Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on Indeterminate Forms and Answers

Engineering Mathematics MCQs focuses on “Indeterminate Forms – 3”.

1. What are Intermediate Forms?
a) Forms(f(x)/g(x)) whose limits x tends to ‘a’ can give rational number directly
b) Forms(f(x)/g(x)) whose limits x tends to ‘a’ can give finite number directly
c) Forms(f(x)/g(x)) whose limits x tends to ‘a’ can not be infinite output or cannot be solved directly
d) Forms(f(x)/g(x)) whose limits x tends to ‘a’ can gives finite output
Answer: c
Explanation:
(lim_{xrightarrow a})= 0⁄0 = 0⁄∞ = ∞⁄0 = ∞⁄∞ are all intermediate forms.
Special Rules are made to solve these forms. They cannot be solved directly.

2. L’Hospital Rule states that
a) If (lim_{xrightarrow a}frac{f(x)}{g(x)}) is an indeterminate form than (lim_{xrightarrow a}frac{f(x)}{g(x)}=lim_{xrightarrow a}frac{f'(x)}{g'(x)}) if (lim_{xrightarrow a} frac{f'(x)}{g'(x)}) has a finite value
b) (lim_{xrightarrow a}frac{f(x)}{g(x)}) always equals to (lim_{xrightarrow a}frac{f'(x)}{g'(x)})
c) (lim_{xrightarrow a}frac{f(x)}{g(x)}) if an indeterminate form than cannot be solved
d) (lim_{xrightarrow a}frac{f(x)}{g(x)}) if an indeterminate form than it is equals to zero.
Answer: a
Explanation: According to L’Hospital Rule, if (lim_{xrightarrow a}frac{f(x)}{g(x)}) is indeterminate and (lim_{xrightarrow a}frac{f'(x)}{g'(x)}) has a finite value then (lim_{xrightarrow a}frac{f(x)}{g(x)} = lim_{xrightarrow a}frac{f'(x)}{g'(x)}). It is helpful in solving limits of indeterminate forms.

3. If f(x) = x2 – 3x + 2 and g(x) = x3 – x2 + x – 1 than find value of (lim_{xrightarrow 1}frac{f(x)}{g(x)})?
a) 0.5
b) 1
c) -.5
d) -1
Answer: c
Explanation:
(lim_{xrightarrow 1}frac{f(x)}{g(x)}=lim_{xrightarrow 1}frac{x^2-3x+2}{x^3-x^2+x-1})=(1-3+2)/(1-1+1-1)=0/0
Hence this gives indeterminent forms hence by L’Hospital Rule
(lim_{xrightarrow 1}frac{f(x)}{g(x)}=lim_{xrightarrow 1}frac{f'(x)}{g'(x)}=lim_{xrightarrow 1}frac{2x-3}{3x^2-2x+1}=frac{2-3}{3-2+1}=-frac{1}{2})
Hence
(lim_{xrightarrow 1}frac{f(x)}{g(x)}=-frac{1}{2})

4. If f(x) = Tan(x)-1 and g(x) = Sin(x) – Cos(x) than find value of limx → π⁄4⁡f(x)⁄g(x)
a) -√2
b) √(-2)
c) √2
d) √3
Answer: c
Explanation:
(lim_{xrightarrow pi/4}frac{f(x)}{g(x)}=lim_{xrightarrow pi/4}frac{tan(x)-1}{sin(x)-cos(x)}=frac{(1-1)}{frac{1}{sqrt{2}}-frac{1}{sqrt{2}}}=0/0)
Hence this gives indeterminate forms hence by L’Hospital Rule
(lim_{xrightarrow pi/4}frac{f(x)}{g(x)}=lim_{xrightarrow pi/4}frac{f'(x)}{g'(x)}=lim_{xrightarrow pi/4}frac{sec^2}{cos(x)+sin(x)}=sqrt{2})
Hence
(lim_{xrightarrow pi/4}frac{f(x)}{g(x)}=sqrt{2})

5. If f(x) = Tan(x) and g(x) = ex – 1 than find value of limx → 0⁡f(x)⁄g(x)
a) 1
b) 0
c) -1
d) 2
Answer: a
Explanation:
(lim_{xrightarrow 0}frac{f(x)}{g(x)}=lim_{xrightarrow 0}frac{tan(x)}{e^x-1}=0/0) (Indeterminate Form)
By L’Hospital’s rule
(lim_{xrightarrow 0}frac{f(x)}{g(x)}=lim_{xrightarrow 0}frac{f'(x)}{g'(x)}=lim_{xrightarrow 0}frac{sec^2(x)}{e^x})=1

6. Find the value of (lim_{xrightarrow infty}(1+frac{1}{x})^x)
a) e-1
b) e
c) e + 1
d) 1
Answer: b
Explanation:
(lim_{xrightarrow infty}(1+frac{1}{x})^x)=1∞ (Indeterminate form)
Hence
Let t=1/x
->y=(lim_{xrightarrowinfty}(1+frac{1}{x})^x=lim_{trightarrow 0}(1+t)^{1/t})
Taking log on both side,
ln(y) = (lim_{trightarrow 0} frac{ln(1+t)}{t}) = 0/0 (Indeterminate form)
Applying L’Hospital Rule again
ln(y) = (lim_{trightarrow 0} frac{1}{1+t}) = 1
=> y = (lim_{xrightarrowinfty}(1+frac{1}{x})^x) = e

7. If f(x) = x3 + 3x2 + Sin(x) and g(x) = ex – 1 than find value of (lim_{xrightarrow 0}f(x)^{frac{1}{g(x)}})
a) e6e
b) e(e/6)
c) e6
d) e(6/e)
Answer: c
Explanation:
y=(lim_{xrightarrow 0}f(x)^{(frac{1}{g(x)})} = lim_{xrightarrow 0}[x^3+3x^2+sin(x)]^{frac{1}{e^x-1}})
Taking ln on both side
ln(y)=(lim_{xrightarrow 0}frac{ln(x^3+3x^2+sin(x))}{e^x-1})=ln(0)/0 (Indeterminate Form)
By L’Hospital Rule
(lim_{xrightarrow 0}frac{3x^+6x+cos(x)}{e^x(x^3+3x^2+sin(x))}=1/0)
Again using L’Hospital rule
ln(y)=(lim_{xrightarrow 0}frac{6x+6+sin(x)}{e^x(x^3+3x^2+sin(x))+e^x(3x^2+6x+cos(x))}=6)
y=e6

8. If f(x) = Sin(x) and g(x) = x than find value of limx → 0⁡f(x)⁄g(x)
a) -1
b) 0
c) 1
d) 2
Answer: c
Explanation: limx → 0⁡f(x)⁄g(x) = limx → 0⁡sin(x)⁄x = ⁡0⁄0 (Indeterminate forms)
By L’Hospital rule
limx → 0⁡cos(x)⁄1 = 1.

9. If f(x) = sin(x)cos(x) and g(x) = x2 than find value of limx → 0⁡f(x)⁄g(x)
a) 2
b) 0
c) -1
d) Cannot be found
Answer: b
Explanation:
(lim_{xrightarrow 0}frac{f(x)}{g(x)}=lim_{xrightarrow 0}frac{sin(x)cos(x)}{x^2}) = 0/0
Henece, by L’Hospital rule
(lim_{xrightarrow 0}frac{f(x)}{g(x)}=lim_{xrightarrow 0}frac{cos(2x)}{2x}=frac{1}{0})
Again by L’Hospital rule
(lim_{xrightarrow 0}frac{f(x)}{g(x)}=lim_{xrightarrow 0}frac{sin(2x)}{1}=0)

10. If f(x) = ex + xcos(x) and g(x) = Sin(x) than find value of limx → 0⁡f(x)⁄g(x)
a) 2
b) 1
c) 3
d) 4
Answer: a
Explanation:
(lim_{xrightarrow 0}frac{f(x)}{g(x)}=lim_{xrightarrow 0}frac{e^x+xcos(x)}{sin(x)}=lim_{xrightarrow 0}frac{e^x}{sin(x)}+lim_{xrightarrow 0}frac{xcos(x)}{sin(x)})=1/0+0/0, i.e both are indeterminate
i.e., applying L’Hospital Rule
(lim_{xrightarrow 0}frac{e^x}{sin(x)}=lim_{xrightarrow 0}frac{e^x}{cos(x)}+lim_{xrightarrow 0}frac{cos(x)-xsin(x)}{cos(x)})=1+1=2

11. Find the value of (lim_{xrightarrow 0}frac{xsin(x-1)}{(x-1)e^x}) is, where {x} is the fractional part of x.
a) 2⁄e
b) 1⁄e
c) 0
d) –1⁄e
Answer: b
Explanation:
(lim_{{x}rightarrow 1^+}frac{{x}sin(x-1)}{(x-1)e^x})
(=lim_{xrightarrow 1^+}{x}lim_{xrightarrow 1^+}{frac{sin(x-1)}{x-1}} lim_{xrightarrow 1^+}frac{1}{e^x}=lim_{xrightarrow 1^+}{x}1.frac{1}{e}=frac{1}{e})
because, (lim_{xrightarrow 1^+}{x}=1).

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250+ TOP MCQs on Partial Differentiation and Answers

Basic Engineering Mathematics Questions and Answers focuses on “Partial Differentiation – 2”.

1. Differentiation of function f(x,y,z) = Sin(x)Sin(y)Sin(z)-Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?
a) f’(x,y,z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)
b) f’(x,y,z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)
c) f’(x,y,z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)
d) f’(x,y,z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)
Answer: b
Explanation: f(x,y,z) = Sin(x)Sin(y)Sin(z)-Cos(x) Cos(y) Cos(z)
Since the function has 3 independent variables hence during differentiation we have to consider x and z as constant and differentiate it w.r.t. y,
f’(x,y,z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z).

2. In euler theorem x ∂z⁄∂x + y ∂z⁄∂y = nz, here ‘n’ indicates?
a) order of z
b) degree of z
c) neither order nor degree
d) constant of z
Answer: a
Explanation: Statement of euler theorem is “if z is an homogeneous function of x and y of order `n` then x ∂z⁄∂x + y ∂z⁄∂y = nz ”.

3. If z = xn f(y⁄x) then?
a) y ∂z⁄∂x + x ∂z⁄∂y = nz
b) 1/y ∂z⁄∂x + 1/x ∂z⁄∂y = nz
c) x ∂z⁄∂x + y ∂z⁄∂y = nz
d) 1/x ∂z⁄∂x + 1/y ∂z⁄∂y = nz
Answer: c
Explanation: Since the given function is homogeneous of order n , hence by euler’s theorem
x ∂z⁄∂x + y ∂z⁄∂y = nz.

4. Necessary condition of euler’s theorem is _________
a) z should be homogeneous and of order n
b) z should not be homogeneous but of order n
c) z should be implicit
d) z should be the function of x and y only
Answer: a
Explanation: Answer `z should be homogeneous and of order n` is correct as statement of euler’s theorem is “if z is an homogeneous function of x and y of order `n` then x ∂z⁄∂x + y ∂z⁄∂y = nz”
Answer `z should not be homogeneous but of order n` is incorrect as z should be homogeneous.
Answer `z should be implicit` is incorrect as z should not be implicit.
Answer `z should be the function of x and y only` is incorrect as z should be the homogeneous function of x and y not non-homogeneous functions.

5. If f(x,y) = x+y⁄y , x ∂z⁄∂x + y ∂z⁄∂y = ?
a) 0
b) 1
c) 2
d) 3
Answer: a
Explanation: Given function f(x,y)=(frac{x+y}{y}) can be written as f(x,y) =(frac{[1+frac{y}{x}]}{frac{y}{x}} = x^0 f(frac{y}{x})),
Hence by euler’s theorem,
(x frac{∂z}{∂x}+y frac{∂z}{∂y})=0

6. Does function f(x,y) = (Sin^{-1} [frac{(sqrt x+sqrt y)}{sqrt{x+y}}]) can be solved by euler’ s theorem.
a) True
b) False
Answer: b
Explanation: No this function cannot be written in form of xn f(y⁄x) hence it does not satisfies euler’s theorem.

7. Value of (x frac{∂u}{∂x}+y frac{∂u}{∂y}) if (u=frac{Sin^{-1} (frac{y}{x})(sqrt x+sqrt y)}{x^3+y^3}) is?
a) -2.5 u
b) -1.5 u
c) 0
d) -0.5 u
Answer: a
Explanation: Since the function can be written as,
(u=x^{-frac{5}{2}} frac{Sin^{-1} (frac{y}{x})(1+sqrt{frac{y}{x}})}{1+(frac{y}{x})^3}=x^n f(frac{y}{x})), by equler’s theorem,
(x frac{∂u}{∂x}+y frac{∂u}{∂y}= -frac{5}{2} u)

8. If u = xx + yy + zz , find du⁄dx + du⁄dy + du⁄dz at x = y = z = 1.
a) 1
b) 0
c) 2u
d) u
Answer: d
Explanation: du⁄dx = xx (1+log⁡(x))
du⁄dy = yy (1+log⁡(x))
du⁄dz = zz (1+log⁡(x))
At x = y = z = 1,
du⁄dx + du⁄dy + du⁄dz = u.

9. If (u=x^2 tan^{-1} (frac{y}{x})-y^2 tan^{-1} (frac{x}{y})) then (frac{∂^2 u}{∂x∂y}) is?
a) (frac{x^2+y^2}{x^2-y^2})
b) (frac{x^2-y^2}{x^2+y^2})
c) (frac{x^2}{x^2+y^2})
d) (frac{y^2}{x^2+y^2})
Answer: b
Explanation:
(frac{∂^2 u}{∂x∂y}=frac{∂}{∂x} frac{∂u}{∂y})
(frac{∂u}{∂y}=-2y ,Tan^{-1} (frac{x}{y})+frac{x^3}{x^2+y^2}+frac{xy^2}{x^2+y^2})
(frac{∂u}{∂y}=-2y ,Tan^{-1} (frac{x}{y})+x)
Now,
(frac{∂^2 u}{∂x∂y}=frac{∂}{∂x} frac{∂u}{∂y}= -frac{2y}{1+frac{x^2}{y^2}} (frac{1}{y})+1=frac{x^2-y^2}{x^2+y^2})

10. If f(x,y)is a function satisfying euler’ s theorem then?
a) (x^2frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
b) (frac{1}{x}^2frac{∂^2 f}{∂x^2}+frac{2}{xy} frac{∂^2 f}{∂x∂y}+frac{1}{y}^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
c) (x^2frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}=nf)
d) (y^2frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+x^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
Answer: a
Explanation: Since f satisfies euler’s theorem,
Since f satisfies euler’s theorem,
(x frac{∂z}{∂x}+y frac{∂z}{∂y})=nz
Differentiating it w.r.t x and y respectively we get,
(xfrac{∂^2 u}{∂x^2}+frac{∂u}{∂x}+y frac{∂^2 u}{∂x∂y}=n frac{∂u}{∂x}),
and
(x frac{∂^2 u}{∂y∂x}+frac{∂u}{∂y}+y frac{∂^2 u}{∂y^2}=n frac{∂u}{∂y})
Multiplying with x and y respectively,
(x^2 frac{∂^2 u}{∂x^2}+x frac{∂u}{∂x}+xy frac{∂^2 u}{∂x∂y}=nx frac{∂u}{∂x}),
and
(xy frac{∂^2 u}{∂y∂x}+y frac{∂u}{∂y}+y^2 frac{∂^2 u}{∂y^2}=ny frac{∂u}{∂y})
Adding above equations we get
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}=n(n-1)u )

11. Find the approximate value of [0.982 + 2.012 + 1.942](1⁄2).
a) 1.96
b) 2.96
c) 0.04
d) -0.04
Answer: b
Explanation: Let f(x,y,z) = (x2 + y2 + z2)(1⁄2) ……………..(1)
Hence, x = 1, y = 2, z = 2 so that, dx = -0.02, dy = 0.01, dz = -0.06
From (1),
∂f⁄∂x = x⁄f
∂f⁄∂y = y⁄f
∂f⁄∂z = z⁄f
And df=(frac{∂f}{∂x} dx+frac{∂f}{∂y} dy+frac{∂f}{∂z} dz=frac{(xdx+ydy+zdz)}{f}=frac{-0.02+0.02-0.12}{3})= -0.04
Hence,
([0.98^2+2.01^2+1.94^2]^{frac{1}{2}})=f(1,2,2)+df=3-0.04=2.96

12. The happiness(H) of a person depends upon the money he earned(m) and the time spend by him with his family(h) and is given by equation H=f(m,h)=400mh2 whereas the money earned by him is also depends upon the time spend by him with his family and is given by m(h)=√(1-h2). Find the time spend by him with his family so that the happiness of a person is maximum.
a) √(1⁄3)
b) √(2⁄3)
c) √(4⁄3)
d) 0
Answer: b
Explanation: Given,H=400mh2 and m=√(1-h2)
Let, ϑ=m2 + h2 + 1 = 0, hence
By lagrange’s method,
∂H⁄∂m + α∂v⁄∂m = 0
400h2 + 2m(α) = 0 …….(1)
Similarly,
∂H⁄∂h + α ∂v⁄∂h=0
800mh + 2h(α) = 0 ………(2)
Multiply by m and h in eq’1’ and eq’2’ respectively and adding them
α=-600mh2
Now from eq. 1 and 2 we get putting value of α,
m = 1⁄√3 and h = √(2⁄3).
Hence, total time spend by a person with his family is √(2⁄3).

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250+ TOP MCQs on Volume of Solid of Revolution and Answers

Differential and Integral Calculus Interview Questions and Answers for freshers focuses on “Volume of Solid of Revolution”.

1. The volume of solid of revolution when rotated along x-axis is given as _____________
a) (int_a^b πy^2 dx )
b) (int_a^b πy^2 dy )
c) (int_a^b πx^2 dx )
d) (int_a^b πx^2 dy )
Answer: a
Explanation: Volume is generated when a 2d surface is revolved along its axis. When revolved along x-axis, the volume is given as (int_a^b πy^2 dx ).

2. The volume of solid of revolution when rotated along y-axis is given as ________
a) (int_a^b πy^2 dx )
b) (int_a^b πy^2 dy )
c) (int_a^b πx^2 dx )
d) (int_a^b πx^2 dy )
Answer: d
Explanation: Volume is generated when a 2d surface is revolved along its axis. When revolved along y-axis, the volume is given as (int_a^b πx^2 dy ).

3. What is the volume generated when the ellipse (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) is revolved about its minor axis?
a) 4 ab cubic units
b) (frac{4}{3} a^2 b ) cubic units
c) (frac{4}{3} ab ) cubic units
d) 4 cubic units
Answer: b
Explanation: y- axis is the minor axis. (x^2 = frac{a^2}{b^2} (b^2 – y^2))
(V = int_a^b πx^2 dy)
(= int_{-b}^b π frac{a^2}{b^2} (b^2 – y^2) ,dy )
(= 2π frac{a^2}{b^2} Big(b^2 y- frac{y^3}{3}Big)_0^b )
(= 2π frac{a^2}{b^2} (b^3- frac{b^3}{3}) )
(= frac{4}{3} a^2 b ) cubic units.

4. What is the volume generated when the region surrounded by y = (sqrt{x}), y = 2 and y = 0 is revolved about y – axis?
a) 32π cubic units
b) (frac{32}{5} ) cubic units
c) (frac{32π}{5}) cubic units
d) (frac{5π}{32} ) cubic units
Answer: c
Explanation: Limits for y -> 0,2 x = y2
(Volume = int_a^b πx^2 dy)
( = int_0^2 πy^4 dy)
( = Big[frac{πy^5}{5}Big]_0^2)
( = frac{32π}{5}) cubic units.

5. What is the volume of the sphere of radius ‘a’?
a) (frac{4}{3} πa )
b) 4πa
c) (frac{4}{3} πa^2 )
d) (frac{4}{3} πa^3 )
Answer: d
Explanation: The equation of a circle is x2 + y2 = a2
When it is revolved about x-axis, the volume is given as
(V = 2 int_a^b πy^2 dy)
(= 2 int_0^a π(a^2-x^2) dx)
(= 2π Big(a^2 x – frac{x^3}{3}Big)_0^a)
(= frac{4}{3} πa^3.)

6. Gabriel’s horn is formed when the curve ____________ is revolved around x-axis for x≥1.
a) y = x
b) y = 1
c) y = 0
d) y = 1/x
Answer: d
Explanation: Gabriel’s horn or Torricelli’s Trumpet is a famous paradox. It has a finite volume but infinite surface area.

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250+ TOP MCQs on First Order Nonlinear Differential Equations and Answers

Ordinary Differential Equations Multiple Choice Questions on “First Order Nonlinear Differential Equations”.

1. Find the solution for (lim_{xto 0}⁡ frac{ax}{ax+x}.)
a) 0
b) a
c) 1
d) (frac{a}{a+1})
Answer: d
Explanation: The given equation can be solved using L-Hospital’s Rule,
(frac{d(ax)}{dx}=a, frac{d(ax+x)}{dx}=a+1)
(lim_{xto 0}⁡frac{ax}{ax+x} → lim_{xto 0}⁡frac{a}{a+1} → frac{a}{a+1})

2. A rectangular frame is to be made of 240 cm long. Determine the value of the length of the rectangle required to maximize the area.
a) 24 cm
b) 60 cm
c) 240 cm
d) 120 cm
Answer: b
Explanation: Let us consider ‘x’ as length and ‘y’ as the breadth of the rectangle.
Given: Perimeter 2(x + y) = 240 cm
x + y = 120
y = 120 – x
Area of the rectangle, a = x*y = x(120-x) = 120x – x2
Finding the derivative, we get, (frac{d(a)}{dx}= frac{d(120x- x^2)}{dx}=120-2x )
To find the value of x that maximizes the area, we substitute (frac{d(a)}{dx}= 0.)
Therefore, we get, 120 – 2x =0
2x = 120
x = 60 cm
To check if x = 60 cm is the value that maximizes the area, we find the second derivative of the area,
(frac{d^2 (a)}{dx^2}= -2) < 0 …………………. (i)
We know that the condition for maxima is (frac{d^2 (f(x))}{dx^2}<0,) which is satisfied by (i), therefore, x = 60 cm maximizes the area of the rectangle.

3. Find the solution of the system using Gauss Elimination method.

x – y + 2z = 8
y – z = 4
2x + 3z = 2

a) x = 18, y = -18, z = -22
b) x = -12, y = -18, z = 22
c) x = 34, y = -18, z = -22
d) x = -18, y = -12, z = 22
Answer: c
Explanation: Augmented Matrix of the given system is,
(
left[ begin{array}{ccc|c}
1 & -1 & 2 & 8 \
0 & 1 & -1 & 4 \
2 & 0 & 3 & 2
end{array} right])
Now, applying the steps for Gauss Elimination method (making the elements below the diagonal zero), we get,
(
left[ begin{array}{ccc|c}
1 & -1 & 2 & 8 \
0 & 1 & -1 & 4 \
2 & 0 & 3 & 2
end{array} right] quad ^{underrightarrow{R3 rightarrow R3 – 2R1}} )
(
left[ begin{array}{ccc|c}
1 & -1 & 2 & 8 \
0 & 1 & -1 & 4 \
0 & 2 & -1 & -14
end{array} right] quad ^{underrightarrow{R3 rightarrow R3 – 2R2}} )
(
left[ begin{array}{ccc|c}
1 & -1 & 2 & 8 \
0 & 1 & -1 & 4 \
0 & 0 & 1 & -22
end{array} right])
Now converting the augmented matrix back to set of equations, we get,
x – y + 2z = 8 …………………………. (i)
y – z = 4 …………………………………. (ii)
z = -22 …………………………………… (iii)
Substituting value of z in (ii), we get,
y + 22 = 4
y = -18
Substituting the value of y and z in (i), we get,
x + 18 + 2(-22) = 8
x – 26 = 8
x = 34
Therefore, the solution of the system is x = 34, y = -18, z = -22.

4. What is the solution of the given equation?
x6y6 dy + (x7y5 +1) dx = 0
a) (frac{(xy)^6}{6} + lnx = c )
b) (frac{(xy)^5}{6} + lny = c)
c) (frac{(xy)^5}{5}+ lnx = c)
d) (frac{(xy)^6}{6}+ lny = c)
Answer: a
Explanation: Given: (x6y6 + 1) dy + x7y5dx = 0, is an example of non-exact differential equation.
Dividing the equation by x we get,
x5y6 dy + x6y5dx + (frac{dx}{x} = 0)
x5y5 (ydy + xdx) + (frac{dx}{x} = 0 )
(xy)5 (d(xy)) + (frac{dx}{x} = 0)
(frac{(xy)^6}{6} + lnx = c )

5. Determine the current i(t) for the circuit shown, if the initial current is zero.

a) i(t) = 9 – 9e8t
b) i(t) = 9 – e-8t
c) i(t) = 9 – 9e-8t
d) i(t) = 8 – 9e-8t
Answer: c
Explanation: The given circuit is an application of First Order Differential equations (R-L Circuit).
Hence, we know that, (Lfrac{di(t)}{dt} + Ri(t) = E(t) )
(frac{1}{2} frac{di(t)}{dt} + 4i(t) = 10)
(frac{di(t)}{dt} + 8i(t) = 20) …………………………. (i)
Integrating Factor, (u(t) = e^{∫8 dt}= e^{8t}) …………………………… (ii)
Applying (ii) on both sides of (i) we get,
(e^{8t}.frac{di(t)}{dt} + e^{8t}. 8i(t) = 72 e^{8t} )
(frac{d(e^{8t}.i(t))}{dt} = 72 e^{8t} )
Integrating on both sides we get,
(∫frac{d(e^{8t}.i(t))}{dt} = ∫72 e^{8t}.dt)
(e^{8t} .i(t) = frac{72}{8} .e^{8t} + c)
i(t) = 9 + ce-8t
Given: i(0) = 0
i(0)= 9 + ce-8(0) = 0
c = -9
Therefore, i(t) = 9 – 9e-8t

6. (xy^3(frac{dy}{dx})^2+yx^2+frac{dy}{dx}=0) is a _____________
a) Second order, third degree, linear differential equation
b) First order, third degree, non-linear differential equation
c) First order, third degree, linear differential equation
d) Second order, third degree, non-linear differential equation
Answer: b
Explanation: Since the equation has only first derivative, i.e. ((frac{dy}{dx}),) it is a first order equation.
Degree is defined as the highest power of the highest order derivative involved. Hence it is 2.
The equation has one/more terms having a variable of degree two/higher; hence it is non-linear.

7. Which of the following is one of the criterions for linearity of an equation?
a) The dependent variable and its derivatives should be of second order
b) The dependent variable and its derivatives should not be of same order
c) Each coefficient does not depend on the independent variable
d) Each coefficient depends only on the independent variable
Answer: d
Explanation: The two criterions for linearity of an equation are:
The dependent variable y and its derivatives are of first degree.
Each coefficient depends only on the independent variable

8. Beta function is not a symmetric function.
a) True
b) False
Answer: b
Explanation: Beta function is a symmetric function, i.e.,
β(x,y) = β(y,x), where x>0 and y>0.

9. Which of the following is the property of error function?
a) erf (0) = 1
b) erf (∞) = 1
c) erf (0) = ∞
d) erf (∞) = 0
Answer: b
Explanation: Error Function is given by, (erf(x) = frac{2}{sqrt π} ∫_0^xe^{-t^2}dt. )
Some of its properties are:
erf (0) = 0
erf (∞) = 1
erf (-x) = -erf(x)

10. The equation (2frac{dy}{dx} – xy = y^{-2},) is an example for Bernoulli’s equation.
a) False
b) True
Answer: b
Explanation: A first order, first degree differential equation of the form,
(frac{dy}{dx} + P(x). y = Q(x). y^a,) is known as Bernoulli’s equation.

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250+ TOP MCQs on Laplace Transform by Properties and Answers

Engineering Mathematics Questions and Answers for Campus interviews focuses on “Laplace Transform By Properties – 2”.

1. Transfer function may be defined as ____________
a) Ratio of out to input
b) Ratio of laplace transform of output to input
c) Ratio of laplace transform of output to input with zero initial conditions
d) None of the mentioned
Answer: c
Explanation: Transfer function may be defined as the ratio of laplace transform of output to input with zero initial conditions.

2. Poles of any transfer function is define as the roots of equation of denominator of transfer function.
a) True
b) False
Answer: a
Explanation: Let transfer function be defined as G(s)/H(s), then poles of transfer function may be defined as H(s)=0.

3. Zeros of any transfer function is define as the roots of equation of numerator of transfer function.
a) True
b) False
Answer: a
Explanation: Let transfer function be defined as G(s)/H(s), then zeros of transfer function may be defined as G(s)=0.

4. Find the poles of transfer function which is defined by input x(t)=5Sin(t)-u(t) and output y(t)=Cos(t)-u(t).
a) 4.79, 0.208
b) 5.73, 0.31
c) 5.89, 0.208
d) 5.49, 0.308
Answer: a
Explanation: Given ,y(t)=Cos(t) – u(t) and x(t) = 5Sin(t) – u(t),
Hence, transfer function H(s)=(frac{[frac{s}{s^2+1}-frac{1}{s}]}{[frac{5}{s^2+1}-frac{1}{s}]} =-frac{1}{frac{s(s^2+1)}{frac{(5s-s^2-1)}{s(s^2+1)}}}=frac{1}{S^2-5S+1})
Roots of equation s2 – 5s + 1 = 0 is s = 4.79, 0.208.

5. Find the equation of transfer function which is defined by y(t)-∫0t y(t)dt + d⁄dt x(t) – 5Sin(t) = 0.
a) (frac{s(e^{-as}-1)}{s-1})
b) (frac{(e^{-as}-s)}{s-1})
c) (frac{s(e^{-as}-s)}{s-1})
d) (frac{s(e^{-as}-s^2)}{s-1})
Answer: c
Explanation:
Given, (y(t)-∫_0^t y(t)dt+frac{d}{dt} x(t)-x(t-a)=0)
Taking Laplace, (Y(s)-frac{Y(s)}{s}+sX(s)-e^{-as} X(s)=0)
H(s)=Y(s)/X(s) =(frac{(e^{-as}-s)}{1-frac{1}{s}}=frac{s(e^{-as}-s)}{s-1})

6. Find the poles of transfer function given by system d2⁄dt2 y(t) – d⁄dt y(t) + y(t) – ∫0t x(t)dt = x(t).
a) 0, 0.7 ± 0.466
b) 0, 2.5 ± 0.866
c) 0, 0 .5 ± 0.866
d) 0, 1.5 ± 0.876
Answer: c
Explanation: We know that,
Given, (frac{d^2}{dt^2} y(t)-frac{d}{dt} y(t)+y(t)-int_0^t x(t)dt=x(t))
Now, Taking laplace, we get, ((s^2-s+1)Y(s)=(1+frac{1}{s})X(s))
H(s)=(frac{s+1}{s(s^2-s+1)})
Roots of s3-s2+s=0, are 0,.5±0.866

7. Find the transfer function of a system given by equation d2⁄dt2 y(t-a) + x(t) + 5 d⁄dt y(t) = x(t-a).
a) (e-as-s)/(1+e-as s2)
b) (e-as-5s)/(e-as s2)
c) (e-as-s)/(2+e-as s2)
d) (e-as-5s)/(1+e-as s2)
Answer: d
Explanation: Given, d2⁄dt2 y(t-a) + x(t) + 5 d⁄dt y(t) = x(t-a).

Taking laplace transform, s2 Y(s) e-sa + X(s) + 5sY(s) = e-as X(s)

Hence, H(s) = Y(s)⁄X(s) =(e-as-5s)/(1+e-as s2).

8. Any system is said to be stable if and only if ____________
a) It poles lies at the left of imaginary axis
b) It zeros lies at the left of imaginary axis
c) It poles lies at the right of imaginary axis
d) It zeros lies at the right of imaginary axis
Answer: a
Explanation: Any system is said to be stable if and only if it poles lies at the left of imaginary axis.

9. The system given by equation 5 d3⁄dt3 y(t) + 10 d⁄dt y(t) – 5y(t) = x(t) + ∫0t x(t)dt, is?
a) Stable
b) Unstable
c) Has poles 0, 0.455, -0.236±1.567
d) Has zeros 0, 0.455, -0.226±1.467
Answer: a
Explanation:
Given, 5 d3⁄dt3 y(t) + 10 d⁄dt y(t) – 5y(t) = x(t) + ∫0t x(t)dt,
By laplace transform, 5s3 Y(s)+10sY(s)-5Y(s)=X(s)+1/s X(s)
Hence,
(frac{Y(s)}{X(s)} =frac{[1+1/s]}{5s^3+10s-5}=frac{[s+1]}{s(5s^3+10s-5)})
Poles of (frac{Y(s)}{X(s)}) are, 0 ,0.455,-0.226±1.467
Hence system is Unstable.

10. Find the laplace transform of input x(t) if the system given by d3⁄dt3 y(t) – 2 d2⁄dt2 y(t) –d⁄dt y(t) + 2y(t) = x(t), is stable.
a) s + 1
b) s – 1
c) s + 2
d) s – 2
Answer: b
Explanation: d3⁄dt3 y(t) – 2 d2⁄dt2 y(t) – d⁄dt y(t) + 2y(t) = x(t),
Taking laplace transform,
(s3 – 2s2 – s + 2)Y(s) = X(s)
H(s) = Y(s)⁄X(s) = 1⁄(s-1)(s+1)(s+2)
For the system to be stable, X(s) = s – 1.

11. The system given by equation y(t – 2a) – 3y(t – a) + 2y(t) = x(t – a) is?
a) Stable
b) Unstable
c) Marginally stable
d) 0
Answer: a
Explanation: Given, y(t-2a)-3y(t-a)+2y(t)=x(t-a), then
Taking Laplace transform, e-2as Y(s)-3e-as Y(s)+2Y(s)=e-as X(s),
Hence, (H(s)=frac{e^{-as}}{(1-e^{-as})(2-e^{-as})})
To find the stability, we should have, (∫_{-∞}^∞ H(s)ds>0)
Hence, (int_{-infty}^infty frac{e^{-as}}{(1-e^{-as})(2-e^{-as})} ds)
Let, (e^{-as}=z=>-ae^{-as} ds=dz)
(frac{1}{a} int_0^∞ frac{1}{(1-z)(2-z)} dz=frac{1}{a} int_0^∞ left [frac{1}{(z-1) }-frac{1}{z-2} right ]dz=frac{1}{a} ln⁡left [frac{z-1}{z-2}right ])
putting, z=0, (frac{1}{a} ln⁡left [frac{z-1}{z-2}right ]=frac{1}{a} ln⁡(frac{1}{2}))
putting, z=∞, (frac{1}{a} ln⁡left [frac{z-1}{z-2}right ]=frac{1}{a} ln⁡[frac{1-1/z}{1-2/z}])=0
hence, (int_{-infty}^infty frac{e^{-as}}{(1-e^{-as})(2-e^{-as})} ds=frac{1}{a} ln⁡(frac{1}{2}))

It is stable.

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250+ TOP MCQs on Eigenvalues and Vectors of a Matrix and Answers

Linear Algebra Multiple Choice Questions on “Eigenvalues and Vectors of a Matrix”.

1. Find the Eigen values for the following 2×2 matrix.
A=(begin{bmatrix}1&8\2&1end{bmatrix}).
a) -3
b) 2
c) 6
d) 4
Answer: a
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
[A-λI]=(begin{bmatrix}1&8\2&1end{bmatrix} -lambda begin{bmatrix}1&0\0&1end{bmatrix})
[A-λI]=(begin{bmatrix}1-lambda &8\2&1-lambda end{bmatrix})
|A-λI|=(1-λ)(1-λ)-16=0
(1-λ)2=16
(1-λ)=±4
λ=-3 or λ=5.

2. Find the Eigenvalue for the given matrix.
A=(begin{bmatrix}4&1&3\1&3&1\2&0&5end{bmatrix}).
a) 13
b) -3
c) 7.1
d) 8.3
Answer: c
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
[A-λI]=(begin{bmatrix}4&1&3\1&3&1\2&0&5end{bmatrix}-lambda begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
The characteristic polynomial is given by-
λ3-(Sum of diagonal elements) λ2+(Sum of minor of diagonal element)λ-|A|=0
λ3-12λ2+40λ-39=0
λ=7.1 or λ=3.

3. Find the Eigen vector for value of λ=-2 for the given matrix.
A=(begin{bmatrix}3&5\3&1end{bmatrix}).
a) (begin{bmatrix}0\-1end{bmatrix})
b) (begin{bmatrix}1\-1end{bmatrix})
c) (begin{bmatrix}-1\-1end{bmatrix})
d) (begin{bmatrix}1\0end{bmatrix})
Answer: b
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
Given that, λ=-2
[A-λI]=(begin{bmatrix}3&5\3&1end{bmatrix}-(-2)begin{bmatrix}1&0\0&1end{bmatrix})
[A-λI]=(begin{bmatrix}3+2&5\3&1+2end{bmatrix})
[A-λI]=(begin{bmatrix}5&5\3&3end{bmatrix})
Since, [A-λI]X=0
(begin{bmatrix}5&5\3&3end{bmatrix}
begin{bmatrix}x\yend{bmatrix} = begin{bmatrix}0\0end{bmatrix})
Thus,
5x+5y=0 and 3x+3y=0
Let x=t,
Then, y=-t
X = (begin{bmatrix}t\-tend{bmatrix} = begin{bmatrix}1\-1end{bmatrix})

4. Find the Eigen vector for value of λ=3 for the given matrix.
A=(begin{bmatrix}3&10&5\-2&-3&-4\3&5&7end{bmatrix}).
a) (begin{bmatrix}-1\-1\2end{bmatrix})
b) (begin{bmatrix}-1\1\2end{bmatrix})
c) (begin{bmatrix}-1\-1\-2end{bmatrix})
d) (begin{bmatrix}-1\-2\2end{bmatrix})
Answer: a
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
Given that, λ=3
[A-λI]=(begin{bmatrix}3&10&5\-2&-3&-4\3&5&7end{bmatrix}-(3)begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
[A-λI]=(begin{bmatrix}3-3&10&5\-2&-3-3&-4\3&5&7-3end{bmatrix})
[A-λI]=(begin{bmatrix}0&10&5\-2&-6&-4\3&5&4end{bmatrix})
Since, [A-λI]X=0
(begin{bmatrix}0&10&5\-2&-6&-4\3&5&4end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})
Using Row transformation

(1) Interchanging R1 and R2/2
(begin{bmatrix}-1&-3&-2\0&10&5\3&5&4end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})
(2) R3=R3+3R1
(begin{bmatrix}0&10&5\0&10&5\0&-4&-2end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})
(3) R2=R2/5 and R3=R3+2R2
(begin{bmatrix}-1&-3&-2\0&2&1\0&0&0end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})
Thus,
-x-3y-2z=0 and 2y+z=0
Let z=t,then y=(frac{-t}{2}) and x=(frac{-t}{2})

X = (begin{bmatrix}-1\-1\2end{bmatrix})

5. Find the Eigen value and the Eigen Vector for the given matrix.
A=(begin{bmatrix}3&4&2\1&6&2\1&4&4end{bmatrix}).
a) 3, (begin{bmatrix}1\1\1end{bmatrix})
b) 9, (begin{bmatrix}1\1\1end{bmatrix})
c) 9, (begin{bmatrix}1\0\1end{bmatrix})
d) 2, (begin{bmatrix}1\0\1end{bmatrix})
Answer: b
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
[A-λI]=(begin{bmatrix}3&4&2\1&6&2\1&4&4end{bmatrix}-lambda begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
The characteristic polynomial is given by-
λ3-(Sum of diagonal elements) λ2+(Sum of minor of diagonal element)λ-|A|=0

λ3-13λ2+40λ-36=0
λ=9 or λ=2
For λ=9,
[A-λI]=(begin{bmatrix}3&4&2\1&6&2\1&4&4end{bmatrix}-9begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
[A-λI]=(begin{bmatrix}3-9&4&2\1&6-9&2\1&4&4-9end{bmatrix})
[A-λI]=(begin{bmatrix}-6&4&2\1&-3&2\1&4&-5end{bmatrix})
[A-λI]X=0
(begin{bmatrix}-6&4&2\1&-3&2\1&4&-5end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})

(1) Interchanging R1 and R2
(begin{bmatrix}1&-3&2\-6&4&2\1&4&-5end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})

(2) R2=R2+6R1 and R3=R3-R1
(begin{bmatrix}1&-3&2\0&-14&14\0&7&-7end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})

(3) R3=R3+R2/2
(begin{bmatrix}1&-3&2\0&-14&14\0&0&0end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})

-14y+14z=0
x-3y+2z=0
Let z=t
Then, y=t and x=t
X=(begin{bmatrix}1\1\1end{bmatrix}).

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