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

1. Mean Value Theorem tells about the

a) Existence of point c in a curve where slope of a tangent to curve is equal to the slope of line joining two points in which curve is continuous and differentiable

b) Existence of point c in a curve where slope of a tangent to curve is equal to zero

c) Existence of point c in a curve where curve meets y axis

d) Existence of point c in a curve where curve meets x axis

View Answer

Explanation: Statement of Mean Value Theorem is that, If function f(x) is 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) = [f(b)-f(a)]/(b-a).

2. If f(a) is euquals to f(b) in Mean Value Theorem, then it becomes

a) Lebniz Theorem

b) Rolle’s 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. Mean Value theorem is applicable to the

a) Functions differentiable in closed interval [a, b] and continuous in open interval (a, 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)

d) Functions differentiable in open interval (a, b) only and having same value at point ‘a’ and ‘b’

View Answer

Explanation: Statement of Mean Value Theorem is that, If function f(x) is 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) = [f(b)-f(a)]/(b-a).

4. Mean Value theorem is also known as

a) Rolle’s Theorem

b) Lagrange’s Theorem

c) Taylor Expansion

d) Leibnitz’s Theorem

View Answer

Explanation: Statement of Mean Value Theorem is that, If function f(x) is 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) = [f(b)-f(a)]/(b-a).

It is also known as Lagrange’s Theorem.

5. Find the point c in the curve f(x) = x^{3} + x^{2} + x + 1 in the interval [0, 1] where slope of a tangent to a curve is equals to the slope of a line joining (0,1)

a) 0.64

b) 0.54

c) 0.44

d) 0.34

View Answer

Explanation: f(x) = x

^{3}+ x

^{2}+ x + 1

f(x) is continuous in given interval [0,1].

f’(x) = 3x

^{2}+2x+1

Since, value of f’(x) is always finite in interval (0, 1) it is differentiable in interval (0, 1).

f(0) = 1

f(1) = 4

By mean value theorem,

f’(c) = 3c

^{2}+ 2c + 1 = (4-1)/(1-0) = 3

⇒ c = 0.548,-1.215

Since c belongs to (0, 1) c = 0.54.

6. Find the equation of curve whose roots gives the point which lies in the curve f(x) = xSin(x) in the interval [0, ^{π}⁄_{2}] where slope of a tangent to a curve is equals to the slope of a line joining (0, ^{π}⁄_{2})

a) c = -Sec(c) – Tan(c)

b) c = -Sec(c) – Tan(c)

c) c = Sec(c) +Tan(c)

d) c = Sec(c) – Tan(c)

View Answer

Explanation: f(x) = xSin(x)

Since f

_{1}(x) = x and f

_{2}(x)=Sin(x) both are continuous in interval [0,

^{π}⁄

_{2}], the curve f(x)=f

_{1}(x)f

_{2}(x) is also continuous.

f’(x) = xCos(x) + Sin(x)

f’(x) always have finite value in interval [0,

^{π}⁄

_{2}] hence it is differentiable in interval (0,

^{π}⁄

_{2}).

f(0) = 0

f(

^{π}⁄

_{2}) =

^{π}⁄

_{2}

By mean value theorem,

f’(c) = cCos(c) + Sin(c) = (^{π}⁄_{2} – 0)/(^{π}⁄_{2} – 0)=1

Hence, c = Sec(c) – Tan(c) is the required curve.

7. Can Mean Value Theorem be applied in the curve

f(x)=\(\begin{cases}3sin(x)&0<x<\pi/4 \\ sin(x)-cos(x)&\pi/4<x<\pi/2\end{cases}\)

a) True

b) False

View Answer

Explanation: Continuity Check

\(\lim_{x\rightarrow π/4-}f(x) = 3Sin(\frac{π}{4}) = 3/\sqrt{2}\)

\(\lim_{x\rightarrow π/4+)}f(x) = Sin(π/4)- Cos(π/4) = 0\)

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

Function f(x) is not continuous hence mean value theorem cannot be applied.

8. Find point c between [2,9] where, the slope of tangent to the function f(x)=1+\(\sqrt[3]{x-1}\) at point c is equals to the slope of a line joining point (2,f(2)) and (9,f(9)).

(Providing given function is continuous and differentiable in given interval).

a) -2.54

b) 4.56

c) 4.0

d) 4.9

View Answer

Explanation: Since the given function is continuous and differentiable in a given interval,

f(2) = 2

f(9) = 3

Applying mean value theorem,

f’(c) = 1/3\(\sqrt[3]{x-1}\)(c-1)

^{2}= [f(9)-f(2)]/(9-2) = 1/7

c = 1 ± (7/3)

^{(3/2)}

c = 4.56,-2.54

Since c lies in (2,9), c = 4.56.

9. Find point c between [-1,6] where, the slope of tangent to the function f(x) = x^{2}+3x+2 at point c is equals to the slope of a line joining point (-1,f(-1)) and (6,f(6)).

(Providing given function is continuous and differentiable in given interval).

a) 2.5

b) 0.5

c) -0.5

d) -2.5

View Answer

Explanation: Since the given function is continuous and differentiable in a given interval,

f(-1) = 0

f(6) = 56

Applying mean value theorem,

f’(c) = 2c+3 = [f(6)-f(-1)]/[6-(-1)] = 56/7 = 8

c = 5/2

c = 2.5.

10. If f(x) = Sin(x)Cos(x) is continuous and differentiable in interval (0, x) then

a) 1<\(\frac{Cos(x)Sin(x)}{x}\) <Sin(2x)

b) 1<\(\frac{Cos(x)Sin(x)}{x}\) <Cos(2x)

c) 1<\(\frac{Cos(x)Sin(x)}{x}\) <xCos(2x)

d) 1<\(\frac{Cos(x)Sin(x)}{x}\) <1+Cos(2x)

View Answer

Explanation: f(x) = sin(x)cos(x)

Given f(x) is continuous and differentiable in interval (0, x),

Applying mean value theorem in interval (0, x)

f’(c) = Cos(2c) = [f(x)-f(0)]/[x-0] = \(\frac{Cos(x)Sin(x)}{x}\) ……………………. (1)

Now, Given

0 < c < x

Multiplying by 2 and taking Cos, We get

1 < Cos(2c) < Cos(2x)

1 < \(\frac{Cos(x)Sin(x)}{x}\) < Cos(2x).

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

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