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In this section, we dive into Chapter 18 of the Class 12 RD Sharma textbook, which focuses on Maxima and Minima. Exercise 18.2 is designed to help students understand how to find the maximum and minimum values of functions using calculus. These concepts are fundamental in optimization problems, which have wide applications in various fields.
This section provides detailed solutions for Exercise 18.2 from Chapter 18 of the Class 12 RD Sharma textbook. These solutions aim to guide students through finding the maxima and minima of functions, ensuring a strong grasp of this important topic in calculus.
Chapter 18 of RD Sharma's Class 12 Mathematics deals with the concepts of maxima and minima. These concepts are essential in calculus and optimization problems where one needs to determine the highest or lowest values of the function within the given range. Understanding how to find these extreme values is crucial for solving problems in various fields such as physics, engineering, and economics.
The Maxima and minima refer to the highest and lowest values of the function respectively. In mathematical terms, the maxima and minima can be classified into the local and global extremes. The Local maxima and minima are points where a function reaches a peak or a valley in the specific interval while global maxima and minima are the overall highest and lowest values over the entire domain of the function.
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
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
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
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
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
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
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 =
⇒ x =
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
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 = ∈ (0, 2π)
Now again differentiate the given function w.r.t. x
f"(x) = -sin x + cos x
<0
>0
Therefore, by second derivative test, is a point of local maxima
Hence, the maximum value is
However, is a point of local minima
Hence, the minimum value is
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).
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 =
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
At x = π/6, f'(x) changes from positive to negative. Hence, x = π/6 is point of local maxima
The maximum value is
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
Solution:
Given function
f(x) = x, x > 0
Now, differentiate the given function w.r.t. x
f'(x) =
=
Now, for local maxima and minima,
Put f'(x) = 0
⇒
⇒ 2 - 3x = 0
⇒ x = 2/3
f "(x) =
=
=
= (3x - 4)/4(1 - x)2
f "(2/3) =
=
=
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
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
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
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
Exercise 18.2 in RD Sharma's Class 12 textbook likely focuses on the application of derivatives to find maxima and minima of functions. This exercise would typically cover the first derivative test, identifying critical points, and determining the nature of these points (maxima, minima, or neither). Students are expected to differentiate various types of functions, including polynomial, trigonometric, and rational functions, to find their critical points. The problems may involve finding local and global extrema, determining intervals of increase and decrease, and sketching graphs based on this information. This exercise builds upon the basic concepts of differentiation and introduces students to the practical applications of derivatives in optimization problems.
