This set of Engineering Mathematics Questions and Answers for Experienced people focuses on “Rolle’s Theorem – 2”.

1. Rolle’s Theorem tells about the

a) Existence of point c where derivative of a function becomes zero

b) Existence of point c where derivative of a function is positive

c) Existence of point c where derivative of a function is negative

d) Existence of point c where derivative of a function is either positive or negative

View Answer

Explanation: Statement of Rolle’s Theorem is that, If function f(x) attains same value at point ‘a’ and ‘b’ [f(a) = f(b)], and continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a,b) and f’(c) = 0.

2. Rolle’s Theorem is a special case of

a) Lebniz Theorem

b) Mean Value Theorem

c) Taylor Series of a function

d) Leibnit’x Theorem

View Answer

Explanation: According to Mean Value Theorem, If a function exist at pt. ‘a’, ‘b’ and continuous in closed interval [a, b] and differentiable in open interval (a, b) then there exists a point ‘c’, such that c∈(a,b), Where,

f’(c)= [f(b)-f(a)]/(b-a).

Hence, By putting f(b) = f(a) in the statement of Mean Value Theorem, we get

f’(c) = [f(b)-f(a)]/(b-a) = 0. Which is a statement of Rolle’s Theorem.

3. Rolle’s theorem is applicable to the

a) Functions differentiable in closed interval [a, b] and continuous in open interval (a, b) only and having same value at point ‘a’ and ‘b’

b) Functions continuous in closed interval [a, b] only and having same value at point ‘a’ and ‘b’

c) Functions continuous in closed interval [a, b] and differentiable in open interval (a, b) only and having same value at point ‘a’ and ‘b’

d) Monotonically Increasing funtions

View Answer

Explanation: Statement of Rolle’s Theorem is that, If function f(x) attains same value at point ‘a’ and ‘b’ [f(a) = f(b)], and continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a,b) and f’(c) = 0.

4. Find the value of c(a point where slope of a atangent to curve is zero) if f(x) = Sin(x) is continuous over interval [0,π] and differentiable over interval (0, π) and c ∈(0,π)

a) π

b) ^{π}⁄_{2}

c) ^{π}⁄_{6}

d) ^{π}⁄_{4}

View Answer

Explanation: Given, f(x)=Sin(x), x ∈ [0,π].

Now f(0) = f(π) = 0

f’(c) = Cos(c) = 0

c =

^{π}⁄

_{2}.

5. Find the value of c if f(x) = x(x-3)e^{3x}, is continuous over interval [0,3] and differentiable over interval (0, 3) and c ∈(0,3)

a) 0.369

b) 2.703

c) 0

d) 3

View Answer

Explanation: f(0) = 0

f(3) = 0

Hence, By rolle’s Theorem

f’(c) = (c-3) e

^{3c}+ c e

^{3c}+ 3c(c-3) e

^{3c}= 0

Hence, c-3 + c + 3c

^{2}-9c = 0

3c

^{2}– 7c – 3 = 0

c = 2.703, -0.369

Now c ∈(0,3), hence, c = 2.703.

6. Find the value of c if f(x) = sin^{3}(x)cos(x), is continuous over interval [0, ^{π}⁄_{2}] and differentiable over interval (0, ^{π}⁄_{2}) and c ∈(0, ^{π}⁄_{2})

a) 0

b) ^{π}⁄_{6}

c) ^{π}⁄_{3}

d) ^{π}⁄_{2}

View Answer

Explanation: f(x) = sin

^{3}(x)cos(x)

f(0) = 0

f(

^{π}⁄

_{2}) = 0

Hence,

f’(c) = 3sin

^{2}(c)cos(c)cos(c) – sin

^{4}(c) = 0

3sin

^{2}(c)cos

^{2}(c) – sin

^{4}(c) = 0

sin

^{2}(c)[3 cos

^{2}(c) – sin

^{2}(c)] = 0

either, sin

^{2}(c)=0 or 3 cos

^{2}(c) – sin

^{2}(c) = 0

Since sin

^{2}(c) cannot be zero because c cannot be 0

Hence, 3 cos

^{2}(c) – sin

^{2}(c)=0

tan

^{2}(c) = 3

tan(c) = √3

c =

^{π}⁄

_{3}.

7. Find value of c where f(x) = sin(x) e^{x} tan(x), c ∈ (0,∞)

a) Tan^{-1}[-(2+c^{2})/(1+c^{2})

b) Tan^{-1}[-(2-c^{2})/(1+c^{2})]

c) Tan^{-1}[(2+c^{2})/(1+c^{2})]

d) Rolle’s Theorem is not applied, Cannot find the value of c

View Answer

Explanation: Since, f(x) = e

^{x}sin(x) tan(x) is not continuous over interval (0,∞), Hnece Rolle’s, theorem is not applied.

8. f(x) = 3Sin(2x), is continuous over interval [0,π] and differentiable over interval (0,π) and c ∈(0,π)

a) π

b) ^{π}⁄_{2}

c) ^{π}⁄_{4}

d) ^{π}⁄_{8}

View Answer

Explanation: f(x) = 3Sin(2x)

f(0)=0

f(π)=0

Hence,

f’(c) = 6Cos(2c) = 0

c=

^{π}⁄

_{2}.

9. Find the value of ‘a’ if f(x) = ax^{2}+32x+4 is continuous over [-4, 0] and differentiable over (-4, 0) and satisfy the Rolle’s theorem. Hence find the point in interval (-2,0) at which its slope of a tangent is zero

a) 2, -2

b) 2, -1

c) 8, -1

d) 8, -2

View Answer

Explanation: Since it satisfies Rolle’s Theorem,

f’(c) = 0 = 2ac+32 ………………(1)

and, f(0) = 4 hence by Rolle’s theorem

and, f(-4) = 4 = 16a-128+4 (because f(0)=f(-4) condition of rolle’s theorem)

⇒ a = 8

from, eq.(1)

⇒ c = -2.

10. Find the value of ‘a’ & ‘b’ if f(x) = ax^{2} + bx + sin(x) is continuous over [0, π] and differentiable over (0, π) and satisfy the Rolle’s theorem at point c = ^{π}⁄_{4}.

a) 0.45,1.414

b) 0.45,-1.414

c) -0.45,1.414

d) -0.45,-1.414

View Answer

Explanation: Since function f(x) is continuous over [0,π] and satisfy rolle’s theorem,

⇒ f(0) = f(π) = 0

⇒ f(π) = a π

^{2}+ b π=0

⇒ a π+b=0 ………………….(1)

Since it satisfies rolle’s theorem at c =

^{π}⁄

_{4}

f’(c) = 2ac + b + Cos(c) = 0

⇒ a(

^{π}⁄

_{2}) + b +

^{1}⁄

_{√2}= 0 ………………..(2)

From eq(1) and eq(2) we get,

⇒ a = 0.45

⇒ b = -1.414.

11. Find value of c(a point in f(x) where slope of tangent to curve is zero) where

f(x) = \(\begin{cases}Tan(x) & 0<x<π/4\\Cos(x) & π/4<x<π/2\end{cases}\), given c ∈(0,π/2)

a) ^{π}⁄_{4}

b) Rolle’s Theorem is not applied, because function is not continuous in interval [0, ^{π}⁄_{2}]

c) Rolle’s Theorem is not applied, because function is not differential in interval (0, ^{π}⁄_{2})

d) Function is both continuous and differentiable but Rolle’s theorem is not applicable as f(0) ≠ f(^{π}⁄_{2})

View Answer

Explanation: Continuity Check.

\(\lim_{x\rightarrowπ/4-}f(x) = \lim_{x\rightarrowπ/4-}Tan(x) = 1\)

\(\lim_{x\rightarrowπ/4+}f(x) = \lim_{x\rightarrowπ/4+}Cos(x) = \frac{1}{\sqrt{2}}\)

\(\lim_{x\rightarrowπ/4-}f(x) ≠ \lim_{x\rightarrowπ/4+}f(x)\)

Hence function is discontinuous in interval (0,

^{π}⁄

_{2}).

Hence Rolle’s theorem cannot be applied.

12. Find value of c(a point in a curve where slope of tangent to curve is zero) where

f(x) = \(\begin{cases}x^2-x & 0<x<1\\3x^3-4x+1 & 1<x<2\end{cases}\), Given c ∈(0,2)

a) 1.5

b) Rolle’s Theorem is not applied, because function is not continuous in interval [0,2]

c) Rolle’s Theorem is not applied, because function is not differential in interval (0,2)

d) Function is both continuous and differentiable but Rolle’s theorem is not applicable as f(0) ≠ f(2)

View Answer

Explanation: Continuity Check

\(\lim_{x\rightarrow1-}f(x) = \lim_{x\rightarrow 1-}x^2-x = 0\)

\(\lim_{x\rightarrow1+}f(x) = \lim_{x\rightarrow 1+}3x^3-4x+1 = 0\)

\(\lim_{x\rightarrowπ/4-}f(x)= \lim_{x\rightarrow π/4+}f(x)\)

Hence function is Continuous.

Differentiability Check

\(\lim_{x\rightarrow 1-}f'(x) = \lim_{x\rightarrow 1-}2x-1 = 1\)

\(\lim_{x\rightarrow 1+}f(x) = \lim_{x\rightarrow 1+}9x^2-4 = 5\)

\(\lim_{x\rightarrow π/4-}f(x) ≠ \lim_{x\rightarrow π/4+}f(x)\)

Hence function is not differentiable so Rolle’s Theorem cannot be applied.

13. f(x) = ln(10-x^{2}), x=[-3,3], find the point in interval [-3,3] where slope of a tangent is zero,

a) 0

b) Rolle’s Theorem is not applied, because function is not continuous in interval [-3,3]

c) Rolle’s Theorem is not applied, because function is not differential in interval (-3,3)

d) 2

View Answer

Explanation: Domain of f(x) = [-√10, +√10]

Hence given f(x) is continuous in interval [-3,3]

f’(x) = \(\frac{-2x}{10-x^2}\)

⇒ x ≠ ±√10

⇒ Domain of f’(x) = (-∞,∞)- ±√10

⇒ Hence f(x) is differential in interval (-3,3)

f’(c) = -2c/(10-c

^{2}) = 0

⇒ c=0.

**Sanfoundry Global Education & Learning Series – Engineering Mathematics.**

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