Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.7 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7

Question 1.

Find intervals of concavity and points of inflection for the following functions:

(i) f(x) = x(x – 4)³

(ii) f(x) = sin x + cos x, 0 < x < 2π

(iii) f(x) = \(\frac { 1 }{ 2 }\)(e^{x} – e^{-x})

Solution:

(i) f(x) = x(x – 4)³

f'(x) = 3x(x – 4)² + (x – 4)³(1)

= (x – 4)² (4x – 4) = 4 (x – 4)² (x – 1)

f'(x) = 4 [(x-4)² (1) + (x – 1) 2 (x – 4)]

= 4(x – 4)(x – 4 + 2x – 2)

= 4(x – 4)(3x – 6) = 12(x – 4)(x – 2)

f'(x) = 0 ⇒ 12 (x – 4) (x – 2) = 0

Critical points x = 2, 4

The intervals are (- ∞, 2), (2, 4) and (4, ∞)

In the interval (-∞, 2), f”(x) > 0 ⇒ Curve is Concave upward

In the interval (2, 4), f”(x) < 0 ⇒ Curve is Concave downward.

In the interval (4, ∞), f”(x) > 0 ⇒ Curve is Concave upward.

The curve is concave upward in

(-∞, 2), (4, ∞) it is concave downward in (2, 4).

f”(x) changes its sign when passing through x = 2 and x = 4

Now f(2) = 2 (2 – 4)³ = -16 and f(4) = 4 (4 – 4)³ = 0

∴ The points of inflection are (2, -16) and (4, 0).

(ii) f(x) = sin x + cos x, 0 < x < 2π

f'(x) = cos x – sin x

f”(x) = – sin x – cos x

f'(x) = 0 ⇒ sin x + cos x = 0

Critical points x = \(\frac { 3π }{ 4 }\), \(\frac { 7π }{ 4 }\)

The intervals are (0, \(\frac { 3π }{ 4 }\)), (\(\frac { 3π }{ 4 }\), \(\frac { 7π }{ 4 }\)) and (\(\frac { 7π }{ 4 }\), 2π)

In the interval (0, \(\frac { 3π }{ 4 }\)), f'(x) < 0 ⇒ curve is concave down.

In the interval (\(\frac { 3π }{ 4 }\), \(\frac { 7π }{ 4 }\)), f'(x) > 0 ⇒ curve is concave up.

In the interval (\(\frac { 7π }{ 4 }\), 2π), f'(x) < 0 ⇒ curve is concave down.

The curve is concave upward in (\(\frac { 3π }{ 4 }\), \(\frac { 7π }{ 4 }\)) and concave downward in (0, \(\frac { 3π }{ 4 }\)) and (\(\frac { 7π }{ 4 }\), 2π)

f'(x) changes its sign when passing through x = \(\frac { 3π }{ 4 }\) and x = \(\frac { 7π }{ 4 }\)

(iii) f(x) = \(\frac { 1 }{ 2 }\) (e^{x} – e^{-x})

f'(x) = \(\frac { 1 }{ 2 }\) (e^{x} + e^{-x})

f”(x) = \(\frac { 1 }{ 2 }\) (e^{x} – e^{-x})

f”(x) = 0 ⇒ \(\frac { 1 }{ 2 }\) (e^{x} – e^{-x}) = 0

Critical point x = 0

The intervals are (-∞, o) and (0, ∞)

In the interval (-∞, 0), f”(x) < 0 ⇒ curve is concave down.

In the interval (0, ∞), f(x) > 0 ⇒ curve is concave up.

∴ The curve is concave up in (0, ∞) and concave down in (-∞, 0).

f'(x) changes its sign when passing through x = 0

Now f(0) = – (e° – e°) = \(\frac { 1 }{ 2 }\) (1 – 1) = 0

∴ The point of inflection is (0, 0).

Question 2.

Find the local extrema for the following functions using second derivative test:

(i) f(x) = -3x^{5} + 5x^{3}

(ii) f(x) = x log x

(iii) f(x) = x² e^{-2x}

Solution:

(i) f(x) = – 3x^{5} + 5x^{3}

f'(x) = 0, f”(x) = -ve at x = a

⇒ x = a is a maximum point

f'(x) = 0, f”(x) = +ve at x = 6

⇒ x = b is a minimum point

f(x) = – 3x^{5} + 5x^{3}

f’ (x) = -15x^{4} + 15x^{2}

f”(x) = -60x^{3} + 30x

f'(x) = 0 ⇒ – 15x^{2} (x^{2} – 1) = 0

⇒ x = 0, +1, -1

at x = 0, f”(x) = 0

at x = 1, f”(x) = -60 + 30 = – ve

at x = -1, f”(x) = 60 – 30 = + ve

So at x = 1, f'(x) = 0 and f”(x) = -ve

⇒ x = 1 is a local maximum point.

and f(1) = 2

So the local maximum is (1, 2)

at x = -1, f'(x) = 0 and f”(x) = +ve

⇒ x = -1 is a local maximum point and f(-1) = -2.

So the local minimum point is (-1, -2)

∴ a local minimum is -2 and the local maximum is 2.

(ii) f(x) = x log x

f'(x) = x – \(\frac { 1 }{ 4 }\) + log x = 1 + log x

For maximum or minimum

f'(x) = 0 ⇒ 1 + log x = 0

⇒ log x = -1

x = e^{-1} = \(\frac { 1 }{ e }\)

f”(x) = \(\frac { 1 }{ x }\)

at x = \(\frac { 1 }{ e }\), f”(x) > 0 ⇒ f(x) attains minimum.

∴ Local minimum f(\(\frac { 1 }{ e }\)) = \(\frac { 1 }{ e }\) log (\(\frac { 1 }{ e }\))

= \(\frac { 1 }{ e }\)(-1) = –\(\frac { 1 }{ e }\)

(iii) f(x) = x^{2} e^{-2x}

f'(x) = x^{2}[-2e^{-2x}] + e^{-2x} (2x)

= 2e^{-2x} (x – x^{2})

f”(x) = 2e^{-2x}(1 – 2x) + (x – ^{2}) (-4e^{-2x})

= 2e^{-2x} [(1 – 2x) + (x – x^{2}) (- 2)]

= 2e^{-2x} [2x^{2} – 4x + 1]

f'(x) = 0 ⇒ 2e^{-2x}(x – x^{2}) = 0

⇒ x (1 – x) = 0

⇒ x = 0 or x = 1

at x = 0, f”(x) = 2 × 1 [0 – 0 + 1] = +ve

⇒ x = 0 is a local minimum point and the minimum value is f(0) = 0 at x = 1,

f”(x) = 2e^{-2} [2 – 4 + 1] = -ve

⇒ x = 1 is a local maximum point and the maximum value is f(1) = \(\frac{1}{e^{2}}\)

Local maxima \(\frac{1}{e^{2}}\) and local minima = 0

Question 3.

For the function f(x) = 4x³ + 3x² – 6x + 1 find the intervals of monotonicity, local extrema, intervals of concavity and points of inflection.

Solution:

(x) = 4x³ + 3x² – 6x + 1

Monotonicity

f(x) = 4x³ + 3x² – 6x + 1

f'(x) = 12x² + 6x – 6

f'(x) = 0 ⇒ 6(2x² + x – 1) = 0

x = -1, \(\frac { 1 }{ 2 }\) (Stationary points)

∴ The intervals of monotonicity are (-∞, -1), (-1, \(\frac { 1 }{ 2 }\)) and (\(\frac { 1 }{ 2 }\), ∞)

In (-∞, -1), f'(x) > 0 ⇒ f(x) is strictly increasing

In (-1, \(\frac { 1 }{ 2 }\)), f'(x) < 0 ⇒ f(x) is strictly decreasing

In (\(\frac { 1 }{ 2 }\), ∞) f'(x) > 0 ⇒ f(x) is strictly increasing

f(x) attains local maximum as f'(x) changes its sign from positive to negative when passing through x = -1

∴ Local maximum f(-1) = -4 + 3 + 6 + 1 = 6

f(x) attains local minimum as f'(x) changes its sign from negative to positive when passing through x = \(\frac { 1 }{ 2 }\)

∴ Local minimum f(\(\frac { 1 }{ 2 }\))

= 4(\(\frac { 1 }{ 8 }\)) + 3(\(\frac { 1 }{ 4 }\)) – 6(\(\frac { 1 }{ 2 }\)) + 1

= \(\frac { 1 }{ 2 }\) + \(\frac { 3 }{ 4 }\) – + 1 = –\(\frac { 3 }{ 4 }\)

f(x) = 4x³ + 3x² – 6x + 1

f'(x) = 12x² + 6x – 6

f”(x) = 24x + 6

f’(x) = 0 ⇒ 24x + 6 = 0

x = –\(\frac { 6 }{ 24 }\) = –\(\frac { 1 }{ 4 }\) (critical points)

∴ The intervals are (∞, \(\frac { 1 }{ 4 }\)) and (\(\frac { 1 }{ 4 }\), ∞) f”(x) > 0

In the interval (-∞, –\(\frac { 1 }{ 4 }\)), f”(x) < 0 ⇒ curve is concave down.

In the interval (-\(\frac { 1 }{ 4 }\), ∞), f”(x) > 0 ⇒ curve is concave up.

The curve is concave upward in (-\(\frac { 1 }{ 4 }\), ∞) and concave downward in (-∞, –\(\frac { 1 }{ 4 }\))

f”(x) changes its sign when passing through x = –\(\frac { 1 }{ 4 }\)