Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f)

I.

Question 1.

Verify Rolle’s theorem for the following functions.

i) x² – 1 on [-1, 1]

Solution:

Let f(x) = x² – 1

f is continuous on [-1,1]

since f(-1) = f(1) = 0 and

f is differentiable on [-1, 1]

∴ By Rolle’s theorem ∃ c ∈ (-1, 1) such that f'(c) = 0

f'(x) = 2x = 0

∴ = f'(c) = 0

2c = 0

c = 0

The point c = 0 ∈ (-1, 1)

Then Rolle’s theorem is verified.

ii) sin x – sin 2x on [0, π].

Solution:

Let f(x) = sin x – sin 2x

f is continuous on [0, π]

since f(0) = f(π)= 0 and

f is differentiable on [0, π]

By Rolle’s theorem ∃ c ∈ (0, π)

such that f'(c) = 0

f'(x) = cos x – 2 cos 2x

f'(c) = 0 ⇒ cosc – 2 cos 2c = 0

⇒ cosc – 2(2cos²c – 1):

cosc – 4 cos²c + 2=0 2

4 cos² c – cosc – 2 = 0

iii) log (x² + 2) – log 3 on [-1, 1]

Solution:

Let f(x) = log (x² + 2) – log 3

f is continuous on [-1, 1]

Since f(-1) = f(1) = 0 and f is

Differentable on [-1, 1]

By Rolle’s theorem ∃ c ∈ (-1, 1)

Such that f'(c) = 0

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

f'(x) = \(\frac{2c}{c^{2}+2}\) = 0

2c = 0

c = 0

c = 0 ∈ (-1, 1).

Question 2.

It is given that Rolle’s theorem holds for the function f(x) = x³ + bx² + ax on [1, 3] with c = 2t + \(\frac{1}{\sqrt{3}}\). Find the values of a and b.

Solution:

Given f(x) = x³ + bx² + ax

f'(x) = 3x² + 2bx + a

∴ f'(c) = 0 6 ⇔ 3c² + 2bc + a = 0

⇔ b = 6 and b² – 3a = 3

36 – 3 = 3a

33 = 3a

a = 11

Hence a = 11 and b = -6.

Question 3.

Show that there is no real number k, for which the equation x² – 3x + k = 0 has two distinct roots in [0, 1].

Solution:

Clearly f(0) = f(c)

0 – 0 + k = 1 – 3 + k

0 = -2

Which is not possible

∴ There is no real number K.

Question 4.

Find a point on the graph of the curve y = (x – 3)², where the tangent is parallel to the chord joining (3, 0) and (4, 1).

Solution:

Given points (3, 0) and (4, 1)

The slope of chord = \(\frac{1-0}{4-3}\) = 1

Given y = (x- 3)²

\(\frac{dy}{dx}\) = 2(x – 3)

⇒ Slope = 2(x – 3)

1 = 2(x – 3)

\(\frac{1}{2}\) = x – 3

x = \(\frac{1}{2}\) + 3 = \(\frac{7}{2}\)

y = (x – 3)² = (\(\frac{7}{2}\) – 3)² = \(\frac{1}{4}\)

∴ The point on the curve is (\(\frac{7}{2}\), \(\frac{1}{4}\))

Question 5.

Find a point on the graph of the curve y = x³, where the tangent is parallel to the chord joining (1, 1) and (3, 27).

Solution:

Given points (1, 1) and (3, 27)

Slope of chord = \(\frac{27-1}{3-1}\) = 13

Given y = x³

\(\frac{dy}{dx}\) = 3x²

⇒ Slope = 3x²

13 = 3x²

∴ The point on the curve is (\(\frac{\sqrt{39}}{3}\), \(\frac{13\sqrt{39}}{9}\))

Question 6.

Find ‘c’, so that f'(c) = \(\frac{f(b)-f(a)}{b-a}\) in the following cases.

i) f(x) = x² – 3x – 1, a = \(\frac{-11}{7}\), b = \(\frac{13}{7}\).

Solution:

f'(x) = 2x – 3

f'(c) = 2c – 3

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

ii) f(x) = e^{x} ; a = 0, b = 1

Solution:

f(b) = f(1) = e’ = e

f(a) = f(0) = e° = 1

Given f(x) = e^{x}

f'(x) = e^{x}

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

Question 7.

Verify the Rolle’s theorem for the function (x² – 1) (x – 2) on [-1, 2]. Find the point in the interval where the derivate vanishes.

Solution:

Let f(x) = (x² – 1) (x- 2) = x³ – 2x² – x + 2

f is continous on [-1, 2]

since f(-1) = f(2) = 0 and f is

Differentiable on [-1, 2]

By Rolle’s theorem ∃ C ∈ (-1, 2)

Let f'(c) = 0

f'(x) = 3x² – 4x – 1

3c² – 4c – 1 = 0

Question 8.

Verify the conditions of the Lagrange’s mean value theorem for the following functions. In each case find a point ‘c’ in the interval as stated by the theorem

i) x² -1 on [2, 3]

Solution:

Solution:

Let f(x) = x² – 1

f is continous on [2, 3]

and f is differentiable

Given f(x) = x² – 1

f'(x) = 2x

By Lagrange’s mean value theorem ∃ C ∈ (2, 3) such there

ii) sin x – sin 2x on [0, π]

Solution:

Let f(x) = sin x – sin 2x

f is continuous on [0, π] and f is differentiable

Given f(x) = sin x – sin 2x

f'(x) = cos x – 2 cos 2x

By Lagrange’s mean value than ∃ C ∈ (0, π) such there

f'(c) = \(\frac{f(\pi)-f(0)}{\pi-0}\)

cosc – 2 cos 2c = 0

cosc 2(2cos² – 1) = 0

cosc – 4 cos²c + 2 = 0

4 cos² c – cos c – 2 = 0

iii) log x on [1, 2].

Solution:

Let f(x) = log x

f is continuous on [1, 2] and f is differentiable

Given f(x) = log x

f'(x) = \(\frac{1}{x}\)

By Lagrange’s mean value theorem ∃ c ∈ (1, 2) such that