# Engineering Mathematics Questions and Answers – Lagrange’s Mean Value Theorem – 2

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

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

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’

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

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) = x3 + x2 + 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

Explanation: f(x) = x3 + x2 + x + 1
f(x) is continuous in given interval [0,1].
f’(x) = 3x2+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) = 3c2 + 2c + 1 = (4-1)/(1-0) = 3
⇒ c = 0.548,-1.215
Since c belongs to (0, 1) c = 0.54.
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6. Find the equation of curve whose roots gives the point which lies in the curve f(x) = xSin(x) in the interval [0, π2] where slope of a tangent to a curve is equals to the slope of a line joining (0, π2)
a) c = -Sec(c) – Tan(c)
b) c = -Sec(c) – Tan(c)
c) c = Sec(c) +Tan(c)
d) c = Sec(c) – Tan(c)

Explanation: f(x) = xSin(x)
Since f1(x) = x and f2(x)=Sin(x) both are continuous in interval [0, π2], the curve f(x)=f1(x)f2(x) is also continuous.
f’(x) = xCos(x) + Sin(x)
f’(x) always have finite value in interval [0, π2] hence it is differentiable in interval (0, π2).
f(0) = 0
f(π2) = π2

By mean value theorem,
f’(c) = cCos(c) + Sin(c) = (π2 – 0)/(π2 – 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

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

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) = x2+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

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)

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).

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