Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on Rolle’s Theorem and Answers

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
Answer: a
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
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. 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
Answer: c
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
Answer: b
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
Answer: b
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
Answer: c
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
Answer: d
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
Answer: b
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
Answer: d
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
Answer: b
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) & 0a) 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)
Answer: b
Explanation: Continuity Check.
(lim_{xrightarrowπ/4-}f(x) = lim_{xrightarrowπ/4-}Tan(x) = 1)
(lim_{xrightarrowπ/4+}f(x) = lim_{xrightarrowπ/4+}Cos(x) = frac{1}{sqrt{2}})
(lim_{xrightarrowπ/4-}f(x) ≠ lim_{xrightarrowπ/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 & 0a) 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)
Answer: c
Explanation: Continuity Check
(lim_{xrightarrow1-}⁡f(x) = lim_{xrightarrow 1-}x^2-x = 0)
(lim_{xrightarrow1+}f(x) = lim_{xrightarrow 1+}3x^3-4x+1 = 0)
(lim_{xrightarrowπ/4-}f(x)= lim_{xrightarrow π/4+}f(x))
Hence function is Continuous.
Differentiability Check
(lim_{xrightarrow 1-}f'(x) = lim_{xrightarrow 1-}2x-1 = 1)
(lim_{xrightarrow 1+}f(x) = lim_{xrightarrow 1+}9x^2-4 = 5)
(lim_{xrightarrow π/4-}f(x) ≠ lim_{xrightarrow 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
Answer: a
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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250+ TOP MCQs on Curvature and Answers

Engineering Mathematics Multiple Choice Questions on “Curvature”.

1. The curvature of a function f(x) is zero. Which of the following functions could be f(x)?
a) ax + b
b) ax2 + bx + c
c) sin(x)
d) cos(x)
Answer: a
Explanation: The expression for curvature is
k=(left |frac{f”(x)}{(1+[f'(X)]^2)^{frac{3}{2}}} right |)
Given that k = 0 we have
f” (x) = 0
f(x) = a + b.

2. The curvature of the function f(x) = x2 + 2x + 1 at x = 0 is?
a) 3⁄2
b) 2
c) (left |frac{2}{5^{frac{3}{2}}} right |)
d) 0
Answer: c
Explanation: The expression as we know is
k=(left |frac{f”(x)}{(1+[f'(X)]^2)^{frac{3}{2}}} right |)
Substituting f(x)=x2+2x+1 we have
k=(left |frac{2}{(1+[2x+2]^2)}^{frac{3}{2}} right |)
Put x=0 to get
k=(left |frac{2}{(1+[2]^2)^{frac{3}{2}}} right |)
k=(left |frac{2}{5^{frac{3}{2}}} right |)

3. The curvature of a circle depends inversely upon its radius r.
a) True
b) False
Answer: a
Explanation:Using parametric form of circle x = r.cos(t) : y = r.sin(t)
Using curvature in parametric form k=(left |frac{y”x’-y’x”} {((x’)^2+(y’)^2)^{frac{3}{2}}}right |) we have
k=(left |frac{(-rsin(t))(-rsin(t))-(-rcos(t))(rcos(t))} {((-rsin(t))^2+(rcos(t))^2)^{frac{3}{2}}}right |)
k=(left |frac{r^2(sin^2(t)+cos^2(t))} {r^3(sin^2(t)+cos^2(t))^{frac{3}{2}}}right |)
k=(left |frac{1}{r}right |)

4. Find the curvature of the function f(x) = 3x3 + 4680x2 + 1789x + 181 at x = -520.
a) 1
b) 0
c) ∞
d) -520
Answer: b
Explanation: For a Cubic polynomial the curvature at x = -b⁄3a is zero because f”(x) is zero at that point.
Looking at the form of the given point we can see that x = -4680⁄3*3 = -520
Thus, curvature is zero.

5. Let c(f(x)) denote the curvature function of given curve f(x). The value of c(c(f(x))) is observed to be zero. Then which of the following functions could be f(x).
a) f(x) = x3 + x + 1
b) f(x)2 + y2 = 23400
c) f(x) = x19930 + x + 90903
d) No such function exist
Answer: b
Explanation: We know that the curvature of a given circle is a constant function. Further, the curvature of any constant function is zero. Thus, we have to choose the equation of circle from the options.

6. The curvature of the function f(x) = x3 – x + 1 at x = 1 is given by?
a) ∣6⁄5∣
b) ∣3⁄5∣
c) (left |frac{6}{5^{frac{3}{2}}} right |)
d) (left |frac{3}{5^{frac{3}{2}}} right |)
Answer: c
Explanation: Using expression for curvature we have
k=(left |frac{f”(x)}{(1+[f'(X)]^2)^{frac{3}{2}}} right |)
Substituting f(x)=x3-x+1 we have
k=(left |frac{2}{(1+[3x^2-1]^2)^{frac{3}{2}}} right |)
Put x=1 to get
k=(left |frac{6}{5^{frac{3}{2}}} right |)

7. The curvature of a function depends directly on leading coefficient when x=0 which of the following could be f(x)?
a) y = 323x3 + 4334x + 10102
b) y = x5 + 232x4 + 232x2 + 12344
c) y = ax5 + c
d) y = 33x2 + 112345x + 8945
Answer: d
Explanation: Using formula for curvature
k=(left |frac{f”(x)}{(1+[f'(X)]^2)^{frac{3}{2}}} right |)
Observe numerator which is f”(x)
Now this second derivative must be non zero for the above condition asked in the question
Looking at all the options we see that only quadratic polynomials can satisfy this.

8. Given x = k1ea1t : y = k2ea2t it is observed that the curvature function obtained is zero. What is the relation between a1 and a2?
a) a1 ≠ a2
b) a1 = a2
c) a1 = (a2)2
d) a2 = (a1)2
Answer: b
Explanation: Using formula for Curvature in parametric form
k=(left |frac{y”x’-y’x”}{((x’)^2+(y’)^2)^{frac{3}{2}}} right |)
Equating y”x’-y’x” to zero (given curvature function is zero) we get
(frac{y”}{y’}=frac{x”}{x’})
Put x=k1ea1t:y=k2ea2t
(frac{k_1(a_1)^2.e^{a_1t}}{k_1.a_1.e^{a_1t}}=frac{k_2(a_2)^2.e^{a_2t}}{k_2.a_2.e^{a_2t}})
a1=a2

9. The curvature function of some function is given to be k(x) = (frac{1}{[2+2x+x^2]^{frac{3}{2}}}) then which of the following functions could be f(x)?
a) x2⁄2 + x + 101
b) x2⁄4 + 2x + 100
c) x2 + 13x + 101
d) x3 + 4x2 + 1019
Answer: a
Explanation:The equation for curvature is
k(x) = (left |frac{f”(x)}{(1+[f'(X)]^2)^{frac{3}{2}}} right |)
Observe that there is no term in the numerator of given curvature function. Hence, the function has to be of the quadratic form. Using f(x) = ax2+bx+c we have
k=(left |frac{2a}{[1+(2a+b)^2]^{frac{3}{2}}} right |=left |frac{1}{[2+2x+x^2]^{frac{3}{2}}} right |)
(left |frac{2a}{[1+(2ax+b)^2]^{frac{3}{2}}} right |=left |frac{1}{[1+(x+1)^2]^{frac{3}{2}}} right |)
2a = 1; b = 1
The values of a, b are
a = 1⁄2; b = 1
c Could be anything.

10. Consider the curvature of the function f(x) = ex at x=0. The graph is scaled up by a factor of and the curvature is measured again at x=0. What is the value of the curvature function at x=0 if the scaling factor tends to infinity?
a) a
b) 2
c) 1
d) 0
Answer: d
Explanation: If the scaling factor is a then the function can be written as f(x) = eax
Now using curvature formula we have
(left |frac{f”(x)}{(1+[f'(X)]^2)^{frac{3}{2}}} right | = left |frac{a^2.e^{ax}}{[1+a^2.e^{2ax}]^{frac{3}{2}}}right |)
Put x=0
=(left |frac{a^2}{[1+a^2]^{frac{3}{2}}}right |)
Now taking the limit as a → ∞ we get
=(lt_{arightarrowinfty}frac{a^2}{a^3.[1+frac{1}{a^2}]^{frac{3}{2}}})
=(lt_{arightarrowinfty}frac{1}{a.[1+frac{1}{a^2}]^{frac{3}{2}}})
=0

250+ TOP MCQs on Differentiation Under Integral Sign and Answers

Differential and Integral Calculus Multiple Choice Questions on “Differentiation Under Integral Sign”.

1. When solved by the method of Differentiation for the given integral i.e (int_0^∞ frac{x^{2}-1}{log⁡x} dx) the result obtained is given by _______
a) log⁡4
b) log⁡3
c) 2log⁡3
d) log⁡8
Answer: b
Explanation: To solve this problem let us assume the given function is dependent on α
Such that α=2 & thus (f (α) = int_0^∞ frac{x^{α}-1}{log⁡x} dx)
(f’(α) = int_0^1frac{∂}{∂α} left(frac{x^{α}-1}{log⁡x}right)dx) …..Leibnitz rule
( =int_0^1frac{x^α.log⁡x}{log⁡x} dx)
( = int_0^1 x^α dx = [frac{x^{α+1}}{α+1}]_0^1 = frac{1}{α+1})
We have (f’(α) = frac{1}{α+1})
Thus (f (α) = intfrac{1}{α+1}dα+c)
f (α) = log(α+1)+c
or f (α) = log(α+1) …… neglecting constant since the function is assumed
thus f (2) = log(2+1) = log(3).

2. Which among the following correctly defines Leibnitz rule of a function given by ( f (α) = int_a^b (x,α)dx) where a & b are constants?
a) (f’(α) = frac{∂}{∂α}int_a^b f (x,α) dx)
b) (f’(α) = frac{d}{dα} int_a^b f (α) dx)
c) (f’(α) = int_a^b frac{∂}{∂α} f (x,α) dx)
d) (f’(α) = int_a^b frac{d}{dα} f (x,α) dx)
Answer: c
Explanation: (f’(α) = int_a^b frac{∂}{∂α} f (x,α) dx = frac{d(f(α))}{dα} = frac{d}{dα} int_a^b f (x,α) dx.)

3. Which among the following correctly defines Leibnitz rule of a function given by
( f (α) = int_a^b (x,α)dx) where a & b are functions of α?
a) (f’(α) = int_a^b frac{∂}{∂α} f(x,α) dx)
b) (f’(α) = frac{d}{dα} int_a^b f(x,α) dx)
c) (f’(α) = int_a^b frac{∂}{∂α} f (x,α) dx + f(b, α) frac{da}{dα} – f(a, α) frac{db}{dα})
d) (f’(α) = int_a^b frac{∂}{∂α} f (x,α) dx + f(b, α) frac{db}{dα} – f(a, α) frac{da}{dα})
Answer: d
Explanation: (f’(α) = int_a^b frac{∂}{∂α} f (x,α) dx + f(b, α) frac{db}{dα} – f(a, α) frac{da}{dα}) when a & b are constants
(frac{da}{dα} & frac{da}{dα} = 0) which reduces the equation d into a

4. Given (f (a) = int_a^{a^2} frac{sin⁡ax}{x} dx ) what is the value of f’(a)?
a) (frac{sin⁡3a}{a})
b) (frac{3 sin⁡ a^3 – 2 sin a^2}{a})
c) (frac{3 sin a^2 – 4 sin a}{a})
d) (frac{3 sin a^3 – 3 sin^2 a}{6a})
Answer: b
Explanation: Applying the Leibniz rule equation given by
( f’ (α) = int_p^q frac{∂}{∂α} f (x,α) dx + f (q, α) frac{dq}{dα} – f (p, α) frac{dp}{dα} …..(1))
(f (x,a) = frac{sin⁡ax}{x}, p=a, q=a^2) & further obtaining
(f(q,a) = f(a^2,a) = frac{sin⁡ a^3}{a^2}, frac{dq}{da} = 2a)
(f(p,a) = f(a,a) = frac{sin⁡ a^2}{a}, frac{dp}{da} = 1)
substituting all these values in (1) we get
(f’(a) = int_a^{a^2} frac{∂}{∂α} (frac{sin⁡ax}{x})dx + frac{sin⁡ a^3}{a^2}.2a – frac{sin⁡ a^2}{a}.1)
(int_a^{a^2} frac{1}{x} (cos(ax))(x)+ frac{2 sin a^3 – sin^2 a}{a})
([frac{sin⁡ax}{x}]_a^{a^2} + frac{2 sin a^3 – sin^2 a}{a} = frac{sin⁡ a^3}{a} – frac{sin⁡ a^2}{a} + frac{2 sin a^3 – sin^2 a}{a})
thus (f’(a) = frac{3 sin⁡ a^3 – 2 sin a^2}{a}.)

5. When solved by the method of Differentiation for the given integral i.e. (int_0^1 frac{x^2-1}{log_2⁡x} dx ) the result obtained is given by _________
a) log⁡5
b) 3 log 3
c) log 4
d) 2 log 3
Answer: a
Explanation: (int_0^1 frac{x^2-1}{log_2⁡x} dx) can also be written as ((log⁡2 int_0^1 frac{x^2-1}{log ⁡x} dx)…….(1).)
Here during integration changing (log_2⁡x = frac{log⁡x}{log⁡2}) and substituting we get (1) since logarithm to base ‘e’ can be easily integratable.
To solve this problem let us assume the given function is dependent on α
Such that α=2 & thus (f (α) = log⁡2 int_0^∞ frac{x^{α}-1}{log⁡x} dx)
(f’(α) = log⁡2 int_0^1frac{∂}{∂α} left(frac{x^{α}-1}{log⁡x}right)dx) …..Leibniz rule
( = log⁡2 int_0^1frac{x^α.log⁡x}{log⁡x} dx)
( = log⁡2 int_0^1 x^α dx = [frac{x^{α+1}}{α+1}]_0^1 = frac{1}{α+1})
We have (f’(α) = log⁡2.frac{1}{α+1})
Thus (f (α) = log⁡2intfrac{1}{α+1}dα+c)
f (α) = log⁡2 .log(α+1)+c
or f (α) = log(α+1+2) neglecting constant since the function is assumed
thus f (2) = log(2+3) = log(5).

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250+ TOP MCQs on Change of Order of Integration: Double Integral and Answers

Differential and Integral Calculus Multiple Choice Questions on “Change of Order of Integration: Double Integral”.

1. Which of the following is not a property of double integration?
a) ∬ af(x,y)ds = a∬ f(x,y)ds, where a is a constant
b) ∬ (f(x,y)+g(x,y))ds = ∬f(x,y)ds+ ∬g(x,y)ds
c) (∬_0^Df(x,y)ds = ∬_0^{D1}f(x,y)ds+ ∬_{D1}^{D2}f(x,y)ds,) where D is union of disjoint domains D1 and D2
d) ∬(f(x,y)*g(x,y))ds = ∬f(x,y)ds*∬g(x,y)ds
View Answer

Answer: d
Explanation: The following are the properties of double integration:

  • ∬af(x,y)ds = a∬f(x,y)ds
  • ∬f(x,y)+g(x,y))ds = ∬f(x,y)ds+ ∬g(x,y)ds
  • (∬_0^Df(x,y)ds = ∬_0^{D1}f(x,y)ds+ ∬_{D1}^{D2}f(x,y)ds)

2. The region bounded by circle is an example of regular domain.
a) False
b) True
View Answer

Answer: b
Explanation: A domain D in the XY plane bounded by a curve c is said to be regular in the Y direction, if straight lines passing through an interior point and parallel to Y axis meets c in two points A and B. Hence, region bounded by circle is an example of regular domain.

3. What is the result of the integration (∫_3^4∫_1^2(x^2+y)dxdy)?
a) (frac{83}{6} )
b) (frac{83}{3} )
c) (frac{82}{6} )
d) (frac{81}{6} )
View Answer

Answer: a
Explanation: Given: (∫_3^4∫_1^2(x^2+y)dxdy)
Integrating with respect to y first, we get,
(∫_3^4(x^2(y)_1^2+(frac{y^2}{2})_1^2)dx= ∫_3^4(x^2+frac{3}{2}) dx)
Next integrating with respect to x, we get,
((frac{x^3}{3})_3^4+frac{3}{2}(x)_3^4= frac{37}{3}+frac{3}{2}=frac{83}{6})

4. Volume of an object expressed in spherical coordinates is given by (V = ∫_0^2π∫_0^frac{π}{3}∫_0^1 r cos∅ ,dr ,d∅ ,dθ.) The value of the integral is _______
a) (frac{√3}{2})
b) (frac{1}{√2} π)
c) (frac{√3}{2}π)
d) (frac{√3}{4} π)
View Answer

Answer: d
Explanation: Given: (V = ∫_0^2π∫_0^frac{π}{3}∫_0^1 r cos∅ ,dr ,d∅ ,dθ.)
( V = ∫_0^2π∫_0^{frac{π}{3}}(frac{r^2}{2})_0^1 cos∅, d∅, dθ)
( V = frac{1}{2} ∫_0^{2π}(sin∅)_0^frac{π}{3} d∅ ,dθ)
(V = frac{1}{2}×frac{√3}{2} ∫_0^2π dθ)
( V = frac{1}{2}×frac{√3}{2} ×2π)
( V = frac{√3}{2} π )

5. Which of the following equation represents Moment of Inertia of a plane region relative to x-axis?
a) ∬x2 f(x,y)dxdy
b) ∬xf(x,y)dxdy
c) ∬y2 f(x,y)dxdy
d) ∬yf(x,y)dxdy
View Answer

Answer: c
Explanation: Moment of Inertia of a plane region,
Relative to x-axis is given by,
Ixx=∬y2 f(x,y)dxdy
Relative to y-axis is given by,
Iyy=∬x2 f(x,y)dxdy

6. What is the mass of the region R as shown in the figure?
” alt=”” width=”360″ height=”311″ data-src=”2020/06/integral-calculus-questions-answers-change-order-integration-double-integral-q6″ data-srcset=”2020/06/integral-calculus-questions-answers-change-order-integration-double-integral-q6 360w, 2020/06/integral-calculus-questions-answers-change-order-integration-double-integral-q6-300×259 300w” data-sizes=”(max-width: 360px) 100vw, 360px” />
a) 8
b) 9
c) (frac{9}{2} )
d) (frac{9}{4} )
View Answer

Answer: b
Explanation: From the above figure, we can see that X-axis ranges from 0 to 3 and Y-axis ranges from 0 to 2.
Therefore, the mass of the region is given by,
(M = ∫_0^2∫_0^3xy ,dx,dy)
( = ∫_0^2y(frac{x^2}{2})_0^3 dy = frac{9}{2}(frac{y^2}{2})_0^2 = 9 )

7. The half-interval method in numerical analysis is also known as __________
a) Newton-Raphson method
b) Regula Falsi method
c) Taylor’s method
d) Bisection method
View Answer

Answer: d
Explanation: The Bisection method, also known as binary chopping or half-interval method, is a starting method which is used, where applicable, for few iterations, to obtain a good initial value.

8. Given (∫_0^8x^frac{1}{3}dx,) find the error in approximating the integral using Simpson’s 1/3 Rule with n=4.
a) 1.8
b) 2.9
c) 0.3
d) 0.35
View Answer

Answer: d
Explanation: Given: (∫_0^8x^frac{1}{3}dx,n = 3,)
Let (f(x)= x^frac{1}{3},)
(∆x = frac{b-a}{2}= frac{8-0}{2}=4) ………………since b=8, a=0 (limits of the given integral)
Hence endpoints xi have coordinates {0, 2, 4, 6, 8}.
Calculating the function values at xi, we get,
(f(0)= 0^frac{1}{3}=0)
(f(2)= 2^frac{1}{3})
(f(4)= 4^frac{1}{3})
(f(6)= 6^frac{1}{3})
(f(8)= 8^frac{1}{3} =2)
Substituting these values in the formula,
(∫_0^8x^frac{1}{3} dx ≈ frac{∆x}{3} [f(0)+4f(2)+2f(4)+4f(6)+f(8)])
( ≈frac{2}{3}[0+4(2^frac{1}{3})+2(4^frac{1}{3})+ 4(6^frac{1}{3})+2] ≈ 11.65)
Actual integral value,
(∫_0^8x^frac{1}{3} dx= left(frac{x^frac{4}{3}}{frac{4}{3}}right)_0^8=12)
Error in approximating the integral = 12 – 11.65 = 0.35

9. A sphere with the dimensions is shown in the figure. What is the error that can be incorporated in the radius such that the volume will not change more than 4%?
” alt=”” width=”197″ height=”195″ data-src=”2020/05/integral-calculus-questions-answers-change-order-integration-double-integral-q9″ />
a) 0.127%
b) 0.0127%
c) 12.7%
d)1.27%
View Answer

Answer: b
Explanation: We know that volume of the sphere is,
(V = frac{4}{3} πR^3 )
Differentiating the above equation with respect to R we get,
(frac{dV}{dR}= frac{4}{3} π×3R^2=4πR^2)
Since the volume of the sphere should not exceed more than 4%,
(dR=frac{dV}{4πR^2}=frac{0.04}{4π(5)^2}=0.000127)
Error in radius = 0.0127%

10. The x-coordinate of the center of gravity of a plane region is given by, (x_c=frac{1}{M}∬xf(x,y)dxdy.)
a) True
b) False
View Answer

Answer: a
Explanation: The coordinates (xc,yc) of the centroid of a plane region with mass M is given by,
(x_c=frac{1}{M} ∬xf(x,y)dxdy)
(y_c=frac{1}{M} ∬yf(x,y)dxdy)

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250+ TOP MCQs on Law of Natural Growth and Decay and Answers

Ordinary Differential Equations Interview Questions and Answers for Experienced people focuses on “Law of Natural Growth and Decay”.

1. The population of the Mysore city of a country was 30 lakhs in the year 2000 and 60 lakhs in the year 2010. What is the estimated population in the year 1990 with the help of the population model?
a) 18 lakhs
b) 15 lakhs
c) 20 lakhs
d) 12 lakhs
Answer: b
Explanation: Here rate of change of population is proportional to the population present at that instant of time i.e (frac{dp}{dt} ∝ P) i.e (frac{dp}{dt}=kP), k is a constant of proportionality.
solving using the variable separable integral form we get (int frac{1}{P} ,dP = int K ,dt + a)
log P=kt+c –> ekt+a = P or P=cekt…where ea is a constant let the year 2000 be the initial reference year –> P(0)=30=cek(0) = c i.e c = 30 & P(10) = cek10
=30ek10 = 60 –> log 60 = log 30 + 10k –> log 6⁄3 =log 2 = 10k –> k = 0.1*0.693 = 0.0693
to find P(-10) = cekt = 30*e-0.693=15 lakhs.

2. Urine culture test is being conducted in the pathology lab from the sample of a patient to prescribe a suitable drug based on bacterial growth. There were 3000 bacteria in the initial sample and it had increased to 3400 in the sample collected after one hour. Based on the exponential growth how long will it take for the bacterial population to reach 10000?
a) 10 hrs
b) 8.5 hrs
c) 9.6 hrs
d) 3.6 hrs
Answer: c
Explanation: As we know according to law of exponential growth, the bacterial population model is
B(t)=cekt at t=0 B(0)=ce0=3000……given –> c=3000 and B(1)=3400=cek=3000*ek
ek=(frac{17}{15}) taking log, k=log 17 – log 15 = 0.125, to find t when B(t)=10000 i.e
10000=3000*ekt = 3000*e0.125t –> 3.33=e0.125t –> log 3.33 = 0.125t –> t=9.62 hrs.

3. Uranium disintegrates at a rate proportional to the amount present at any time instant. If a and b grams of uranium are present at times t’ & t’’ respectively. What is the expression for the half life of uranium?
a) (frac{(t”+t’)log2}{log⁡(frac{b}{a})} )
b) (frac{(t”-2t’)log2}{2log(frac{a}{b})} )
c) (frac{(t”+t’)}{2log(frac{a}{b})} )
d) (frac{(t”-t’)log2}{log⁡(frac{b}{a})} )
Answer: d
Explanation: Let P grams of uranium be present at any time t. According to law of decay, (frac{dp}{dt} = -kP…k) is a constant its solution is P=ce-kt given that at t=t’ P=a
a=ce-kt‘…..(1) and b=ce-kt‘….(2) dividing (1) & (2) we get (frac{a}{b}=e^{k(t^{”}-t’)})
taking log on both side (log(frac{a}{b})) = k(t’’-t’)…(3) at t=0 ‘c’ is the initial mass of uranium present in the sample thus at half life i.e t=T ,P=c/2 P = P=ce-kt becomes (frac{c}{2}) = ce-kT
log 2=kT substituting the value of k from (3) we get (T = frac{(t”-t’)log2}{log⁡(frac{b}{a})} ) is the equation for half life period.

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250+ TOP MCQs on Solution of DE With Constant Coefficients using the Laplace Transform and Answers

Ordinary Differential Equations Multiple Choice Questions & Answers focuses on “Solution of DE With Constant Coefficients using the Laplace Transform”.

1. While solving the ordinary differential equation using unilateral laplace transform, we consider the initial conditions of the system.
a) True
b) False
Answer: a
Explanation: When bilateral laplace transformation is used in solving differential equations, we don’t consider the initial conditions as the transformation is from -∞ to +∞. But when we consider unilateral laplace transformation, the integral is from 0 to ∞. So, the initial conditions are considered.

2. With the help of _____________________ Mr.Melin gave inverse laplace transformation formula.
a) Theory of calculus
b) Theory of probability
c) Theory of statistics
d) Theory of residues
Answer: d
Explanation: Let f(t) be the function in time. The laplace transformation of the function is L[f(t)] = F(s). So, the inverse laplace transform of F(s) comes out to be the function f(t) in time. The formula for laplace transform is derived using the theory of residues by Mr.Melin.

3. What is the laplce tranform of the first derivative of a function y(t) with respect to t : y’(t)?
a) sy(0) – Y(s)
b) sY(s) – y(0)
c) s2 Y(s)-sy(0)-y'(0)
d) s2 Y(s)-sy'(0)-y(0)
Answer: b
Explanation: Let (f(t) = y(t) )
(L[f’(t)] = int_0^∞ e^{-st} f'(t)dt )
( = e^{-st} f(t)(from , 0 , to , infty) – int_0^∞ (-s) e^{-st} f(t)dt )
( = -f(0) + sint_0^{infty} e^{-st} f(t)dt )
( = -f(0) + sF(s) )
( = sY(0) -y(0) ).

4. Solve the Ordinary Differential Equation by Laplace Transformation y’’ – 2y’ – 8y = 0 if y(0) = 3 and y’(0) = 6.
a) (3e^t cos(3t)+tsint(3t) )
b) (3e^t cos(3t)+te^{-t} sint(3t) )
c) (2e^{-t} cos(3t)-2 frac{t}{3} sint(3t) )
d) (2e^{-t} cos(3t)-2 frac{te^{-t}}{3} sint(3t) )
Answer: a
Explanation: L[y’’ – 2y’ – 8y ] = 0
s2 Y(s) – sy(0) – y'(0) – 2sY(s) + 2y(0) – 8Y(s) = 0
(s2 – 2s – 8)Y(s) = 2s
(L[y(t)] = 2 frac{s}{(s^2-2s-8)} )
Therefore, y(t) = 3et cos(3t) + tsint(3t).

5. Solve the Ordinary Differential Equation y’’ + 2y’ + 5y = e-t sin(t) when y(0) = 0 and y’(0) = 1.(Without solving for the constants we get in the partial fractions).
a) (e^t [Acost+A1sint+Bcos(2t)+frac{(B1)}{2} sin(2t)] )
b) (e^{-t} [Acost+A1sint+Bcos(2t)+B1sin(2t)] )
c) (e^{-t} [Acost+A1sint+Bcos(2t)+frac{(B1)}{2} sin(2t)] )
d) (e^t [Acost+A1sint+Bcos(2t)+(B1)sin(2t)] )
Answer: c
Explanation: (L[y’’+2y’ +5y = e^{-t} sin(t)] )
(s^2 Y(s)-sy(0)-y'(0)+ 2sY(s) -2y(0) + 5Y(s) = frac{1}{(s+1)^2+1} )
((s^2+2s+5)Y(s)= frac{1}{(s+1)^2+1}+1 )
((s^2+2s+5)Y(s)= frac{(s^2+2s+3)}{(s^2+2s+2)} )
( Y(s) = frac{(s^2+2s+3)}{(s^2+2s+2)(s^2+2s+5)} )
( = frac{(s+1)^2+2}{((s+1)^2+1)((s+1)^2+4)} )
( y(t) = e^{-t} L^{-1} [frac{(As+A1)}{(s^2+1)}+frac{(Bs+B1)}{(s^2+4)}] )
( = e^{-t} [Acost+A1sint+Bcos(2t)+frac{(B1)}{2} sin(2t)]).

6. Solve the Ordinary Diferential Equation using Laplace Transformation y’’’ – 3y’’ + 3y’ – y = t2 et when y(0) = 1, y’(0) = 0 and y’’(0) = 2.
a) (2e^t frac{t^5}{720}+e^t+2e^t frac{t}{6}+4e^t frac{t^2}{24} )
b) (e^t frac{t^5}{720}+2e^{-t}+2e^t frac{t}{6}+4e^t frac{t^2}{24} )
c) (e^{-t} frac{t^5}{720}+e^{-t}+2e^{-t} frac{t}{6}+4e^{-t} frac{t^2}{24} )
d) (2e^{-t} frac{t^5}{720}+e^{-t}+2e^{-t} frac{t}{6}+4e^{-t} frac{t^2}{24} )
Answer: a
Explanation: L[y’’’ – 3y’’ + 3y’ – y = t2 et]
s3 Y(s) – s2 y(0) – sy'(0) – y”(0) – 3s2 Y(s) + 3sy(0) + 3y'(0) + 3sY(s) – 3y(0) – Y(s) = (frac{2}{(s-1)^3} )
(Y(s) = frac{2}{(s-1)^6} +frac{(s^2+3s+5)}{(s-1)^3} )
(y(t) = 2e^t frac{t^5}{720}+e^t+2e^t frac{t}{6}+4e^t frac{t^2}{24}).

7. Take Laplace Transformation on the Ordinary Differential Equation if y’’’ – 3y’’ + 3y’ – y = t2 et if y(0) = 1, y’(0) = b and y’’(0) = c.
a) ((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=frac{2}{(s-1)^3} )
b) ((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)+(-3a-c)s)=frac{2}{(s-1)^3} )
c) ((s^3-3s^2+3s)Y(s)+(-as+(3a-b)s+(-3a-c))=frac{2}{(s-1)^3} )
d) ((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=frac{2}{(s-1)^3} )
Answer: a
Explanation: L[y’’’ – 3y’’ + 3y’ – y = t2 et]
s3 Y(s) – s2 y(0) – sy'(0) – y”(0) – 3s2 Y(s) + 3sy(0) + 3y'(0) + 3sY(s) – 3y(0) – Y(s) = (frac{2}{(s-1)^3} )
((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=frac{2}{(s-1)^3}.)

8. What is the inverse Laplace Transform of a function y(t) if after solving the Ordinary Differential Equation Y(s) comes out to be (Y(s) = frac{s^2-s+3}{(s+1)(s+2)(s+3)} ) ?
a) (frac{1}{2} e^{-t}+frac{9}{2} e^{-3t}-3e^{-2t} )
b) (frac{-1}{2} e^{-t}+frac{9}{2} e^{-2t}-3e^{-3t} )
c) (frac{1}{2} e^{-t}-frac{3}{2} e^{-2t}-3e^{-3t} )
d) (frac{-1}{2} e^{t}+frac{9}{2} e^{2t}-3e^{3t} )
Answer: b
Explanation: Taking inverse Laplace Transformation for
(Y(s) = frac{(s^2-s+3)}{(s+1)(s+2)(s+3)} )
Solving the partial fractions we get,
(Y(s) = frac{-1}{2} frac{1}{(s+1)}+frac{9}{2} frac{1}{(s+2)}-3 frac{1}{(s+3)} )
Therefore, (y(t) = frac{-1}{2} e^{-t}+frac{9}{2} e^{-2t}-3e^{-3t}. )

9. For the Transient analysis of a circuit with capacitors, inductors, resistors, we use bilateral Laplace Transformation to solve the equation obtained from the Kirchoff’s current/voltage law.
a) True
b) False
Answer: b
Explanation: For the transient analysis of the circuit with capacitors, inductors, resistors, we have to know the initial condition of the components used. So, the unilateral Laplace Transform is used to solve the equations obtained from the Kirchoff’s current/voltage law.

10. While solving an Ordinary Differential Equation using the unilateral Laplace Transform, it is possible to solve if there is no function in the right hand side of the equation in standard form and if the initial conditions are zero.
a) True
b) False
Answer: b
Explanation: It is not possible to solve an equation if the input and the initial conditions are zero becase Y(s) becomes zero where Y(s) is the Laplace Transform of y(t) function.

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