-3, 1] is 3 at x = 1.

### (iii) f(x) = 3x4 – 8x3 + 12x2 – 48x + 25 in [0, 3]

Solution:

Given: f(x) = 3x4 – 8x3 + 12x2 – 48x + 25 in [0,3]

On differentiation, we get,

fâ€™(x) = 12x3 – 24x2 + 24x – 48

= 12(x3 – 2x2 + 2x – 4)

= 12(x – 2)(x2 + 2)

Now, for local minima and local maxima we have fâ€²(x) = 0

x = 2 or  x2 + 2 = 0 having no real roots

therefore we consider only x = 2 âˆˆ [0, 3]

Let’s evaluate the value of f at the critical point x = 2 and at the interval [0, 3]

f(2) = 3(2)4 – 8(2)3 + 12(2)2 – 48(2) + 25

= 3(16) – 8(8) + 12(4) – 96 + 25

= 48 – 64 + 48 – 96 + 25

= -39

f(0) = 3(0)4 – 8(0)3 + 12(0)2 – 48(0) + 25 = 25

f(3) = 3(3)4 – 8(3)3 + 12(3)2 – 48(3) + 25

= 3(81) – 8(27) + 12(9) – 144 + 25

= 243 – 216 + 108 – 144 + 25

= 16

Hence, the absolute maximum value of f in [0, 3] is 25 at x = 0 and

the absolute minimum value of f in [0, 3] is -39 at x = 2.

### (iv)  in [1, 9]

Solution:

Given: in [1, 9]

On differentiation, we get,

Now, for local minima and local maxima we have fâ€²(x) = 0

= x = 4/3

Let’s evaluate the value of f at the critical point x = 4/3 and at the interval [1, 9]

Hence, the absolute maximum value of f in [1, 9] is 14âˆš2 at x = 9 and

the absolute minimum value of f in [1, 9] is  at x = 4/3

### Question 2.  Find the maximum value of 2x3 – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].

Solution:

Let f(x) = 2x3 – 24x + 107

âˆ´ f'(x) = 6x2 – 24 = 6(x2 – 4)

Now, for local maxima and local minima we have f'(x) = 0

â‡’ 6(x2 – 4) = 0

â‡’ x2 = 4

â‡’ x = Â±2

We first consider the interval [1, 3].

Let’s evaluate the value of f at the critical point x = 2 âˆˆ [1, 3] and at  the interval [1, 3].

f(2) = 2(2)3 – 24(2) + 107 = 75

f(1) = 2(1)3 – 24(1) + 107 = 85

f(3) = 2(3)3 – 24 (3) + 107 = 89

Hence, the absolute maximum value of f in the interval [1, 3] is 89 occurring at x = 3,

Next, we consider the interval [â€“ 3, â€“ 1].

Evaluate the value of f at the critical point x = â€“ 2 âˆˆ [-3, -1]

f(âˆ’3) = 2(-3)3 – 24(-3) + 107 = 125

f(-2) = 2(-2)3 – 24(-3) + 107 = 139

f(-1) = 2(-1)3 – 24(-2) + 107 = 129

Hence, the absolute maximum value of f is 139 when x = -2.

### Question 3. Find the absolute maximum and minimum values of the function f given by f(x) = cos2x + sinx, x âˆˆ [0, Ï€].

Solution:

Given: f(x) = cos2x + sinx, x âˆˆ [0, Ï€]

On differentiation, we get,

fâ€²(x) = 2cosx(âˆ’sinx) + cosx

= âˆ’2sinxcosx + cosx

Now, for local minima and local maxima we have fâ€²(x) = 0

âˆ’2sinxcosx + cosx = 0

â‡’ cosx(âˆ’2sinx + 1) = 0

â‡’ sinx = 1â€‹/2 or cos x = 0

â‡’ x = Ï€/6â€‹, Ï€/2â€‹ as x âˆˆ [0, Ï€]

Let’s evaluate the value of f at the critical points x=6/Ï€ and x = 2/Ï€ and at the interval [0, Ï€]

f(Ï€/6â€‹) = cos2Ï€/6 + sinÏ€/6 = (âˆš3/2â€‹â€‹)2+1/2 â€‹= 5/4â€‹

f(Ï€/2â€‹) = cos2Ï€/2â€‹ + sinÏ€/2â€‹ = 0 + 1 = 1

f(0) = cos20 + sin0 = 1 + 0 = 1

f(Ï€) = cos2Ï€ + sinÏ€ = (âˆ’1)2 + 0 = 1

Hence, the absolute maximum value of f in [0, Ï€] is 5/4 at x = Ï€/6â€‹   and

the absolute minimum value of f in [0, Ï€] is 1 at x = 0, Ï€/2â€‹, Ï€.

### Question 4. Find the absolute maximum and minimum values of the function f given by f(x) = 12x4/3â€‹ – 6x1/3â€‹, x âˆˆ [âˆ’1, 1].

Solution:

Given: f(x) = 12x4/3â€‹ – 6x1/3â€‹, x âˆˆ [âˆ’1, 1].

On differentiation, we get,

Now, for local minima and local maxima we have fâ€²(x) = 0

â‡’ x = 1/8

Let’s evaluate the value of f at the critical points x = 1/8 and at the interval [-1, 1]

f(1/8) = 12(1/8)4/3 – 6(1/8)1/3 = -9/4

f(âˆ’1) = 12(âˆ’1)4/3– 6(âˆ’1)1/3 â€‹= 18

f(1) = 12(1)4/3– 6(1)1/3 â€‹= 6

Hence, the absolute maximum value of f in [-1, 1] is 18 at x = -1 and

the absolute minimum value of f in [-1, 1] is -9/4 at x = 1/8.

### Question 5. Find the absolute maximum and minimum values of the function f given by f(x) = 2x3 – 15x2+ 36x + 1 in the interval [1, 5].

Solution:

Given: f(x) = 2x3 – 15x2+ 36x + 1 in the interval [1, 5]

On differentiation, we get,

fâ€²(x) = 6x2 – 30x + 36 = 6(x2 – 5x + 6) = 6(x – 2)(x – 3)

Now, for local minima and local maxima we have fâ€²(x) = 0

6(x – 2)(x – 3) = 0

â‡’ x = 2 and x = 3

Let’s evaluate the value of f at the critical points x = 2 and x = 3 and at the interval [1, 5]

f(1) = 2(1)3 – 15(1)2 + 36(1) + 1 = 24

f(2) = 2(2)3 – 15(2)2 + 36(2) + 1 = 29

f(3) = 2(3)3 – 15(3)2 + 36(3) + 1 = 28

f(5) = 2(5)3 – 15(5)2 + 36(5) + 1 = 56

Hence, the absolute maximum value of f in [1, 5] is 56 at x = 5 and

the absolute minimum value of f in [1, 5] is 24 at x = 1.

My Personal Notes arrow_drop_up
Related Articles