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Class 12 RD Sharma Solutions- Chapter 18 Maxima and Minima – Exercise 18.4
  • Last Updated : 25 Jan, 2021

Question 1. Find the absolute maximum and absolute minimum values of the following functions in the given intervals:

(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

⇒ 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

⇒ 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(−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,

f’(x) = 12x3 – 24x2 + 24x – 48

= 12(x3 – 2x2 + 2x – 4)

= 12(x – 2)(x2 + 2)

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

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



therefore we consider only x = 2 ∈ [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) f(x)=(x-2) \sqrt{x-1}      in [1, 9]

Solution:

Given:  f(x)=(x-2) \sqrt{x-1} in [1, 9]

On differentiation, we get,

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

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

\sqrt{x-1} + \frac {x-2} {2\sqrt {x-1}}=0

\frac {2(x-1) +  {x-2}} {2\sqrt {x-1}}=0

\frac {3x-4} {2\sqrt {x-1}}=0

= x = 4/3

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

f(1)=(1-2)\sqrt {1-1}=(-1)\sqrt {0}=0

f(\frac 4 3)= (\frac 4 3 -2)\sqrt {\frac 4 3 -1}=\frac {4-6} {3\sqrt3}=\frac {-2} {3\sqrt3}=\frac {-2\sqrt3} 9

f(9)=(9-2)\sqrt{9-1}=7\sqrt8=14\sqrt2

Hence, the absolute maximum value of f in [1, 9] is 14√2 at x = 9 and 

the absolute minimum value of f in [1, 9] is  \frac {-2\sqrt3} 9  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 

∴ f'(x) = 6x2 – 24 = 6(x2 – 4) 

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

6(x2 – 4) = 0

x2 = 4

x = ±2

We first consider the interval [1, 3].

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

Evaluate the value of f at the critical point x = – 2 ∈ [-3, -1]

f(−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 ∈ [0, π].

Solution:

Given: f(x) = cos2x + sinx, x ∈ [0, π]

On differentiation, we get,

f′(x) = 2cosx(−sinx) + cosx

= −2sinxcosx + cosx

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

−2sinxcosx + cosx = 0

⇒ cosx(−2sinx + 1) = 0

⇒ sinx = 1​/2 or cos x = 0

⇒ x = π/6​, π/2​ as x ∈ [0, π]

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

f(π/6​) = cos2π/6 + sinπ/6 = (√3/2​​)2+1/2 ​= 5/4​

f(π/2​) = cos2π/2​ + sinπ/2​ = 0 + 1 = 1

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

f(π) = cos2π + sinπ = (−1)2 + 0 = 1

Hence, the absolute maximum value of f in [0, π] is 5/4 at x = π/6​   and

the absolute minimum value of f in [0, π] is 1 at x = 0, π/2​, π.

Question 4. Find the absolute maximum and minimum values of the function f given by f(x) = 12x4/3​ – 6x1/3​, x ∈ [−1, 1].

Solution:

Given: f(x) = 12x4/3​ – 6x1/3​, x ∈ [−1, 1].

On differentiation, we get,

f'(x)=16x^{\frac 1 3}-\frac 2 {x^{\frac 2 3}}=\frac {2(8x-1)} {x^{\frac 2 3}}

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

\frac {2(8x-1)} {x^{\frac 2 3}}=0

⇒ 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(−1) = 12(−1)4/3– 6(−1)1/3 ​= 18

f(1) = 12(1)4/3– 6(1)1/3 ​= 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,

f′(x) = 6x2 – 30x + 36 = 6(x2 – 5x + 6) = 6(x – 2)(x – 3)

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

6(x – 2)(x – 3) = 0

⇒ 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.

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