250+ TOP MCQs on Lagrange’s Mean Value Theorem and Answers

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
Answer: a
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
Answer: b
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’
Answer: c
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
4) Leibnitz’s Theorem
Answer: 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).
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
Answer: b
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.

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)
Answer: d
Explanation: f(x) = xSin(x)
Since f₁(x) = x and f₂(x)=Sin(x) both are continuous in interval [0, ^π⁄₂], 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, ^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.