Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on Rolle’s Theorem and Answers

Engineering Mathematics Multiple Choice Questions on “Rolle’s Theorem – 1”.

1. For y = -x2 + 2x there exist a c in the interval [- 19765, 19767] Such that f'(c) = 0.
a) True
b) False
Answer: a
Explanation: The key here is to rewrite the function as y = -(x – 1)2 + 1
Observe here that on substituting – 19765 and 19767 in the equation we get
(-19766)2 + 1 and (-19766)2 + 1 respectively.
As we are dealing with their squared values they have to be equal
We have
f(-19765) = f(19767)
Polynomial functions are continuous and differentiable over the whole domain and hence by Rolles Theorem we must have a c such that f'(c) = 0 in the interval [-19765, 19767]
Hence, the claim is true.

2. For the function f(x) = (frac{sin(x)}{x^2}) How many points exist in the interval [0, 7π] Such that f'(c) = 0.
a) 8
b) 0
c) 7
d) 6
Answer: d
Explanation: We know that sine is a periodic function and it is divided by x2.
Observe that the sine takes the value of zero at integral arguments of π hence at every interval of the form [nπ, (n + 1)π]
We have f(nπ) = f((n + 1) π)
The sine and the polynomial combination is continuous and differentiable at every point except x = 0
Every such interval has a point such that f'(c) = 0
Hence, by Rolles theorem, in every interval of the form [nπ, (n + 1)π] we must have a point such that f'(c) = 0
Leaving the interval [0, π] we are left with six such intervals from 0 to 7π.

3. f(x) = (frac{sin(x)}{x}), How many points exist such that f'(c) = 0 in the interval [0, 18π].
a) 18
b) 17
c) 8
d) 9
Answer: a
Explanation: We have the sine function that takes the value of zero at Integral multiples of π
But for (frac{sin(x)}{x}) we have the exceptional value of (lim_{x rightarrow 0}frac{sin(x)}{x}) reaching one.
So leaving the first interval [0, π], for every other interval of the form [nπ (n + 1)π] we must have f(nπ) = f((n + 1)π)
By Rolles theorem we have
f’ (c) = 0 For every interval of the form [nπ (n + 1)π]
There are 17 such intervals.

4. Let f(x) = x + sin(x) Every point on the graph is rotated by 45 degree with respect to the origin along the radius equal to the radius vector at that point. How many c that belong to [0, 11π] exist Such that f'(c) = 0.
a) 10
b) 11
c) 110
d) 9
Answer: b
Explanation: The function has the beautiful property that the sine function is built over the line y = x
In other words the axis for the sine graph is no more on the x axis but is the line y = x.
When this graph is rotated by 45 degrees which Is equal to the slope of the line y = x we must get the original graph sin(x) again. The number of c in the interval [0, 11π] is then obviously equal to eleven.

5. A Function f(x) has the property f(a) = f(b) for ∀a,b…∊….I and a + b = 20 then which of the following even degree polynomials could be f(x)?
a) x4⁄4 – 10x3 + 150x2 – 1000x + 10131729
b) x2 + 5x + 6
c) x2 + x + 1
d) Polynomial functions are inadequate representations
Answer: a
Explanation: Given that f(a) = f(b) for all integral values of a, b, one additional constraint is that (frac{a+b}{2}) = 10
This means that the graph is symmetric about the line x = 10
This means that the even degree polynomial has a Rolle point(the only Rolle point) at x = 10
We need to differentiate the functions given in the options and observe which of these has a single(OR repeated root), which is equal to 10.
The first Option when differentiated yields
f'(x) = x3 – 30x2 + 300x – 1000
Equating to zero, we see that x = 10
satisfies the equation
(10)3 – 30(10)2 + 300(10) – 1000 = 0.

6. For some function f(x) we have f(a) = f(b) for ∀a,b…∊….I and a + b = 2 then which of the following even degree polynomials could f(x) be
a) x2 + 3x +1
b) 5x2⁄2 – 5x + 101
c) x2 + 2x + 1
d) Even degree polynomials of such kind cannot exist
Answer: b
Explanation: Given that f(a) = f(b) for all integral values of a, b with the condition (frac{a+b}{2}) = 1
We know that the function is symmetric about the line x = 1
Thus we must have a Rolle point at 1 for any function that satisfies these conditions.
Differentiating the given functions in the options we have for a
2x + 3 = 0
x = -3⁄2 ≠ 1
Doing the same for (b) we get
5x – 5 = 0
x = 1.

7. For all second degree polynomials with y = ax2 + bx + k, it is seen that the Rolles’ point is at c = 0. Also the value of k is zero. Then what is the value of b?
a) 0
b) 1
c) -1
d) 56
Answer: a
Explanation: Given that k = 0 we must rewrite the function to get
y = ax2 + bx
Now differentiating the function yields
y’ = 2ax + b = 0
Equating it to zero we get the Rolle point which is also zero
2a(0) + b = 0
b = 0.

8. For second degree polynomial it is seen that the roots are equal. Then what is the relation between the Rolles point c and the root x?
a) c = x
b) c = x2
c) They are independent
d) c = sin(x)
Answer: a
Explanation: For a polynomial with equal roots we can write the polynomial as f(x) = (x – a)2
Where a is the repeated root.
Differentiating it and equating the function to zero we get the Rolles point,
f'(x) = 2(x – a) = 0
x = a = c.

9. For any second degree polynomial with two real unequal roots. The relation between Rolles point r1 and the two roots r2 is
a) They are independent
b) c = r1 – r2
c) c = r1 * 1⁄r2
d) c = (frac{r_1 + r_2}{2})
Answer: d
Explanation: Given the polynomial has two real and unequal roots, we can write the polynomial in the following form
y = (x – a)(x – b)
Where a ≠ b are the two, unequal, real roots.
Now differentiation and equating to zero yields
y = (x – b) + (x -a ) = 0
Put x = c (because is the Rolles point here)
(c – a) + (c – b) = 0
2c – (a + b) = 0
c = a + b / 2 = (frac{r_1 + r_2}{2}).

10. If the domain of a function can be broken into infinite number of disjoint subsets such that every subset has a Rolles point then the function cannot be in a polynomial structure.
a) True
b) False
Answer: b
Explanation: The function could be an infinite degree polynomial (similar to Taylor series), such that after differentiating once we get another polynomial with infinite degree.
Hence, we may conclude that it has infinite distinct roots (with proper choice of polynomial) and hence, the domain can be broken into epsilon (very small) intervals around the roots of the differentiated polynomial to get infinite intervals containing the Rolles point.

250+ TOP MCQs on Derivatives of Area and Answers

Differential and Integral Calculus Multiple Choice Questions on “Derivatives of Area”.

1. What is the maximum area of the rectangle with perimeter 620 mm?
a) 24,025 mm2
b) 22,725 mm2
c) 24,000 mm2
d) 24,075 mm2
View Answer

Answer: a
Explanation: Let x be the length of the rectangle and y be the width of the rectangle. Then, Area A is,
A=x*y …………………………………………………. (1)
Given: Perimeter of the rectangle is 620 mm. Therefore,
P=2(x+y)
620=2(x+y)
x+y=310
y=310-x
We can now substitute the value of y in (1)
A=x*(310-x)
A=310x-x2
To find maximum value we need derivative of A,
dA/dx=310-2x
To find maximum value, (frac{dA}{dx}=0 )
310-2x=0
2x=310
x=155 mm
Therefore, when the value of x=155 mm and the value of y=310-155=155 mm, the area of the rectangle is maximum, i.e., A=155*155=24,025 mm2

2. Power Rule of the derivative states that, (frac{d(x^n)}{dx}=nx^{n-1}.)
a) True
b) False
View Answer

Answer: b
Explanation: Power Rule is given by, (frac{d(x^n)}{dx}=nx^{n-1}.)
For example, (frac{d(x^4)}{dx}=4x^3. )

3. Which of the following trigonometric function derivatives is correct?
a) (frac{d(sinx)}{dx}=-cosx)
b) (frac{d(secx)}{dx}=secxtanx)
c) (frac{d(tanx)}{dx}=cot^2 x)
d) (frac{d(cosx)}{dx}=sinx)
View Answer

Answer: b
Explanation: Correct forms of Trigonometric Derivative Functions

  • (frac{d(sinx)}{dx}=cosx )
  • (frac{d(cosx)}{dx}=-sinx)
  • (frac{d(secx)}{dx}=secxtanx)
  • (frac{d(tanx)}{dx}=sec^2 x )

4. Product Rule of the derivative is given by ________
a) (frac{d(fg)}{dx}=frac{d(g)}{dx}*frac{d(f)}{dx})
b) (frac{df}{dx}*frac{dg}{dx}=frac{d(g)}{dx}+frac{d(f)}{dx})
c) (frac{d(fg)}{dx}=f frac{d(g)}{dx}*g frac{d(f)}{dx})
d) (frac{d(fg)}{dx}=f frac{d(g)}{dx}+g frac{d(f)}{dx})
View Answer

Answer: d
Explanation: Product Rule of derivative is given by,
( frac{d(fg)}{dx}=f frac{d(g)}{dx}+g frac{d(f)}{dx})

5. What is the derivative of (frac{3x^2}{(1-sinx)})?
a) (frac{6xsinx-3x^2 cosx}{(1-sinx)^2} )
b) (frac{6x}{1-sinx} )
c) (frac{6xsinx-3x^2 cosx-6x}{(1-sinx)^2} )
d) (frac{6xsinx-3cosx-6}{(1-sinx)^2} )
View Answer

Answer: c
Explanation: Given: y = (frac{3x^2}{(1-sinx)})
(frac{dy}{dx}=frac{3x^2(frac{d(1-sinx)}{dx})-(frac{d(3x^2)}{dx})(1-sinx)}{(1-sinx)^2})
(frac{dy}{dx}=frac{3x^2(-cosx)-6x(1-sinx)}{(1-sinx)^2})
(frac{dy}{dx}=frac{6xsinx-3x^2 cosx-6x}{(1-sinx)^2})

6. Which of the following is a type of Iterative method of solving non-linear equations?
a) Graphical method
b) Interpolation method
c) Trial and Error methods
d) Direct Analytical methods
View Answer

Answer: b
Explanation: There are 2 types of Iterative methods, (i) Interpolation methods (or Bracketing methods) and (ii) Extrapolation methods (or Open-end methods).

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

Answer: d
Explanation: Given: (∫_0^8 x^{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 8%?
” alt=”” width=”197″ height=”195″ data-src=”2020/05/differential-calculus-questions-answers-derivatives-area-q9″ />
a) 0.127%
b) 0.02546%
c) 0.002546%
d) 0.2546%
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 8%,
(dR=frac{dV}{4πR^2}=frac{0.08}{4π(5)^2}=0.0002546)
Error in radius = 0.02546%

10. Half Range Fourier Series contains either sine or cosine terms.
a) True
b) False
View Answer

Answer: a
Explanation: A series which contains only sine or cosine terms is called Half Range Fourier Sine Series or Cosine Series respectively.

Global Education & Learning Series – Differential and Integral Calculus.

To practice all areas of Differential and Integral Calculus, here is complete set of 1000+ Multiple Choice Questions and Answers.

 

250+ TOP MCQs on Jacobians and Answers

Differential and Integral Calculus Multiple Choice Questions on “Jacobians”.

1. The jacobian of p,q,r w.r.t x,y,z given p=x+y+z, q=y+z, r=z is ________
a) 0
b) 1
c) 2
d) -1
Answer: b
Explanation: We have to find
(J = frac{∂(p,q,r)}{∂(x,y,z)} = begin{vmatrix}
frac{∂p}{∂x} & frac{∂p}{∂y} &frac{∂p}{∂z}\
frac{∂q}{∂x} &frac{∂q}{∂y} &frac{∂q}{∂z}\
frac{∂r}{∂x} &frac{∂r}{∂y} &frac{∂r}{∂z}\
end{vmatrix})
But p=x+y+z, q=y+z, r=z (taking partial derivative)
(J=begin{vmatrix}
1&1&1\
0&1&1\
0&0&1\
end{vmatrix}(frac{∂p}{∂x}=1, frac{∂p}{∂y}=1, frac{∂p}{∂z}=1, frac{∂q}{∂x}=0, frac{∂q}{∂y}=1, frac{∂q}{∂z}=1, frac{∂r}{∂x}=0, \
frac{∂r}{∂y}=0, frac{∂r}{∂z}=1))
On expanding we get
J = 1(1 – 0) = 1
Thus j = 1.

2. Given (u=frac{yz}{x}, v=frac{zx}{y}, w=frac{xy}{z}) then the value of (frac{∂(u,v,w)}{∂(x,y,z)}) is ________
a) 4
b) -4
c) 0
d) 1
Answer: a
Explanation: By Data (u=frac{yz}{x}, v=frac{zx}{y}, w=frac{xy}{z})

(J = frac{∂(u,v,w)}{∂(x,y,z)} = begin{vmatrix}
frac{∂u}{∂x} & frac{∂u}{∂y} &frac{∂u}{∂z}\
frac{∂v}{∂x} &frac{∂v}{∂y} &frac{∂v}{∂z}\
frac{∂w}{∂x} &frac{∂w}{∂y} &frac{∂w}{∂z}\
end{vmatrix} = begin{vmatrix}
frac{-yz}{x^2} & frac{z}{x} &frac{y}{x}\
frac{z}{y} &frac{-zx}{y^2} &frac{x}{y}\
frac{y}{z} &frac{x}{z} &frac{-xy}{z^2}\
end{vmatrix})
(=frac{-yz}{x^2}Big{(frac{-zx}{y^2})(frac{-xy}{y^2}) – (frac{x}{z})(frac{x}{y})Big} – (frac{z}{x})Big{(frac{z}{y})(frac{-xy}{z^2})-(frac{y}{z})(frac{x}{y})Big} \
+ frac{y}{x}Big{(frac{z}{y})(frac{x}{z})-(frac{y}{z})(frac{-zx}{y^2})Big})
(=frac{-yz}{x^2} Big{frac{x^2}{yz} – frac{x^2}{yz}Big} – frac{z}{x}Big{frac{-x}{z} – frac{x}{z}Big} + frac{y}{x}Big{frac{x}{y} + frac{x}{y}Big} = 0 + 1 + 1 + 1 = 4)
Therefore (frac{∂(u,v,w)}{∂(x,y,z)} = 4.)

3. If u=x+3y2-z3, v=4x2 yz, w=2z2-xy then (frac{∂(u,v,w)}{∂(x,y,z)} ) at (1,1,1).
a) -184
b) -90
c) 20
d) 40
Answer: a
Explanation: Given that u=x+3y2-z3, v=4x2 yz, w=2z2-xy
(frac{∂(u,v,w)}{∂(x,y,z)} = begin{vmatrix}
frac{∂u}{∂x} & frac{∂u}{∂y} &frac{∂u}{∂z}\
frac{∂v}{∂x} &frac{∂v}{∂y} &frac{∂v}{∂z}\
frac{∂w}{∂x} &frac{∂w}{∂y} &frac{∂w}{∂z}\
end{vmatrix} = begin{vmatrix}
1 & 6y & -3z^2\
8xyz &4x^2 z &4x^2 y\
-y &-x &4z\
end{vmatrix})
(frac{∂(u,v,w)}{∂(x,y,z)} ,at, (1,1,1) = begin{vmatrix}
1 & 6 & -3\
8 &4&4\
-1 &-1 &4\
end{vmatrix})
=1(16+4) – 6(32+4) – 3(-8+4) = -184.

4. If x=rcos⁡θ, y=rsin⁡θ then the value of (frac{∂(x,y)}{∂(r,θ)}) is ________
a) 1
b) 0
c) r
d) (frac{1}{r})
Answer: c
Explanation: Wkt, (frac{∂(x,y)}{∂(r,θ)} = begin{vmatrix}
frac{∂x}{∂r} & frac{∂x}{∂θ}\
frac{∂y}{∂r} & frac{∂y}{∂θ}\
end{vmatrix} ,but, x=rcos⁡θ, y=rsin⁡θ)
Thus ( begin{vmatrix}
cos⁡θ & -rsin⁡θ\
sin⁡θ & rcos⁡θ\
end{vmatrix} = rcos^2 θ + rsin^2 θ = r.)

5. The application of Jacobians is significant in the evaluation of double integral of the form ∬f(x,y) dx dy and triple integral of the form ∭f(x,y,z)dx dy dz by transformation from one system of coordinate to the other.
a) True
b) False
Answer: a
Explanation: The principle of evaluation is analogous with the evaluation of ∫f(x)dx by taking a suitable substitution.

6. If u+v=ex cos⁡y and u-v=ex sin y the value of (J(frac{u,v}{x,y})) is ________
a) e2x
b) (frac{e^2x}{2} )
c) (frac{-e^2x}{2} )
d) 0
Answer: c
Explanation: wkt (J(frac{u,v}{x,y}) = frac{∂(u,v)}{∂(x,y)} = begin{vmatrix}
frac{∂u}{∂x} & frac{∂u}{∂y}\
frac{∂v}{∂x} & frac{∂v}{∂y}\
end{vmatrix})
But we have to find u and v first using given equation
u+v=ex cos ⁡y………(1)
u-v=ex sin⁡ y……..(2)
solving (1)&(2)
we get u=(frac{e^x}{2} )(cos⁡y+sin⁡y)
v=(frac{e^x}{2} )(cos⁡y-sin⁡y)
(frac{∂u}{∂x} = frac{e^x}{2}(cos⁡y+sin⁡y) )
(frac{∂v}{∂x} = frac{e^x}{2}(cos⁡y-sin⁡y) )
(frac{∂u}{∂y} = frac{e^x}{2}(cos⁡y-sin⁡y) )
(frac{∂v}{∂y} = frac{e^x}{2}(-cos⁡y-sin⁡y) )
(J(frac{u,v}{x,y}) = begin{vmatrix}
frac{e^x}{2}(cos⁡y+sin⁡y) & frac{e^x}{2}(cos⁡y-sin⁡y)\
frac{e^x}{2}(cos⁡y-sin⁡y) & frac{e^x}{2}(-cos⁡y-sin⁡y)\
end{vmatrix})
((frac{e^x}{2})(frac{e^x}{2})){-(cos⁡y+sin⁡y)2 -(cos⁡y-sin⁡y)2}…….(expanding & solving by taking cos2 y+sin2 y=1)
=(frac{-e^2x}{2} ).

7. Which among the following is the definition of Jacobian of u and v w.r.t x and y?
a) (J(frac{x,y}{u,v}))
b) (J(frac{u,v}{x,y}))
c) (frac{∂(x,y)}{∂(u,v)} )
d) (frac{∂(u,x)}{∂(v,y)} )
Answer: b
Explanation: (J(frac{u,v}{x,y}) = frac{∂(u,v)}{∂(x,y)} = begin{vmatrix}
frac{∂u}{∂x} & frac{∂u}{∂y}\
frac{∂v}{∂x} & frac{∂v}{∂y}\
end{vmatrix})

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250+ TOP MCQs on Application of Double Integrals and Answers

Engineering Mathematics Multiple Choice Questions on “Application of Double Integrals”.

1. Distance travelled by any object is _____________
a) Double integral of its acceleration
b) Double integral of its velocity
c) Double integral of its Force
d) Double integral of its Momentum
Answer: b
Explanation: We know that,
x(t) = (intint a(t) ,dtdt).

2. Find the distance travelled by a car moving with acceleration given by a(t)=t2 + t, if it moves from t = 0 sec to t = 10 sec, if velocity of a car at t = 0sec is 40 km/hr.
a) 743.3km
b) 883.3km
c) 788.3km
d) 783.3km
Answer: d
Explanation: Add constant automatically
We know that,
a=(intint(t^2+t)dtdt=intleft [(frac{t^3}{3}+frac{t^2}{2})+cright ]dt)
=(int_0^{10}left [(frac{t^3}{3}+frac{t^2}{2})+40right ]dt=frac{1000}{3}+50+400)=783.3km

3. Find the distance travelled by a car moving with acceleration given by a(t)=Sin(t), if it moves from t = 0 sec to t = π/2 sec, if velocity of a car at t=0sec is 10 km/hr.
a) 10.19 km
b) 19.13 km
c) 15.13 km
d) 13.13 km
Answer: d
Explanation: Add constant automatically
We know that,
a=(intint Sin(t)dtdt=int[-Cos(t)+c]dt=int_0^{π/2}[-Cos(t)+9]dt)
(=[-Sin(t)+9t]=-1+frac{9π}{2})=13.13 km

4. Find the distance travelled by a car moving with acceleration given by a(t)=t2 – t, if it moves from t = 0 sec to t = 1 sec, if velocity of a car at t = 0sec is 10 km/hr.
a) 119⁄22 km
b) 119⁄15 km
c) 129⁄12 km
d) 119⁄12 km
Answer: b
Explanation: Add constant automatically
We know that,
a=(intint[t^2-t] dtdt=int [frac{t^3}{3}-frac{t^2}{2}]dt=int_0^1left [frac{t^3}{3}-frac{t^2}{2}+10 right ] dt=left [frac{t^4}{12}-frac{t^3}{6}+10tright ])
=(frac{1}{12}-frac{1}{6}+10=frac{1}{6} [frac{1}{2}-1]+10 km=-frac{1}{12}+10=frac{119}{12})km

5. Find the value of ∫∫ xydxdy over the area bpunded by parabola y=x2 and x = -y2, is?
a) 1⁄67
b) 1⁄24
c) –1⁄6
d) –1⁄12
Answer: b
Explanation:
(int_0^1 int_{-√y}^{-y^2} y.x dxdy=frac{1}{2} int_0^1 y[y^4-y]dy=frac{1}{2} [frac{1}{6}-frac{1}{3}]=-frac{1}{12})

6. Find the value of (intint ,xydxdy) over the area b punded by parabola x = 2a and x2 = 4ay, is?
a) a4⁄4
b) a4⁄3
c) a5⁄3
d) a2⁄3
Answer: b
Explanation:
(int_0^2a int_0^{x^2/4a} x.y dxdy=int_0^{2a} xdx int_0^{x^2/4a} ydy=int_0^{2a}frac{xx^4}{32a^2}dx)
=(frac{1}{32a^2} int_0^{2a}x^5 dx=frac{a^4}{3})

7. Find the integration of (intint0x (x2 + y2) ,dxdy), where x varies from 0 to 1.
a) 4⁄3
b) 5⁄3
c) 2⁄3
d) 1
Answer: c
Explanation: Add constant automatically
Given, f(x)=(int_0^1 int_0^x (x^2+y^2) dydx = int_0^1 (x^3+frac{x^3}{3})dx=1+frac{1}{3}=frac{2}{3})

8. Evaluate the value of (intint_0^y frac{2xy^5}{sqrt{1+x^2 y^2-y^4}} dxdy), where y varies from 0 to 1.
a) 11⁄12
b) 14⁄6
c) 11⁄6
d) 11⁄7
Answer: c
Explanation:
Given, f(x)=(int_0^{2a} int_0^y frac{2xy^5}{sqrt{1+x^2 y^2-y^4}} dxdy)
=(int_0^1 int_0^y frac{1}{y} frac{2xy^5}{sqrt{(frac{1-y^4}{y^2})+x^2)}} dxdy=int_0^1 2y^4 |(frac{1-y^4}{y^2})+x^2]|_0^y dy)
=(2int_0^1 [y^3-sqrt{1-y^4}y^3]dy=2left [frac{y^4}{4}-frac{2}{3} (1-y^4)^{frac{3}{2}}right ] _1^0=2[frac{1}{4}+frac{2}{3}]=frac{11}{6})

9. Evaluate ∫∫[x2 + y2 – a2 ]dxdy where, x and y varies from –a to a.
a) –2⁄3 a4
b) –4⁄3 a4
c) –4⁄3 a5
d) –2⁄3 a5
Answer: b
Explanation:
Equation of circle is given by x2 + b2 = a2
(int_{-a}^a int_{-a}^a [x^2+y^2-a^2 ]dxdy)
=(int_{-a}^a [frac{x^3}{3}+y^2 x-a^2 x]_a^{-a}dy=int_{-a}^a [frac{a^3}{3}+ay^2-a^2 a+frac{a^3}{3}+y^2 a-a^2 a]dy)
(int_{-a}^a frac{2a^3}{3}+2ay^2- 2a^3 dy=[frac{2a^3 y}{3}+frac{2ay^3}{3}-2a^3 y] _a^{-a}=frac{4a^4}{3}+frac{4a^4}{3}-4a^4=-frac{4}{3} a^4)

10. Find the area inside a ellipse of minor-radius ‘b’ and major-radius ‘a’.
a) –4⁄3 a2
b) –4⁄3 ab2
c) 4⁄3 ab
d) –4⁄3
Answer: c
Explanation:
Equation of ellipse is given by (frac{x^2}{a^2} +frac{y^2}{b^2}=1)
(int_{-b}^b int_{-a}^a [frac{x^2}{a^2}+frac{y^2}{b^2}-1]dxdy=int_{-b}^b [frac{x^3}{3a^2}+frac{xy^2}{b^2}-x]_a^{-a} dy)
(int_{-b}^b [frac{a}{3}+frac{ay^2}{b^2} -a+frac{a}{3}+frac{ay^2}{b^2}-a] dy)
=(int_{-b}^b [frac{2a}{3}+frac{2ay^2}{b^2}-2a] dy=[frac{2ay}{3}+frac{2ay^3}{3b^2}-2ay] _b^{-b})
=(frac{4ab}{3}+frac{4ab}{3}-4ab=-frac{4}{3} ab)

11. Find the value of (int_0^{1-y} xysqrt{1-x-y} ,dxdy) where, y varies from 0 to 1.
a) 16⁄946
b) 8⁄945
c) 16⁄45
d) 16⁄945
Answer: d
Explanation:
Given, f(x)=(int_1^0∫_0^{1-y} xysqrt{1-x-y} ,dxdy)
putting, (t=frac{x}{1-y})=>x=t(1-y)=>dx=(1-y)dt
(int_1^0int_0^1 t(1-y)ysqrt{1-t(1-y)-y} (1-y)dtdy)
=(int_1^0int_0^1 y(1-y)^{5/2} t(1-t)^{1/2} dtdy)
=(int_1^0y(1-y)^{5/2} dy int_0^1 t(1-t)^{1/2} dt)
=(int_0^1 y^{2-1} (1-y)^{7/2-1} dy int_0^1 t^{2-1} (1-t)^{3/2-1} dt=β(2,frac{7}{2})β(2,frac{3}{2})=frac{16}{945})

12. Find the value of integral (int_0^1int_{x^2}^x xy(x+y)dydx).
a) 3⁄15
b) 2⁄15
c) 2⁄30
d) 1⁄15
Answer: b
Explanation: Given, F(x)=(int_0^1int_{x^2}^x xy(x+y)dydx=int_0^1 int_{x^2}^x(x ^2 y+xy^2)dydx)
=(int_0^1 [frac{x^2 y^2}{2}+frac{xy^3}{3}] _x^{x^2}dx=int_0^1 [frac{x^3}{2}+frac{x^4}{3}-frac{x^4}{2}-frac{x^5}{3} ]dx=frac{1}{2}+frac{1}{3}-frac{1}{2}-frac{1}{5}=frac{2}{15})

250+ TOP MCQs on Orthogonal Trajectories and Answers

Ordinary Differential Equations Multiple Choice Questions on “Orthogonal Trajectories”.

1. Find the orthogonal trajectories of the family of parabolas y2=4ax.
a) 2x2+y2=k
b) 2y2+x2=k
c) x2-2y2=k
d) 2x2-y2=k
Answer: a
Explanation: Consider (frac{y^2}{x} = 4a)……..differentiating w.r.t x we get (frac{x.2y frac{dy}{dx} – y^2.1}{x^2} = 0)
(2xy frac{dy}{dx} -y = 0)…..DE of the family of curve y2=4ax replacing (frac{dy}{dx}) by (-frac{dx}{dy})
since (frac{dy}{dx}) is the slope of the tangent to the curve –> (-frac{dx}{dy}) is the slope of orthogonal line we get (2x(-frac{dx}{dy}) – y = 0 , or, 2x ,dx + y ,dy = 0 rightarrow int2x ,dx + int y ,dy = c rightarrow x^2 + frac{y^2}{2} = c.)
or 2x2+y2=k is the required orthogonal trajectory.

2. Find the orthogonal trajectories of the family of curves (frac{x^2}{a^2} + frac{y^2}{b^2+k} = 1) where k is the parameter.
a) x2-y2-3a2 log⁡x-k = 0
b) x2+2y2–(frac{a^2}{2}) log⁡x-k = 0
c) x2+y2-2a2 log⁡x-k = 0
d) 2x2-y2–(frac{a^2}{3}) log⁡x-k = 0
Answer: c
Explanation: (frac{x^2}{a^2} + frac{y^2}{b^2+k} = 1)…….(1) differentiating w.r.t x we have (frac{2x}{a^2} + frac{2yy’}{b^2+k} = 0 ,where, y’ = frac{dy}{dx})
i.e (frac{x^2}{a^2} = frac{-yy’}{b^2+k} …..(2) ,from, (1), frac{x^2}{a^2} – 1 = frac{-y^2}{b^2+k} rightarrow frac{x^2-a^2}{a^2} = frac{-y^2}{b^2+k})…..(3) divide (2)&(3)
we get (frac{x}{x^2-a^2} = frac{yy’}{y^2} rightarrow frac{x}{x^2-a^2} = frac{y’}{y})
now (y’=frac{dy}{dx}) is replaced by (-frac{dx}{dy} ,i.e, frac{x}{x^2-a^2} = frac{1}{y}(-frac{dx}{dy}))
separating the variable we get (y ,dy = -frac{(x^2-a^2)}{x} dx) integrating this equation
(int y ,dy = int-x ,dx + int a^2 frac{1}{x} ,dx + c rightarrow frac{y^2}{2} = frac{-x^2}{2} + a^2 log⁡ ,x + c)
–> x2+y2-2a2 log⁡x – 2c=0 or x2+y2-2a2 log⁡x-k=0 where k=2c is the required orthogonal trajectory.

3. The Orthogonal DE for family of parabola y2=4a(x+a) is same as _______(where DE stands for Differential equation)
a) DE of parabola y2=4a(x+a)
b) DE of parabola y2=4ax
c) DE of parabola x2=4ay
d) DE of parabola x2=4a(y+a)
Answer: a
Explanation: y2=4a(x+a)……….differentiating w.r.t x we get (2y frac{dy}{dx} = 4a rightarrow a =frac{yy’}{2})
substituting the ‘a’ in given equation i.e : (y^2 = 2yy’(x + frac{yy’}{2}), y=2xy’+yy’^2)…(1)
replacing y’ by (frac{-1}{y’} ,i.e, y=2xfrac{-1}{y’} + y(frac{-1}{y’})^2) or on solving yy’2+2xy’=y…(2)
comparing (2) & (1) we observe that DE of the orthogonal family is same as the DE of the given family of curves thus the family of parabola is self orthogonal.

4. Find the orthogonal trajectories of the family r=a(1+sin θ).
a) r=k(sin θ)
b) r2=k(cos θ)2
c) r=k(1-cos θ)
d) r=k(1-sin θ)
Answer: d
Explanation: r=a(1+sin θ) taking log we get log r=log a+log(1+sin θ)…differentiating
w.r.t θ we have, ( frac{1}{r} frac{dr}{dθ} = frac{cos⁡θ}{1+sin⁡θ}) since given equation is polar slope of tangent is given by ( frac{1}{r} frac{dr}{dθ}) and perpendicular line has a slope of (-rfrac{dθ}{dr} , replacing, frac{1}{r} frac{dr}{dθ} ,by, -rfrac{dθ}{dr})
we have (-rfrac{dθ}{dr} = frac{cos⁡θ}{1+sin⁡θ}) …..separating the variables and integrating it
(int frac{dr}{r} + int frac{1+sin⁡θ}{cos⁡θ} ,dθ = c rightarrow log r + int sec⁡θ ,dθ + int tan⁡θ , dθ = c)
log r + log(sec⁡θ + tan⁡θ) + log(sec⁡θ) = c = log k
(rightarrow log(r(sec⁡θ + tan⁡θ) (sec⁡θ)) = log k rightarrow rleft(frac{1}{cos⁡θ} + frac{sin⁡θ}{cos⁡θ}right) frac{1}{cos⁡θ} = k)
(frac{r(1+sin⁡θ)}{cos^2 θ} = frac{r(1+sin⁡θ)}{1-sin^2 θ} rightarrow r = k(1-sin θ)) is the required orthogonal trajectory.

5. Which among the following is true for the curve rn = a sin⁡nθ?
a) Given family of curve is Self orthogonal
b) Orthogonal trajectory is rn=k cos⁡nθ where k is an constant
c) Orthogonal trajectory is rn=k cosec⁡nθ where k is an constant
d) Orthogonal trajectory is rn=k sin⁡nθ where k is an constant
Answer: b
Explanation: Consider rn = a sin⁡nθ..(1) –> n log r = log a + log(sin⁡nθ) …differentiating w.r.t θ
( frac{n}{r} frac{dr}{dθ} = frac{n cos⁡nθ}{sin⁡nθ} ,or, frac{1}{r} frac{dr}{dθ} = frac{cos⁡nθ}{sin⁡nθ} = cot ,nθ)
replacing (frac{1}{r} frac{dr}{dθ} ,by, -rfrac{dθ}{dr} ,we,, get, -rfrac{dθ}{dr} = cot ,nθ)
separating the variables and integrating (int frac{dr}{r} int tan⁡,nθ ,dθ = c )
( rightarrow log ,r + frac{log⁡(sec⁡nθ)}{n} = c) or n log r + log⁡(sec nθ) = nc = log⁡k
log(rn sec⁡nθ) = log k –> rn sec⁡ nθ=k or rn=k cos⁡nθ ..(2) is the required orthogonal trajectory since (1)&(2) are not same the given family of curve is not self orthogonal.

Global Education & Learning Series – Ordinary Differential Equations.

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250+ TOP MCQs on Convolution and Answers

Ordinary Differential Equations Multiple Choice Questions on “Convolution”.

1. Find the (L^{-1} (frac{1}{s(s^2+4)})).
a) (frac{1-sin⁡(t)}{4})
b) (frac{1-cos⁡(t)}{4})
c) (frac{1-sin⁡(2t)}{4})
d) (frac{1-cos⁡(2t)}{4})
View Answer

Answer: d
Explanation: In the given question
Let p1(s) = (frac{1}{s^2+4}) and p2(s) = (frac{1}{s})
(f_1 (t) = L^{-1} (frac{1}{s^2+4}) = frac{sin⁡(2t)}{2})
(f_2 (t) = L^{-1} (frac{1}{s}) = 1)
By Convolution Theorem,
(L^{-1} (p_1(s)×p_2(s)) = int_{0}^{t}f_1(u) f_2(t-u) dt)
(L^{-1} (frac{1}{s(s^2+4)}) = int_{0}^{t}frac{1}{2} sin⁡(2u) du)
=(frac{1-cos⁡(2t)}{4})
Thus, the answer is (frac{1-cos⁡(2t)}{4}).

2. Find the (L^{-1} (frac{1}{s(s+4)^frac{1}{2}})), give the answer in terms of error function.
a) (frac{1}{2} erf⁡(2t))
b) (frac{1}{2} erf⁡(sqrt{t}))
c) (frac{1}{2} erf⁡(2sqrt{t}))
d) (frac{1}{2} erf⁡(4sqrt{t}))
View Answer

Answer: c
Explanation: In the given question,
Let (p_1(s) = frac{1}{(s+4)^{frac{1}{2}}}) and (p_2(s) = frac{1}{s})
(f_1 (t) = L^{-1} (frac{1}{(s+4)^{frac{1}{2}}}) = frac{e^{-4t}×sqrt{t}}{sqrtpi})
(f_2 (t) = L^{-1} frac{1}{s} = 1)
By Convolution Theorem,
(L^{-1} (p_1(s)×p_2(s)) = int_{0}^{t}f_1(u) f_2(t-u) dt)
(L^{-1} (frac{1}{s(s+4)^{frac{1}{2}}}) = int_{0}^{t}frac{sqrt{u}}{sqrtpi} e^{-4u} ,du)
(=frac{1}{sqrtpi} int_{0}^{t}sqrt{u} e^{-4u} du)
We know that error function is given by –
(erf⁡(x)=frac{2}{sqrtpi} int_{0}^{x}e^{-z^2} dz)
Applying, 4u=z2 And setting the limits of u in terms of z, we get
(L^{-1} (frac{1}{s(s+4)^{frac{1}{2}}}) = frac{1}{2} erf⁡(2sqrt{t}))
Thus, the answer is (frac{1}{2} erf⁡(2sqrt{t})).

3. Find the (L^{-1} frac{s}{(s^2+4)^2}).
a) (frac{1}{4} tcos(2t))
b) (frac{1}{4} tsin(t))
c) (frac{1}{4} tsin(2t))
d) (frac{1}{2} tsin(2t))
View Answer

Answer: c
Explanation: In the given question,
Let (p_1(s) = frac{1}{s^2+4}) and (p_2(s) = frac{1}{s})
(f_1 (t) = L^{-1} (frac{1}{s^2+4}) = frac{sin⁡(2t)}{2})
(f_2 (t) = L^{-1} (frac{s}{s^2+4}) = cos⁡(2t))
By Convolution Theorem,
(L^{-1} (p_1(s)×p_2(s)) = int_{0}^{t} f_1(u) f_2(t-u) dt)
(L^{-1} left (frac{s}{(s^2+4^2)^2}right )=int_{0}^{t} sin⁡(2u)×frac{1}{2}×cos⁡(2(t-u))du)
=(frac{1}{4}[tsin(2t)-frac{cos⁡(2t)}{4}+frac{cos⁡(2t)}{4}])
=(frac{1}{4} tsin(2t))
Thus, the correct answer is (frac{1}{4} tsin(2t))
.

Global Education & Learning Series – Ordinary Differential Equations.

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