Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.3 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.3

Question 1.

Explain why Rolle’s theorem is not applicable to the following functions in the respective intervals.

(i) f(x) = |\(\frac { 1 }{ x }\)|, x ∈ [-1, 1]

(ii) f(x) = tan x, x ∈ [0, π]

(iii) f(x) = x – 2 log x, x ∈ [2, 7]

Solution:

(i) f(x) = |\(\frac { 1 }{ x }\)|, x ∈ [-1, 1]

f(-1) = 1

f(1) = 1

⇒ f(-1) = f(1) = 1

But f(x) is not differentiable at x = 0

∴ Rolle’s theorem is not applicable.

(ii) f (x) = tan x

f(x) is not continuous at x = \(\frac{\pi}{2}\).

So Rolle’s Theorem is not applicable.

(iii) f(x) = x – 2 log x

f(x) = x – 2 log x

f(2) = 2 – 2 log 2 = 2 – log 4

f(7) = 7 – 2 log 7 = 7 – log 49

f(2) ≠ f(7)

So Rolle’s theorem is not applicable.

Question 2.

Using the Rolle’s theorem, determine the values of x at which the tangent is parallel to the x-axis for the following functions:

(i) f(x) = x² – x, x ∈ [0, 1]

(ii) f(x) = \(\frac { x^2-2x }{ x+2 }\), x ∈ [-1, 6]

(iii) f(x) = √x – \(\frac { x }{ 3 }\), x ∈ [0, 9]

Solution:

(i) f(x) = x² – x, x ∈ [0, 1]

f(0) = 0, f(1) = 0

⇒ f(0) = f(1) = 0

f(x) is continuous on [0, 1]

f(x) is differentiable on (0, 1)

Now, f'(x) = 2x – 1

Since, the tangent is parallel to x-axis then

f'(x) = 0 ⇒ 2x – 1 = 0

x = \(\frac { x }{ 3 }\) ∈ (0, 1)

(ii) f(x) = \(\frac { x^2-2x }{ x+2 }\), x ∈ [-1, 6]

f(-1) = \(\frac { 1+2 }{ -1+2 }\) = 3

f(6) = \(\frac { 36-12 }{ 8 }\) = \(\frac { 24 }{ 8 }\) = 3

⇒ f (-1) = 3 = f(6)

f(x) is continuous on [- 1, 6]

f(x) is differentiable on (- 1, 6)

Now, f'(x)

Since the tangent is parallel to the x-axis.

f'(x) = 0

⇒ x² + 4x – 4 = 0

⇒ x = –\(\frac { 4±\sqrt{16+16} }{ 2 }\)

x = –\(\frac { 4±4√2 }{ 2 }\) = -2 ± 2√2

x = -2 ± 2√2 ∈ (-1, 6)

(iii) f(x) = √x – \(\frac { x }{ 3 }\), x ∈ [0, 9]

f(0) = 0, f(9) = √9 – \(\frac { 9 }{ 3 }\) = 3 – 3 = 0

⇒ f(0) = 0 = f(9)

f(x) is continuous on [0, 9]

f(x) is differentiable on (0, 9)

Now f'(x) = \(\frac { 1 }{ 2√x }\) – \(\frac { 1 }{ 3 }\)

Since, the tangent is parallel to x-axis.

f'(x) = 0

Question 3.

Explain why Lagrange’s mean value theorem is not applicable to the following functions in the respective intervals:

(i) f(x) = \(\frac { 1 }{ 2√x }\), x ∈ [-1, 2]

(ii) f(x) = |3x + 1|, x ∈ [-1, 3]

Solution:

(i) f(x) = \(\frac { 1 }{ 2√x }\), x ∈ [-1, 2]

f(0) = undefined

∴ f(x) is not continuous at x = 0

Hence, Lagrange’s mean value theorem is not applicable.

(ii) f(x) =|3x + 1|, x ∈ [-1, 3]

The function is not differentiable at x = \(\frac{-1}{3}\).

So Lagrange’s mean value theorem is not applicable in the given interval.

Question 4.

Using the Lagrange’s mean value theorem determine the values of x at which the tangent is parallel to the secant line at the end points of the given interval:

(i) f(x) = x³ – 3x + 2, x ∈ [-2, 2]

(ii) f(x) = (x – 2) (x – 7), x ∈ [3, 11]

Solution:

f(x) = x³ – 3x + 2, x ∈ [-2, 2]

f(x) is continuous in [- 2, 2]

f(x) is differentiable in (- 2, 2)

f(-2) = (-2)³ – 3 (-2) + 2 = – 8 + 6 + 2 = 0

f(2) = (2)³ -3(2) + 2 = 8 – 6 + 2 = 4

∴ f(x) is defined in the given interval.

Given that tangent is parallel to the secant line of the curve between x = -2 and x = 2.

(ii) f(x) = (x – 2)(x – 7), x ∈ [3, 11]

f(x) is continuous in [3, 11]

f(x) is differentiable in (3, 11)

f(3) = (3 – 2) (3 – 7) = (1) (-4) = -4

f(11) = (11 – 2) (11 – 7) = (9) (4) = 36

∴ f(x) is defined in the given interval.

Given that the tangent is parallel to the secant line ofthe curve between x = 3 and x = 11.

∴ f'(c) = \(\frac { f(b)-f(a) }{ b-a }\)

2c – 9 = \(\frac { 36+4 }{ 11-3 }\) where f'(x) = 2x – 9

2x – 9 = \(\frac { 40 }{ 8 }\) = 5

2c = 14 ⇒ c = 7 ∈ (3, 11)

∴ x = 7.

Question 5.

Show that the value in the conclusion of the mean value theorem for

(i) f(x) = \(\frac { 1 }{ x }\) on a closed interval of positive numbers [a, b] is \(\sqrt { ab }\)

(ii) f(x) = Ax² + Bx + C on any interval [a, b] is \(\frac { a+b }{ 2 }\)

Solution:

(ii) f(x) = Ax² + Bx + C, x ∈ [a, b]

f'(x) = 2Ax + B

By Mean Value Theorem,

Question 6.

A race car driver is racing at 20th km. If his speed never exceeds 150 km/hr, what is the maximum distance he can cover in the next two hours?

Solution:

Let a = 0, b = 2 and the interval is [0, 2] and f(0) = 20 (given)

We need to find f(2)

By Lagrange’s Mean Value Theorem,

f(b) – f(a) ≤ f'(c) [b – a]

f(b) – 20 ≤ 150(2 – 0)

f(b) ≤ 300 + 20

f(b) ≤ 320

∴ Maximum distance f(2) = 320 km.

Question 7.

Suppose that for a function f(x), f'(x) ≤ 1 for all 1 ≤ x ≤ 4. Show that f(4) – f(1) ≤ 3.

Solution:

Given: For f(x), f'(x) ≤ 1 for all 1 ≤ x ≤ 4

∴ a = 1, b = 4.

By Lagrange’s Mean Value Theorem,

f(b) – f(a) ≤ f'(c) (b – a)

f(4) – f(1) ≤ 1(4 – 1)

f(4) – f(1) ≤ 3

Hence Proved.

Question 8.

Does there exist a differentiable function f(x) such that f(0) = -1, f(2) = 4 and f (x) ≤ 2 for all x. Justify your answer.

Solution:

Given:For f(x), f'(x) ≤ 2, f(0) = -1, f(2) = 4

∴ a = 0, b = 2

By Lagrange’s Mean Value Theorem,

f(b) – f(a) ≤ f'(c)(b – a)

f(2) – f(0) ≤ f'(c) (2 – 0)

\(\frac { 4+1 }{ 2 }\) ≤ f'(c) ⇒ \(\frac { 5 }{ 2 }\) ≤ f'(c) ≤ 2 (given)

f(x) cannot be a differentiable function in (0, 2) as f'(x) cannot be 2.5.

Question 9.

Show that there lies a point on the curve f(x) = x(x + 3)e^{-π/2}, -3 ≤ x ≤ 0 where tangent drawn is parallel to the x-axis.

Solution:

f(x) = x(x + 3)e^{-π/2}, -3 ≤ x ≤ 0

f(x) is continuous in [-3, 0]

f(x) is differentiable in (- 3, 0)

f(-3) = -3 (-3 + 3)e^{-π/2} = 0

f(0) = 0

⇒ f(-3) = f(0) = 0

Since the tangent is parallel to x-axis.

f'(c) = 0

e^{-π/2} (2c + 3) = 0 where f'(x) = e^{-π/2} (2x + 3)

2c + 3 = 0

c = –\(\frac { 3 }{ 2 }\) ∈ (-3, 0)

∴ The point lies on the curve.

Question 10.

Using Mean Value Theorem prove that for, a > 0, b > 0, |e^{-a} – e^{-b}| < |a – b|

Solution:

Let f(x) = e^{-x}

f'(x) = e^{-x}

By Lagrange’s Mean Value Theorem,

Hence Proved.