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By Karan Singh Bisht
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Updated on 11 Jun 2025, 18:07 IST
RD Sharma Solutions for Class 12 Maths Chapter 18 – Maxima and Minima are designed to help students develop a clear understanding of key calculus concepts. These step-by-step solutions are created by experienced subject experts to address common doubts and make problem-solving easier.
Practicing the RD Sharma solutions from this chapter allows students to strengthen their grasp of mathematical principles and improve overall performance in board exams. Chapter 18 focuses on identifying the maximum and minimum values of a function within its domain—an essential concept in Class 12 calculus.
By regularly working through these exercises, students can gain confidence and accuracy in tackling related problems. You can download the complete PDF of RD Sharma Chapter 18 solutions using the link provided below for easy and effective exam preparation.
RD Sharma Class 12 Maths Solutions Chapter 18 Maxima and Minima cover all the questions from the textbook, crafted by expert Mathematics teachers at Infinity Learn. Download our free PDF of Chapter 18 Maxima and Minima RD Sharma Solutions for Class 12 to boost your performance in board exams and competitive exams.
Q. Find the maximum and the minimum values, if any, without using derivatives of the following functions:
1. f (x) = –(x – 1)2 + 2 on R
Solution:
Given f(x) = – (x – 1)2 + 2
It can be observed that (x – 1)2 ≥ 0 for every x ∈ R
Therefore, f(x) = – (x – 1)2 + 2 ≤ 2 for every x ∈ R
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The maximum value of f is attained when (x – 1) = 0
(x – 1) = 0, x = 1
Since, Maximum value of f = f (1) = – (1 – 1)2 + 2 = 2
Hence, function f does not have minimum value.
2. f (x) = sin 2x + 5 on R
Solution:
Given f (x) = sin 2x + 5 on R
We know that – 1 ≤ sin 2x ≤ 1
⇒ – 1 + 5 ≤ sin2x + 5 ≤ 1 + 5
⇒ 4 ≤ sin 2x + 5 ≤ 6
Therefore, the maximum value and minimum value of f are 6 and 4 respectively.
Q. f (x) = 4x2 – 4x + 4 on R
Solution:
Given f (x) = 4x2 – 4x + 4 on R
= 4x2 – 4x + 1 + 3
By grouping the above equation we get,
= (2x – 1)2 + 3
Since, (2x – 1)2 ≥ 0
= (2x – 1)2 + 3 ≥ 3
= f(x) ≥ f (½)
Thus, the minimum value of f(x) is 3 at x = ½
Since, f(x) can be made large. Therefore maximum value does not exist.
Q. f (x) = |x + 2| on R
Solution:
Given f (x) = |x + 2| on R
⇒ f(x) ≥ 0 for all x ∈ R
So, the minimum value of f(x) is 0, which attains at x = -2
Thus, f(x) = |x + 2| does not have the maximum value.
2. Find the points of local maxima or local minima and corresponding local maximum and local minimum values of each of the following functions. Also, find the points of inflection, if any:
Q. f (x) = x3 – 6x2 + 9x + 15
Solution:
Given f (x) = x3 – 6x2 + 9x + 15
Differentiating f with respect to x
∴ f'(x) = 3x2 – 12x + 9 = 3(x2 – 4x + 3)
f” (x) = 6x – 12 = 6(x – 2)
For maxima and minima, f'(x) = 0
3(x2 – 4x + 3) = 0
So roots will be x = 3, 1
Now, f” (3) = 6 > 0
x = 3 is point of local minima
f”(1) = – 6 < 0
x = 1 is point of local maxima
Local max value = f (1) = 19 and local min value = f (3) = 15
Q. f (x) = x ex
Solution:
Given f(x) = x ex
f'(x) = ex + x ex = ex(x + 1)
f”(x) = ex(x + 1) + ex
= ex(x + 2)
For maxima and minima,
f'(x) = 0
ex(x + 1) = 0
x = – 1
Now f’’ (– 1) = e – 1 = 1/e > 0
x = – 1 is point of local minima
Hence, local min = f (– 1) = – 1/e
Q. f(x) = x4 – 62x2 + 120x + 9
Solution:
Given f (x) = x4 – 62x2 + 120x + 9
∴ f'(x) = 4x3 – 124x + 120 = 4(x3 – 31x + 30)
f”(x) = 12x2 – 124 = 4(3x2 – 31)
For maxima and minima, f'(x) = 0
4(x3 – 31x + 30) = 0
So roots will be x = 5, 1, – 6
Now, f”(5) = 176 > 0
x = 5 is point of local minima
f”(1) = – 112 < 0
x = 1 is point of local maxima
f”(– 6) = 308 > 0
x = – 6 is point of local minima
Local max value = f (1) = 68
Local min value = f (5) = – 316 and f (– 6) = – 1647
3. Find the local extremum values of the following functions:
Q. f (x) = – (x – 1)3(x + 1)2
Solution:
Given f (x) = – (x – 1)3(x + 1)2
f’(x) = – 3(x – 1)2(x + 1)2 – 2(x – 1)3(x + 1)
= – (x – 1)2(x + 1) (3x + 3 + 2x – 2)
= – (x – 1)2(x + 1) (5x + 1)
f’’(x) = – 2(x – 1)(x + 1)(5x + 1) – (x – 1)2(5x + 1) – 5(x – 1)2(x + 1)
For maxima and minima, f'(x) = 0
– (x – 1)2(x + 1) (5x + 1) = 0
x = 1, – 1, – 1/5
Now f’’ (1) = 0
x = 1 is inflection point
f’’(– 1) = – 4× – 4 = 16 > 0
x = – 1 is point of minima
f’’ (– 1/5) = – 5(36/25) × 4/5 = – 144/25 < 0
x = – 1/5 is point of maxima
Hence local max value = f (– 1/5) = 3456/3125
Local min value = f (– 1) = 0
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Chapter 18 of RD Sharma Class 12 Maths focuses on Maxima and Minima, explaining how to find the maximum and minimum values of a function using first and second derivative tests. It includes detailed examples and exercises aligned with CBSE and JEE syllabus.
Yes, RD Sharma Chapter 18 solutions are extremely useful for board exam preparation. They offer step-by-step methods, clear explanations, and cover all question types expected in CBSE Class 12 Maths exams.
You can download the RD Sharma Solutions for Chapter 18 – Maxima and Minima in PDF format from trusted educational platforms like Infinity Learn. These solutions are based on the latest CBSE guidelines and are easy to follow.
RD Sharma solutions simplify the process of learning Maxima and Minima by providing clear derivations, graphical interpretations, and real-world application examples. They make complex calculus concepts easier to understand and apply in exams.