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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6

Question 1.

Find the absolute extrema of the following functions on the given closed interval.

(i) f(x) = x^{3} – 12x + 10; [1, 2]

(ii) f(x) = 3x^{4} – 4x^{3} ; [-1, 2]

Solution:

∴ Absolute maximum is – 1 and absolute minimum is – 6

(ii) f(x) = 3x^{4} – 4x^{3}

f'(x) = 12x^{3} – 12x^{2}

f'(x) = 0 ⇒ 12x^{2}(x – 1) = 0

⇒ x = 0 or x = 1

[Here x = 0, 1 ∈ [-1, 2]]

Now f (-1) = 4

f(0) = 0

f(1) = -1

f(2) = 16

so absolute maximum = 16 and absolute minimum = -1

(iii)

(iv) f(x) = 2 cos x + sin 2x

f'(x) = -2 sin x + 2 cos 2x

f'(x) = 0 ⇒ cos 2x = sin x

Question 2.

Find the intervals of monotonicities and hence find the local extremum for the following functions

Solution:

(i) f(x) = 2x^{3} + 3x^{2} – 12x

f'(x) = 6x^{2} + 6x – 12

f'(x) = 0 ⇒ 6(x^{2} + x – 2) = 0

(i.e.,) 6(x + 2)(x – 1) = 0

⇒ x = -2 or 1

Taking the points in the number line

The intervals are (-∞, -2), (-2, 1), (1, ∞)

when x ∈ (-∞, -2), f'(x) = 6 (-1) (-4) = +ve

say x = – 3

⇒ f(x) is strictly increasing in the interval (-∞, -2) when x ∈ (-2, 1), f’ (x) = 6 (2) (-1) = -ve

say x = 0

⇒ f(x) is strictly decreasing in the interval (-2, 1)

when x ∈ (1, ∞), f'(x) = 6 (4) (+1) = + ve say x = 2

⇒ f(x) is strictly increasing in (1, ∞)

Since f(x) changes from +ve to – ve when passing through -2, the first derivative, test tells us there is a local maximum at x = -2 and the local maximum value is f(-2) = 20.

Again f'(x) changes from – ve to +ve when passing through 1 ⇒ there is a local minimum at x = 1 and the local minimum value is f( 1) = -7. So (1 )f(x) is strictly increasing on (-∞, -2) and (1, ∞). And (2)f(x) is strictly decreasing on (-2, 1)

The local maximum = 20 and the local minimum = -7

f(x) is strictly decreasing on (-∞, 5) and (5, ∞)

And there is no local extremum

(iii)

For all x values, so f(x) is strictly increasing in (-∞, ∞) and there is no local extremum.

⇒ f(x) is strictly decreasing in (0, 1)

say x = 2

⇒ f(x) is strictly increasing in (1, ∞)

since f'(x) changes from -ve to +ve at x = 1, there is a local minimum at x = 1 and the local minimum values is f(1) = \(\frac{1}{3}-0=\frac{1}{3}\)

So the function is strictly decreasing on (0, 1) and strictly increasing on (1, ∞) and the local minimum value is \(\frac{1}{3}\)

(v)

### Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 Additional Problems

Find the absolute maximum and absolute minimum values of f on the given interval.

Question 1.

f(x) = 1 – 2x – x^{2}, [-4, 1]

Solution:

f(x) = 1 – 2x – x^{2}

f'(x) = -2 – 2x

f'(x) = 0 ⇒ 2x = -2 ⇒ x = -1; The points are -4, -1, 1

At x = -4, f(x) = 1 – 2(-4) – (-4)^{2} = 1 + 8 – 16 = -7

At x = -1, f(x) = 1 – 2 (-1) – (-1)^{2} = 1 + 2 – 1 = 2

At x = 1, f(x) = 1 – 2 (1) – (1)^{2} = 1 – 2 – 1 = – 2

Therefore, the absolute minimum is -7 and the absolute maximum is 2.

Question 2.

f(x) = x^{3} – 12x + 1, [-3, 5]

Solution:

f(x) = x^{3} – 12x + 1

f'(x) = 3x^{2} – 12

f'(x) = 0 ⇒ 3x^{2} – 12 = 0

3x^{2} = 12 ⇒ x^{2} = 4 ⇒ x = ±2

The x values are -3, -2, 2, 5

f(x) (at x = -3) = (-3)^{3} – 12 (-3) + 1 = -27 + 36 + 1 = 10

f(x) (at x = -2) = (-2)^{3} – (12)(-2) + 1 = -8 + 24 + 1 = 17

f(x) (at x = 2) = 2^{3} – 12 (2) + 1 = 8 – 24 + 1 = -15

f(x) (at x = 5) = 53 – 12 (5) + 1 = 125 – 60 + 1 = 66

From the above four values 10,17,-15 and 66, we see that absolute maximum is 66 and the absolute minimum is -15.

Question 3.

Solution:

So, the absolute maximum is \(\frac{2}{3}\) and the absolute minimum is \(\frac{1}{2}\)

Question 4.

f(x) = sin x + cos x, [0, π/3]

f'(x) = sin x + cos x

f'(x) = cos x – sin x

f’ (x) = 0 ⇒ cos x – sin x = 0

Question 5.

f(x) = x – 2 cos x, [-π, π]

Solution:

f(x) = x – 2 cos x

f'(x) = 1 + 2 sin x

f'(x) = 0 ⇒ 1 + 2 sin x = 0

From the above values, absolute maximum is π + 2 and the absolute minimum is \(\frac{-\pi}{6}-\sqrt{3}\)

From the local maximum and minimum values of the following functions.

Question 6.

2x^{3} + 5 x^{2} – 4x

Solution:

Local maximum = 12

Question 7.

t + cos t

Solution:

t = \(\frac{\pi}{2}\) is neither a maximum point nor a minimum point. So there is no maximum or minimum.

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