RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.2 (2023-24)

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Access RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.2

Find the points of local maxima or local minima, if any, of the following functions using the first derivative test. Also, find the local maximum or local minimum values, as the case may be:

1. f (x) = (x – 5)⁴

Solution:

Given f (x) = (x – 5)⁴

Differentiate with respect to x

f’(x) = 4(x – 5)³

For local maxima and minima

f‘(x) = 0

= 4(x – 5)³ = 0

= x – 5 = 0

x = 5

f‘(x) changes from negative to positive as it passes through 5.

So, x = 5 is the point of local minima

Thus, the local minima value is f (5) = 0

2. f (x) = x³ – 3x

Solution:

Given, f (x) = x³ – 3x

Differentiate with respect to x then we get,

f’ (x) = 3x² – 3

Now, f‘(x) =0

3x² = 3 ⇒ x = ±1

Again differentiate f’(x) = 3x² – 3

f’’(x)= 6x

f’’(1)= 6 > 0

f’’ (– 1)= – 6 < 0

By the second derivative test, x = 1 is a point of local minima and local minimum value of f at

x = 1 is f (1) = 1³ – 3 = 1 – 3 = – 2

However, x = – 1 is a point of the local maxima and local maxima value of f at

x = – 1 is

f (– 1) = (– 1)³ – 3(– 1)

= – 1 + 3

= 2

Hence, the value of minima is – 2 and maxima is 2.

3. f (x) = x³ (x – 1)²

Solution:

Given, f(x) = x³(x – 1)²

Differentiate with respect to x, we get,

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)

For all maxima and minima,

f ’(x) = 0

= x²(x – 1) (5x – 3) = 0

By solving the above equation, we get

x =0, 1, 3/5

At x = 3/5, f’(x) changes from negative to positive

Since x = 3/5 is a point of Minima

At x =1, f‘(x) changes from positive to negative

Since x =1 is the point of maxima.

4. f (x) = (x – 1) (x + 2)²

Solution:

Given, f(x) = (x – 1) (x + 2)²

Differentiate with respect to x, we get,

f‘(x) = (x + 2)² + 2(x – 1)(x + 2)

= (x + 2) (x + 2 + 2x – 2)

= (x + 2) (3x)

For all maxima and minima,

f’(x) = 0

(x + 2) (3x) = 0

By solving the above equation, we get

x = 0, – 2

At x = – 2, f’(x) changes from positive to negative

Since x = – 2 is a point of Maxima

At x = 0, f‘(x) changes from negative to positive

Since x = 0 is a point of Minima.

Hence, local min value = f (0) = – 4

Local max value = f (– 2) = 0.

Therefore x = 0, now for the values close to x = 0 and to the left of 0, f'(x) > 0

Also for values x = 0 and to the right of 0, f’(x) < 0

Therefore, by the first derivative test, x = 0 is a point of local maxima and the local minima value of f (x) is ½.

6. f (x) = x³ – 6x² + 9x +15

Solution:

Given, f(x) = x³ – 6x² + 9x + 15

Differentiate with respect to x, we get, f‘(x) = 3x² – 12x + 9 = 3(x² – 4x + 3)

= 3 (x – 3) (x – 1)

For all maxima and minima,

f’(x) = 0

= 3(x – 3) (x – 1) = 0

= x = 3, 1

At x = 1, f’(x) changes from positive to negative

Since x = 1 is a point of Maxima

At x = 3, f‘(x) changes from negative to positive

Since x = 3 is a point of Minima.

Hence, local maxima value f (1) = (1)³ – 6(1)² + 9(1) + 15 = 19

Local minima value f (3) = (3)³ – 6(3)² + 9(3) + 15 = 15

7. f (x) = sin 2x, 0 < x < π

Solution:

Given f (x) = sin 2x

Differentiate w.r.t x, we get

f'(x) = 2 cos 2x, 0 < x < π

For the point of local maxima and minima, f’(x) = 0

2 cos 2x = 0

cos 2x = 0

2x = π/2, 3π/2

x = π/4, 3π/4

Now, at x = π/4, f’(x) changes from positive to negative

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RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.2: Important Topics From The Chapter

Some of the crucial topics of RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.2 are enlisted here.

• Maximum and minimum values of a function in its domain
• Definition and meaning of maximum
• Definition and meaning of minimum
• Local maxima
• Local minima
• First derivative test for local maxima and minima
• Higher-order derivative test
• Theorem and algorithm based on higher derivative test
• Point of inflection
• Point of inflection
• Properties of maxima and minima
• Maximum and minimum values in a closed interval
• Applied problems on maxima and minima

We have included all the information regarding CBSE RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.2. If you have any queries feel free to ask in the comment section. 

FAQ: RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.2