300+ REAL TIME Engineering Mathematics Objective Questions & Answers 2023

250+ TOP MCQs on Indeterminate Forms and Answers

Engineering Mathematics Multiple Choice Questions on “Indeterminate Forms – 2”.

1. Find (lt_{xrightarrow -2}frac{sin(frac{1+(frac{(x+2)^2(x^2+1)}{x^3+3})}{x+2})}{(x+2)})
a) ∞
b) 0
c) 2
d) -∞
Answer: c
Explanation: First evaluate
(=lt_{xrightarrow -2}frac{ln(1+frac{(x+2)^2(x^2+1)}{x^3+3})}{x+2})
(=lt_{xrightarrow -2}(frac{1}{x+2})times(frac{(x+2)^2(x^2+1)}{(x^3+3)}-frac{(x+2)^3(x^3+1)^2}{2.(x^3+3)^2}+…infty))
(=lt_{xrightarrow -2}times(frac{(x+2)^2(x^2+1)}{(x^3+3)}-frac{(x+2)^3(x^3+1)^2}{2.(x^3+3)^2}+…infty))
We hence the form for the totle limit as
(lt_{xrightarrow a}frac{sin(f(x))}{g(x)})=1
Where f(x)→0 : g(x)→0 as x→a
This is true for the above problem
Thus, we can deduce the limit as
= 1
Hence, 2 is the right answer.

2. Find (lt_{xrightarrow 0}frac{(3e^x-2e^{2x}-e^{3x})}{(e^x+e^{2x}-2e^{3x})})
a) ³⁄₂
b) 0
c) ⁴⁄₃
d) –⁴⁄₃
Answer: c
Explanation: Form is 0 / 0
Applying L hospitals rule we have
(=lt_{xrightarrow 0}frac{3e^x-4e^{2x}-3e^{3x}}{e^x+2e^{2x}-6e^{3x}})
(=frac{3-4-3}{1+2-6})
(=frac{4}{3})

3. Find relation between a and b such that the following limit is got after a single application of L hospitals Rule (lt_{xrightarrow 0}frac{ae^x+be^{2x}}{be^x+ae^{2x}})
a) ^b⁄_a = 2
b) ^a⁄_b = 2
c) a = b
d) a = -b
Answer: d
Explanation: Given differentiation is applied once we get
ae⁰ + be⁰ = 0 = a + b (numerator → zero)
be⁰ + ae⁰ = 0 = a + b (denominator → zero)
Thus the relation between (a, b) and is
a + b = 0
OR
a = -b.

4. Find (lt_{xrightarrow 0}frac{2cos(2x)+3cos(5x)-5cos(19x)}{cos(4x)-cos(3x)})
a) -76
b) -6
c) -7
d) 0
Answer: a
Explanation: Form here ⁰⁄₀
Applying L hospitals rule we have
(=lt_{xrightarrow 0}frac{4cos(2x)+15cos(5x)-95cos(19x)}{4cos(4x)-3cos(3x)})
(frac{4+15-95}{4-3}) = -76.

5. Find how many rounds of differentiation are required to have finite limit for (lt_{xrightarrow 0}frac{cos(ax)+cos(bx)-2cos(cx)}{cos(ax)+2cos(bx)-3cos(cx)}) given that a ≠ b ≠ c
a) 3
b) 0
c) 2
d) 4
Answer: c
Explanation: Applying L hospitals rule
(=lt_{xrightarrow 0}frac{a.cos(ax)+b.cos(bx)-2c.cos(cx)}{a.cos(ax)+2b.cos(bx)-3c.cos(cx)}=lt_{xrightarrow 0}frac{a+b-2c}{a+2b-3c})
Assume now that
a + b + 2c = 0 and a + 2b – 3c = 0
We must have
a = c = b but given a ≠ c ≠ b
Thus, our assumption is false and a finite limit exists after first round of differentiation.
Hence, 2 is the right answer.

6. Find (lt_{prightarrowinfty}frac{p^5.p!}{5.6…(5+p)})
a) 4!
b) 5!
c) 0
d) ∞
Answer: a
Explanation: (=lt_{prightarrowinfty}frac{1.2.3.4}{1.2.3.4}times frac{p^5.p!}{5.6…(5+p)})
(=lt_{prightarrowinfty}frac{4!.p^5.p!}{(p+5)!})
(=lt_{prightarrowinfty}frac{4!.p^5}{(p+5)(p+4)(p+3)(p+2)(p+1)})
(=lt_{prightarrowinfty}(4!)timesfrac{p^5}{(p+5)(p+4)(p+3)(p+2)(p+1)}=4!times(1))
= 4!

7. Find (=lt_{xrightarrow 0}frac{sin(x)}{tan(x)})
a) 0
b) 1
c) ∞
d) 2
Answer: b
Explanation: (=lt_{xrightarrow 0}frac{frac{sin(x)}{x}}{frac{tan(x)}{x}})
(=frac{lt_{xrightarrow 0}frac{sin(x)}{x}}{lt_{xrightarrow 0}frac{tan(x)}{x}})
=1/1=1

8. Find (lt_{xrightarrow 1012345}(frac{[sinh(x)]^2-[cosh(x)]^2}{[sinh(x)]^2+[cosh(x)]^2}))
a) ¹⁄_{cosh(1012345)}
b) 90987
c) 1012345
d) ∞
Answer: a
Explanation: (lt_{xrightarrow 1012345}left ( frac{1}{(frac{(e^x+e^{-2x})^2+(e^x-e^{-x})^2}{4})}right ))
(=lt_{xrightarrow 1012345}frac{1}{frac{(e^{2x}+e^{-2x})}{2}}=frac{1}{cosh(1012345)})

9. Let f oⁿ (f(x)) denote the composition of f(x) with itself n number of times then the value of lt_{n → ∞} f oⁿ (sin(x)) =
a) -1
b) 2
c) ∞
d) 0
Answer: d
Explanation: Drawing the graph of y = x and y = sin(x) we can write the limit value as 0.

10. Find (lt_{xrightarrow 0}frac{sin(x^2)}{x})
a) ∞
b) -1
c) 0
d) 2²
Answer: c
Explanation: Expand into Taylor Series
(=lt_{xrightarrow 0}(frac{1}{x})times(frac{x^2}{1!}-frac{x^6}{3!}+..infty))
(=lt_{xrightarrow 0}times(frac{x}{1!}-frac{x^5}{3!}+..infty))
=0

11. Find (lt_{xrightarrow -33}frac{ln(x^3+68x^2+1222x+2179)-ln(x+1)}{(x^2+66x+1089)})
a) -33
b) ¹⁄₂
c) 0
d) ³¹⁄₃₂
Answer: d
Explanation: (=lt_{xrightarrow -33}frac{ln(1+frac{(x+33)^2(x+2)}{(x+1)})}{(x+33)^2})
(=lt_{xrightarrow -33}(frac{1}{(x+33)^2})times(frac{(x+33)^2(x+2)}{(x+1)}-frac{(x+33)^4(x+2)^2}{2(x+1)^2}….infty))
(=lt_{xrightarrow -33}(frac{(x+2)}{(x+1)}-frac{(x+33)^2(x+2)^2}{2(x+1)^2}….infty))
(=lt_{xrightarrow -33}(frac{(x+2)}{(x+1)})=frac{(-33+2)}{(-33+1)})

12. Find (lt_{prightarrowinfty}frac{p^{frac{1}{2}}.p!}{frac{1}{2}.frac{3}{2}…(p+frac{1}{2})})
a) √π
b) ∞
c) √^π⁄₂
d) 0
Answer: c
Explanation: Using the Gauss definition of the Gamma function we have
(tau(x)=lt_{prightarrowinfty}frac{p^x.p!}{x.(x+1)…(x+p)})
Where τ(x) is the Gamma function Using formula
(tau(x) times tau(1-x)=frac{pi.x}{sin(pi.x)})
put x=1/2 to get
((tau(frac{1}{2}))^2=frac{pi}{2.sin(frac{pi}{2})}=frac{pi}{2})
(tau(frac{1}{2})=sqrt{frac{pi}{2}})

13. Find (lt_{nrightarrowinfty}(1+frac{1}{n})^n)
a) e
b) e – 1
c) 0
d) ∞
Answer: a
Explanation: (lt_{nrightarrowinfty}(1+frac{1}{n})^n=e^{lt_{nrightarrowinfty}frac{n}{n}})
(lt_{nrightarrowinfty}frac{n}{n}) = 1
= e

250+ TOP MCQs on Partial Differentiation and Answers

Engineering Mathematics Multiple Choice Questions on “Partial Differentiation – 1”.

1. f(x, y) = x² + xyz + z Find f_x at (1,1,1)
a) 0
b) 1
c) 3
d) -1
Answer: c
Explanation: f_x = 2x + yz
Put (x,y,z) = (1,1,1)
f_x = 2 + 1 = 3.

2. f(x, y) = sin(xy) + x² ln(y) Find f_yx at (0, ^π⁄₂)
a) 33
b) 0
c) 3
d) 1
Answer: d
Explanation: f_y = xcos(xy) + ^x²⁄_y
f_yx = cos(xy) – xysin(xy) + ^2x⁄_y
Put (x,y) = (0, ^π⁄₂)
= 1.

3. f(x, y) = x² + y³ ; X = t² + t³; y = t³ + t⁹ Find ^df⁄_dt at t=1.
a) 0
b) 1
c)-1
d) 164
Answer: d
Explanation: Using chain rule we have
(frac{df}{dt}=f_x.frac{dx}{dt}+f_y.frac{dy}{dt})
=(2x).(2t + 3t²) + (3y²).(3t² + 9t⁸)
Put t = 1; we have x = 2; y = 2
=4.(5) + 12.(12) = 164.

4. f(x, y) = sin(x) + cos(y) + xy²; x = cos(t); y = sin(t) Find ^df⁄_dt at t = ^π⁄₂
a) 2
b)-2
c) 1
d) 0
Answer: b
Explanation:Using chain rule we have
(frac{df}{dt}=f_x.frac{dx}{dt}+f_y.frac{dy}{dt})
= (cos(x) + y²).(-sin(t)) + (-sin(y) + 2xy).(cos(t))
Put t= ^π⁄₂; we have x=0; y=1
=(1 + 1).(-1) + 0 = -2.

5. f(x, y, z, t) = xy + zt + x² yzt; x = k³ ; y = k²; z = k; t = √k
Find ^df⁄_dt at k = 1
a) 34
b) 16
c) 32
d) 61
Answer: b
Explanation: Using Chain rule we have
(frac{df}{dt}=f_x.frac{dx}{dk}+f_y.frac{dy}{dk}+f_z.frac{dz}{dk}+f_t.frac{dt}{dk})
= (y + 2xyzt).(3k²) + (x + x²zt).(2k) + (t + x²yt).(1) + (z + x²yz).((frac{1}{2sqrt{k}})
Put k=1; we have x=y=z=t=1
9 + 4 + 2 + 1 = 16.

6. The existence of first order partial derivatives implies continuity.
a) True
b) False
Answer: b
Explanation: The mere existence cannot be declared as a condition for contnuity because the second order derivatives should also be continuous.

7. The gradient of a function is parallel to the velocity vector of the level curve.
a) True
b) False
Answer: b
Explanation: The gradient is perpendicular and not parallel to the velocity vector of the level curve.

8. f(x, y) = sin(y + yx²) / 1 + x² Value of f_xy at (0,1) is
a) 0
b) 1
c) 67
d) 90
Answer: a
Explanation: First find
f_y = cos(y + yx²)
Hence
f_yx = f_xy = – (2xy).sin(y + yx²)
Now put (x,y) = (0,1)
= 0.

9. f(x, y) = sin(xy + x³y) / x + x³ Find f_xy at (0,1).
a) 2
b) 5
c) 1
d) undefined
Answer: c
Explanation: First find
f_y = sin(xy + x³y)
Hence
f_yx = f_xy = (cos(xy + x³y)) . (y + 3x²³y)
Now put (x,y) = (0,1)
= 1.

250+ TOP MCQs on Rectification in Polar and Parametric Forms and Answers

Differential and Integral Calculus Multiple Choice Questions on “Rectification in Polar and Parametric Forms”.

1. Find the length of the curve given by the equation.
(x^{frac{2}{3}}+y^frac{2}{3}=a^frac{2}{3})
a) (frac{3a}{2})
b) (frac{-7a}{2})
c) (frac{-3a}{4})
d) (frac{-3a}{2})
View Answer

Answer: d
Explanation: We know that,
S=(int_{x1}^{x2}sqrt{1+frac{dy}{dx}^2})
(y^frac{2}{3}=a^frac{2}{3}-x^frac{2}{3})
Differentiating on both sides
(frac{2}{3} y^{frac{2}{3}-1}= frac{-2}{3} x^{frac{2}{3}-1})
(frac{dy}{dx} = -frac{y}{x}^{frac{1}{3}})
((frac{dy}{dx})^2 = (frac{y}{x})^{frac{1}{3}})
(1+(frac{dy}{dx})^2=1+(frac{y}{x})^frac{2}{3})
Substituting from the original equation-
(1+(frac{dy}{dx})^2=(frac{a}{x})frac{2}{3})
(sqrt{1+frac{dy^2}{dx}}=(frac{a}{x})^{frac{1}{3}})
(S=int_{a}^{0}(frac{a}{x})^{frac{1}{3}} dx )
(s=frac{-3a}{2})
Thus, length of the given curve is (frac{-3a}{2}).

2. Find the length of one arc of the given cycloid.

x=a(θ-sinθ)
y=a(1+cosθ)

a) a
b) 4a
c) 8a
d) 2a
View Answer

Answer: c
Explanation: We know that
(s=int_{theta1}^{theta2}sqrt{(frac{dx}{dtheta})^2+(frac{dy}{dtheta})^2})
(frac{dx}{dtheta}=a(1-costheta))
(frac{dy}{dtheta}=a(-sintheta))
((frac{dx}{dtheta})^2+(frac{dy}{dtheta})^2=a^2(1-costheta)^2+a^2 sin^2theta)
((frac{dx}{dtheta})^2+(frac{dy}{dtheta})^2=4a^2 sin^2frac{theta}{2})
(s=int_{0}^{2}pisqrt{4a^2 sin^2frac{theta}{2}} dtheta)
On solving the given integral, we get
s=8a
Thus length of one arc of the given cycloid is 8a.

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 Bernoulli Equations and Answers

Ordinary Differential Equations Multiple Choice Questions on “Bernoulli Equations”.

1. Solution of the differential equation (frac{dy}{dx} + frac{y}{x} = y^2) x is ________
a) (frac{1}{y} = -x + c)
b) (frac{1}{xy} = -x + c)
c) (frac{1}{(xy)^2} = -y + c)
d) (frac{1}{xy} = -x^2 + c)
Answer: b
Explanation: Given equation is of the form (frac{dy}{dx} + Py = Qy^n) …divide by y²
where P and Q are functions of x hence this is Bernoulli’s equation in (y, frac{1}{y^2} frac{dy}{dx} + frac{1}{yx} = x)
put (frac{1}{y} = t rightarrow frac{-1}{y^2} frac{dy}{dx} = frac{dt}{dx}) substituting we get – (frac{dt}{dx} + frac{t}{x} = x ,or, frac{dt}{dx} – frac{t}{x} = -x)
this equation is linear in t i.e it is of the form
(frac{dt}{dx} + Pt = Q, I.F = e^{int P ,dx} =e^{int frac{-1}{x} ,dx} = e^{-log⁡x} = frac{1}{x})
its solution is
(te^{int P ,dx} = int Q e^{int P ,dx} ,dx + c rightarrow t frac{1}{x} = int -x * frac{1}{x} ,dx + c rightarrow frac{t}{x} = -x + c ,but, t = frac{1}{y})
thus solution is given by (frac{1}{xy} = -x + c).

2. Solution of the differential equation (frac{dy}{dx} -y ,tan⁡x = frac{sin⁡x cos^2⁡x}{y^2}) is ______
a) (y ,cos^2 x = frac{-cos^4 xsin^2 x}{2} + c)
b) (y^2 cos^2 x = frac{sin^6 x}{2} + c )
c) (y^3 cos^3 x = frac{-cos^6 x}{2} + c )
d) (y^4 cos^5 x = frac{sinx cos⁡x}{2} + c )
Answer: c
Explanation: (frac{dy}{dx} -y ,tan⁡x = frac{sin⁡x cos^2⁡x}{y^2}) multiplying by
(y^2 rightarrow y^2 frac{dy}{dx} – y^3 tan⁡x = sin⁡x ,cos^2⁡x )
put (t = y^3rightarrow3y^2 frac{dy}{dx} = frac{dt}{dx} ,or, y^2 frac{dy}{dx} = frac{1}{3} frac{dt}{dx})
substituting (frac{dt}{dx}-3t ,tan⁡x = 3sin⁡x ,cos^2⁡x )
this equation is linear in t i.e it is of the form (frac{dt}{dx} + Pt = Q, e^{int P ,dx} = e^{int -3tan⁡x ,dx})
(e^{-3 log⁡, sec⁡x} = cos^3 ,x) its solution is (te^{int P ,dx} = int Q ,e^{int P ,dx} dx + c)
t cos³ x=∫3sin⁡x cos²x cos³ x dx + c = ∫3sin⁡x cos⁵⁡x dx+c, put v=cos x
dv=-sin x dx i.e (int 3sin⁡x ,cos^5, ⁡x ,dx = int -3v^5 ,dx = frac{-v^6}{2} = frac{-cos^6 x}{2}) hence its solution becomes (t cos^3 x = frac{-cos^6 x}{2} + c rightarrow y^3 cos^3 ,x = frac{-cos^6 x}{2} + c.)

3. Solution of the differential equation 6y² dx – x(x³ + 2y)dy = 0 is ________
a) (frac{y}{x^3} = frac{-log⁡y}{2} + c )
b) (frac{y^2}{x^3} = frac{-log⁡x}{4} + c )
c) (frac{x}{y^3} = frac{-log⁡x}{2} + c )
d) (frac{x}{y^2} = frac{-log⁡y}{4} + c )
Answer: a
Explanation: Equation is reduced to (frac{dy}{dx} = frac{x(x^3+2y)}{6y^2} ,i.e, frac{dx}{dy}-frac{x}{3y} = frac{x^4}{6y^2}) …divide by x⁴
we get (frac{1}{x^4} frac{dy}{dx} – frac{1}{3x^3 y} = frac{1}{6y^2} ,put, frac{1}{x^3} = t rightarrow frac{-3}{x^4} frac{dx}{dy} = frac{dt}{dy} ,or, frac{1}{x^4} frac{dy}{dx} = frac{-1}{3} frac{dt}{dy} )
substituting (frac{-1}{3} frac{dt}{dy} – frac{t}{3y} = frac{1}{6y^2} ,or, frac{dt}{dy} + frac{t}{y} = frac{-1}{2y^2}) this equation is linear in t i.e it is of the form
(frac{dt}{dy} + Pt = Q ,where, P=frac{1}{y}, Q=frac{-1}{2y^2} ,I.F, = e^{int P ,dy} = e^{int frac{1}{y} ,dy} = e^{log⁡y} = y) its solution is (te^{int P ,dy} = int Q ,e^{int P ,dy} + c rightarrow ty = int y * frac{-1}{2y^2} ,dy + c = frac{-log⁡y}{2} + c)
(frac{y}{x^3} = frac{-log⁡y}{2} + c ) is the required solution.

Global Education & Learning Series – Ordinary Differential Equations.

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250+ TOP MCQs on Laplace Transform by Properties and Answers

Engineering Mathematics Multiple Choice Questions on “Laplace Transform by Properties – 1”.

1. Laplace of function f(t) is given by?
a) F(s)=(int_{-infty}^infty f(t)e^{-st} ,dt)
b) F(t)=(int_{-infty}^infty f(t)e^{-t} ,dt)
c) f(s)=(int_{-infty}^infty f(t)e^{-st} ,dt)
d) f(t)=(int_{-infty}^infty f(t)e^{-t} ,dt)
Answer: a
Explanation: Laplace of function f(t) is given by
F(s)=(int_{-infty}^infty f(t)e^{-st} ,dt).

2. Laplace transform any function changes it domain to s-domain.
a) True
b) False
Answer: a
Explanation: Laplace of function f(t) is given by F(s)=(int_{-infty}^infty f(t)e^{-st} ), hence it changes domain of function from one domain to s-domain.

3. Laplace transform if sin⁡(at)u(t) is?
a) s ⁄ a²+s²
b) a ⁄ a²+s²
c) s² ⁄ a²+s²
d) a² ⁄ a²+s²
Answer: b
Explanation: We know that,
F(s)=(int_{-infty}^infty sin⁡(at)u(t) e^{-st} dt=int_0^∞ sin⁡(at)e^{-st} dt)
=(left [frac{e^{-st}}{a^2+s^2}[-ssin(at)-acos⁡(at)]right ]_∞^0)
=(frac{a}{a^2+s^2})

4. Laplace transform if cos⁡(at)u(t) is?
a) s ⁄ a²+s²
b) a ⁄ a²+s²
c) s² ⁄ a²+s²
d) a² ⁄ a²+s²
Answer: a
Explanation: We know that,
F(s)=(int_{-infty}^infty cos(at)u(t) e^{-st} dt=int_0^∞ cos⁡(at)e^{-st} dt)
=(left [frac{e^{-st}}{a^2+s^2}[-scos(at)-asin⁡(at)]right ]_∞^0)
=(frac{a}{a^2+s^2})

5. Find the laplace transform of e^t Sin(t).
a) (frac{a}{a^2+(s+1)^2})
b) (frac{a}{a^2+(s-1)^2})
c) (frac{s+1}{a^2+(s+1)^2})
d) (frac{s+1}{a^2+(s+1)^2})
Answer: b
Explanation:
F(s)=(int_{-infty}^infty e^t sin⁡(at)u(t) e^{-st} dt=∫_0^∞ sin⁡(at)e^{-(s-1)t} dt)
=(left [frac e^{-st}{a^2+(s-1)^2} [-(s-1)sin(at)-acos⁡(at) ]right ]_0^∞)
=(frac{a}{a^2+(s-1)^2})

6. Laplace transform of t² sin⁡(2t).
a) (left [frac{12s^2-16}{(s^2+4)^4}right ])
b) (left [frac{3s^2-4}{(s^2+4)^3}right ])
c) (left [frac{12s^2-16}{(s^2+4)^6}right ])
d) (left [frac{12s^2-16}{(s^2+4)^3}right ])
Answer: d
Explanation: We know that,
(L(t^n f(t))=(-1)^n frac{d^n F(s)}{ds^n}),
Here, f(t)=sin⁡(2t)=>F(s)=(frac{2}{s^2+4}),
Hence, (L(t^2 sin⁡(2t))=frac{d^2}{ds^2} (frac{2}{s^2+4})=frac{d}{ds} frac{(s^2+4).0-2(2s)}{(s^2+4)^2})
=(-4left [frac{(s^2+4)^2-2s(s^2+4)2s}{(s^2+4)^4} right ]=left [frac{12s^2-16}{(s^2+4)^3}right ])

7. Find the laplace transform of t^⁵⁄₂.
a) (frac{15}{8} frac{√π}{s^{5/2}})
b) (frac{15}{8} frac{√π}{s^{7/2}})
c) (frac{9}{4} frac{√π}{s^{7/2}})
d) (frac{15}{4} frac{√π}{s^{7/2}})
Answer: b
Explanation:
(g(t)=t^{5/2}=frac{5}{2} int_0^t t^{frac{3}{2}} dt=frac{15}{4} int_0^t int_0^t √t dt dt)
let f(t)=√t, hence, F(s)=(frac{sqrt{π}}{2s^{frac{3}{2}}})
hence, G(s)=(frac{15}{4} ,frac{1}{s^2} ,F(s)=frac{15}{8} frac{√π}{s^{7/2}})

8. Value of (int_{-infty}^infty e^t ,Sin(t)Cos(t)dt) = ?
a) 0.5
b) 0.75
c) 0.2
d) 0.71
Answer: c
Explanation: L(Sin(2t)) = (int_{-infty}^infty e^{-st} ,Sin(2t)dt) = 2/(s² + 4)
Putting s=-1
(int_{-infty}^infty e^t ,Sin(2t)dt) = 0.4
hence,
(int_{-infty}^infty e^{-st} ,Sin(t)Cos(t)dt) = 0.2.

9. Value of (int_{-infty}^infty e^t ,Sin(t) ,dt) = ?
a) 0.50
b) 0.25
c) 0.17
d) 0.12
Answer: a
Explanation: L(Sin(2t)) = (int_{-infty}^infty e^{-st} ,Sin(t)dt) = 1/(s² + 1)
Putting s = -1
(int_{-infty}^infty e^t ,Sin(t)dt) = 0.5.

10. Value of (int_{-infty}^infty e^t ,log(1+t)dt) = ?
a) Sum of infinite integers
b) Sum of infinite factorials
c) Sum of squares of Integers
d) Sum of square of factorials
Answer: b
Explanation:
(int_{-infty}^infty e^t (t-t^2/2+t^3/3-….)dt)
(int_{-infty}^infty te^t dt=0.5 int_{-infty}^infty te^t dt)
Now,
(int_{-infty}^infty te^t dt– 1/2 int_{-infty}^infty t^2 e^t dt + (1/3) int_{-infty}^infty t^3 e^t dt-………)
Now, (int_{-infty}^infty t^n e^t dt=n!/(-1)^{n+1})
Hence,
(int_{-infty}^infty t^n e^t dt = 1 – (1/2)(2!/(-1)^3) + (1/3)(3!/)-…….)
(int_{-infty}^infty t^n e^t dt) = 0! + 1! + 2! + 3! +…. = Sum of infinite factorials.

11. Find the laplace transform of y(t)=e^t.t.Sin(t)Cos(t).
a) (frac{4(s-1)}{[(s-1)^2+4]^2})
b) (frac{2(s+1)}{[(s+1)^2+4]^2})
c) (frac{4(s+1)}{[(s+1)^2+4]^2})
d) (frac{2(s-1)}{[(s-1)^2+4]^2})
Answer: d
Explanation:
y(t)=(frac{1}{2} t.e^t Sin(2t))
Laplace transform of Sin(2t)=(frac{2}{s^2+4})
Laplace transform of tSin(2t)=(-frac{d}{dt} frac{2}{s^2+4}=frac{2(2s)}{(s^2+4)^2}=frac{4s}{(s^2+4)^2})
Laplace transform of te^t Sin(2t)=(frac{4(s-1)}{[(s-1)^2+4]^2})
Laplace transform of 1/2 te^t Sin(2t)=(frac{2(s-1)}{[(s-1)^2+4]^2})

12. Find the value of (int_0^{infty} tsin(t)cos(t)).
a) s ⁄ s²+2²
b) a ⁄ a²+s⁴
c) 1
d) 0
Answer: d
Explanation:
y(t)=(frac{1}{2} t Sin(2t)u(t))
Laplace transform of Sin(2t)=(frac{2}{s^2+4})
Laplace transform of tSin(2t)=(-frac{d}{dt} frac{2}{s^2+4}=frac{2(2s)}{(s^2+4)^2}=frac{4s}{(s^2+4)^2})
Laplace transform of (frac{1}{2}tsin(2t)=int_{-0}^{infty} e^{-st} tsin(t)cos(t)dt=frac{2s}{[s^2+4]^2})
Putting, s = 0, (int_0^{infty} tsin(t)cos(t)dt=0)

13. Find the laplace transform of y(t)=e^|t-1| u(t).
a) (frac{2s}{1-s^2} e^s)
b) (frac{2s}{1+s^2} e^{-s})
c) (frac{2s}{1+s^2} e^s)
d) (frac{2s}{1-s^2} e^{-s})
Answer: d
Explanation:
y(t)=(e^{|t-1|})
Laplace transform of e^|t| =(int_{-infty}^infty e^{|t|} e^{-st} dt)
=(int_0^∞ e^t e^{-st} dt-int_{-∞}^0 e^{-t} e^{-st} dt)
=(int_0^∞ e^{-(s-1)t} dt-∫_{-∞}^0 e^{(-s-1)t} dt)
Now,(int_0^∞ e^{-(s-1)t} dt=left [-frac{1}{s-1} [e^{-(s-1)t}]right ]_∞^0)
=(left [-frac{1}{s-1} [e^{-(s-1)t} ]right ]_∞^0=frac{-1}{s-1})
Now, (∫_{-∞}^0 e^{(-s-1)t} dt=left [frac{1}{-(s+1)} [e^{(-s-1)t}]right ]_0^{-∞})
=(left [-frac{1}{s+1} [e^{(-s-1)t} ]right ]_0^{-∞}=-frac{1}{(s+1)})
Laplace transform of |t| e^|t| =(int_{-infty}^infty e^{|t|} e^{-st} dt=-left [frac{1}{s-1}+frac{1}{s+1}right ]=-left [frac{2s}{s^2-1}right ])
Laplace transform of |t| e^|t| = (int_{-infty}^infty e^{|t-1|} e^{-st} dt=frac{2s}{1-s^2} e^{-s})

250+ TOP MCQs on Derogatory and Non-Derogatory Matrices and Answers

Matrices Multiple Choice Questions on “Derogatory and Non-Derogatory Matrices”.

1. Identify the type of Matrix
(begin{bmatrix}7&4&-1\4&7&-1\-4&-4&4end{bmatrix})
a) Identity Matrix
b) Non Derogatory Matrix
c) Derogatory Matrix
d) Symmetrical Matrix
Answer: c
Explanation: In this we have,

A=(begin{bmatrix}7&4&-1\4&7&-1\-4&-4&4end{bmatrix})

Finding the Characteristic Equation of the matrix
γ³-18γ²+81γ-108=0
Solving the Characteristic equation, we get the Eigenvalues
γ=3,3,12
Since Eigenvalues are repeated, this is a Derogatory Matrix.

2. Find the Eigenvalues and the type of the given matrix.
A=(begin{bmatrix}3&10&5\-2&-3&-4\3&5&7end{bmatrix})
a) 3, 1, 3 Non Derogatory
b) 2, 2, 2 Derogatory
c) 3, 2, 2 Derogatory
d) 1, 2, 3 Non Derogatory
Answer: c
Explanation: In this we have,
A=(begin{bmatrix}3&10&5\-2&-3&-4\3&5&7end{bmatrix})
Finding the characteristic equation of the Matrix,
-γ³+3γ²+24γ-141=0
Solving the Characteristic equation, we get the Eigenvalues
γ=3, 2, 2
Since Eigenvalues are repeated, this is a Derogatory Matrix.

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