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)e3x, 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) e3c + c e3c + 3c(c-3) e3c = 0
Hence, c-3 + c + 3c2 -9c = 0
3c2 – 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) = sin3(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) = sin3(x)cos(x)
f(0) = 0
f(π⁄2) = 0
Hence,
f’(c) = 3sin2(c)cos(c)cos(c) – sin4(c) = 0
3sin2(c)cos2(c) – sin4(c) = 0
sin2(c)[3 cos2(c) – sin2(c)] = 0
either, sin2(c)=0 or 3 cos2(c) – sin2(c) = 0
Since sin2(c) cannot be zero because c cannot be 0
Hence, 3 cos2(c) – sin2(c)=0
tan2(c) = 3
tan(c) = √3
c = π⁄3.
7. Find value of c where f(x) = sin(x) ex tan(x), c ∈ (0,∞)
a) Tan-1[-(2+c2)/(1+c2)
b) Tan-1[-(2-c2)/(1+c2)]
c) Tan-1[(2+c2)/(1+c2)]
d) Rolle’s Theorem is not applied, Cannot find the value of c
View Answer
Explanation: Since, f(x) = exsin(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) = ax2+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) = ax2 + 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 π2 + 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-x2), 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-c2) = 0
⇒ c=0.
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