Here we provide RD Sharma Class 12 Ex 18.4 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 18.4 Solutions Chapter 18 Maxima and Minima book pdf download. Now you will get step-by-step solutions to each question.
| Textbook | NCERT |
| Class | Class 12th |
| Subject | Maths |
| Chapter | 18 |
| Exercise | 18.4 |
| Category | RD Sharma Solutions |
RD Sharma Class 12 Ex 18.4 Solutions Chapter 18 Maxima and Minima
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) 
in [1, 9]
Solution:
Given: 
in [1, 9]
On differentiation, we get,
Now, for local minima and local maxima we have f′(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 14√2 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
∴ 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,
Now, for local minima and local maxima we have f′(x) = 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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