Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on Leibniz Rule and Answers

Engineering Mathematics Interview Questions and Answers for freshers focuses on “Leibniz Rule – 3”.

1. Leibniz rule gives the
a) Nth derivative of addition of two function
b) Nth derivative of division of two functions
c) Nth derivative of multiplication of two functions
d) Nth derivative of subtraction of two function
Answer: c
Explanation: Leibniz rule is
(frac{d^n}{dx^n}(uv))
(= n_{C_0} u_n v+ n_{C_1} u_{(n-1)} v_1+ n_{C_2} u_{n-2} v_2+…+ n_{C_r} u_{n-r} v_r+…+ n_{C_n} u_0 v_n)
Hence Leibniz theorem gives nth derivative of multiplication of two functions u and v.
.

2. Leibniz theorem is applicable if n is a
a) Rational Number
b) Negative Integer
c) Positive Integer
d) Decimal Number
Answer: c
Explanation: Leibniz rule is
(frac{d^n}{dx^n}(uv))
(= n_{C_0} u_n v+ n_{C_1} u_{n-1} v_1+ n_{C_2} u_{n-2} v_2+…+ n_{C_r} u_{n-r} v_r+…+ n_{C_n} u_0 v_n)
For all n > 0, i.e n should be positive
Hence Leibniz theorem gives nth derivative of multiplication of two functions u and v if n is a positive integer.

3. If nth derivative of xy3 + x2y2 + x3y0 = 0 then order of its nth differential equation is,
a) n
b) n+1
c) n+2
d) n+3
Answer: d
Explanation:
1. If we differentiate this equation n times then terms comes in nth order differential equation is yn+3, yn+2, yn+1, yn, yn-1, yn-2, yn-3. Hence order of differential equation becomes n+3.
2. By Leibniz rule differentiating it n times, we get
Xyn+3 + nyn+2 + x2yn+2 + 2nxyn+1 + 2n(n-1)yn + x3yn + 3nx2yn-1 + 6n(n-1)xyn-2 + 6n(n-1)(n-2)yn-3 = 0
Hence order of differential equation becomes n+3.

4. Find nth derivative of x2y2 + xy1 + y = 0
a) x2yn+2 + (2n+1)xyn+1 + (n2+1)yn = 0
b) x2yn+2 + nxyn+1 + (n2+1)yn = 0
c) x2yn+2 + (2n+1)xyn+1 + n2yn = 0
d) x2yn+2 + 2nxyn+1 + n2yn = 0
Answer: a
Explanation: x2y2 + xy1 + y = 0
By Leibniz theorem
x2yn+2 + n(2x)(yn+1) + n(n-1)(2)(yn) + xyn+1 + nyn + yn= 0

x2yn+2 + xyn+1(2n+1) + yn(n2+1) = 0.

5. The nth derivative of x2y2 + (1-x2)y1 + xy = 0 is,
a) x2yn+2 + yn(2n2-2n-2nx+x) – yn-1(2n2-3n)=0
b) x2yn+2 + yn+1(2nx-x2) + yn(2n2-2n-2nx+x) – yn-1(2n2-3n)=0
c) x2yn+2 + yn+1(2nx+1-x2) + yn(2n2-2n-2nx+x)-yn – 1(2n2-3n)=0
d) x2yn+2 + yn+1(2nx+1-x2) + xyn2n2-yn – 1(2n2-3n)=0
Answer: c
Explanation: x2y2 + (1-x2)y1 + xy = 0
Differentiating n times by Leibniz Rule
x2yn+2 + 2nxyn+1 + 2n(n-1)yn + (1-x2)yn+1 – 2nxyn – 2n(n-1)yn-1 + xyn + nyn-1 = 0
x2yn+2 + yn+1(2nx+1-x2) + yn(2n2-2n-2nx+x) – yn-1(2n2-3n)=0.

6. Find nth derivative of xnSin(nx)
a) (n!Sin(nx) + n_{C_1} n_{P_{n-1}}x^{n-1}nCos(nx) – n_{C_2}n_{P_{n-2}}x^{n-2}n^2Sin(nx))
(-n_{C_3}n_{P_{n-3} }x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+x^nn^nSin(nx+nπ/2))
b) (Sin(nx) + n_{C_1} n_{P_{n-1}}x^{n-1}nCos(nx) – n_{C_2}n_{P_{n-2}}x^{n-2}n^2Sin(nx))
(-n_{C_3}n_{P_{n-3}}x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+n!x^nn^nSin(nx+nπ/2))
c) (nSin(nx) + n_{C_1} n_{P_{n-1}}x^{n-1}nCos(nx) – n_{C_2}n_{P_{n-2}}x^{n-2}n^2Sin(nx))
(-n_{C_3}n_{P_{n-3}}x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+nx^nn^nSin(nx+nπ/2))
d) ((n-1)!Sin(nx) + n_{C_1} n_{P_{n-1}}x^{n-1}nCos(nx) – n_{C_2}n_{P_{n-2}}x^{n-2}n^2Sin(nx))
(-n_{C_3}n_{P_{n-3}}x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+x^nn^{n-1}Sin(nx+nπ/2))
Answer: a
Explanation: Y = xnSin(nx)
By Leibniz Rule , put u = xn and v = Sin(nx), we get
(n!Sin(nx) + n_{C_1} n_(P_{n-1})x^{n-1}nCos(nx) – n_{C_2}n_(P_{n-2})x^{n-2}n^2Sin(nx))
(-n_{C_3}n_{P_{n-3}}x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+x^nn^nSin(nx+nπ/2))

7. If y(x) = tan-1x then
a) (yn+1)(0) = (n-1)(yn-1)0
b) (yn+1)(0) = n(n-1)(yn-1)0
c) (yn+1)(0) = -(n-1)(yn-1)0
d) (yn+1)(0) = -n(n-1)(yn-1)0
Answer: d
Explanation: Y = tan-1x
Y1 = 1/(1+x2)
(1+x2)y1 = 1
By Leibniz Rule,
(1+x2)yn+1 + 2nxyn + n(n-1)yn-1 = 0
Put x=0, gives
→ (yn+1)(0) = -n(n-1)(yn-1)(0).

8. If y = sin-1x, then
a) (1-x2)yn+2 – xyn+1(2n-1) = nyn(2n-1)
b) x2yn+2 – xyn+1(2n-1) = nyn(2n-1)
c) (1-x2)yn+2 – 2nxyn+1 = nyn(2n-1)
d) (1-x2)yn+2 – xyn+1(2n-1) +nyn(2n-1)=0
Answer: a
Explanation: Y = sin-1x
Differentiating it
Y1 = (frac{1}{sqrt{1-x^2}})
(1-x2)(y1)2= 1
Again Differentiating we get
(1-x2)2y1y2 – 2x(y1)2 = 0
(1-x2)y2 = xy1
By Leibniz Rule, Diff it n times,
(1-x2)yn+2 – 2xnyn+1 – 2n(n-1)yn = xyn+1 + nyn
(1-x2) yn+2 – xyn+1(2n-1) = nyn(2(n-1)+1)
(1-x2) yn+2 – xyn+1(2n-1) = nyn(2n-1).

9. If y = cos-1x, then
a) (yn+2)(0) = -n(2n-1) yn(0)
b) (yn+2)(0) = n(2n-1) yn(0)
c) (yn+2)(0) = n(n-2) yn(0)
d) (yn+2)(0) = n(n-3) yn(0)
Answer: b
Explanation: y = cos-1x
Differentiating it
Y1 = (frac{1}{sqrt{1-x^2}})
(1-x2)(y1)2= 1
Again Differentiating we get
(1-x2)2y1y2 – 2x(y1)2 = 0
(1-x2)y2 = xy1
By Leibniz Rule, Diff it n times,
(1-x2)yn+2 – 2xnyn+1 – 2n(n-1)yn = xyn+1 + nyn
(1-x2) yn+2 – xyn+1(2n-1) = nyn(2(n-1)+1)
(1-x2) yn+2 – xyn+1(2n-1) = nyn(2n-1)
At x=0, we get
(yn+2)(0) = n(2n-1) yn(0).

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250+ TOP MCQs on Derivative of Arc Length and Answers

Differential Calculus Question Bank focuses on “Derivative of Arc Length”.

1. For the cartesian curve y=f(x) with ‘s’ as arc length which of the following condition holds good?
a) (frac{ds}{dx} = sqrt{1+(frac{dy}{dx})^2})
b) (frac{d^2s}{dx^2} = sqrt{1-(frac{dy}{dx})^2})
c) (frac{dy}{dx} = sqrt{1+(frac{ds}{dx})^2})
d) (sqrt{(frac{ds}{dx})^2 + (frac{dy}{dx})^2} = 1)
View Answer

Answer: a

2. For the curve (y=a ,log ,sec(frac{x}{a})) what is the value of ( frac{ds}{dx})? (where φ is the angle made by tangent to the curve with x-axis)?
a) cos φ
b) sec φ
c) tan φ
d) cot φ
View Answer

Answer: b
Explanation: w.k.t (frac{ds}{dx} = sqrt{1+(frac{dy}{dx})^2})
(frac{dy}{dx} = a frac{tan frac{x}{a}.secfrac{x}{a}}{secfrac{x}{a}} frac{1}{a} = tan frac{x}{a} )
substituting we get
(frac{ds}{dx} = sqrt{1+(tan frac{x}{a})^2}=secfrac{x}{a}, ,but ,w.k.t, frac{dy}{dx} = tan φ = tanfrac{x}{a} ,thus, φ=frac{x}{a})
(frac{ds}{dx} = sec⁡φ.)

3. If the parametric equation of the curve is given by x=aet sin⁡t & y=aet cos⁡ t the value of (frac{ds}{dt}) is given by _____
a) aet
b) 2aet
c) √3 aet
d) √2 aet
View Answer

Answer: d
Explanation: w.k.t (frac{ds}{dt} = sqrt{(frac{dx}{dt})^2 + (frac{dy}{dt})^2}…(1))
(frac{dx}{dt} = ae^t (cos t + sin t), frac{dy}{dt} = ae^t (-sin⁡ t + cos⁡t))
substituting in (1) we get
(frac{ds}{dt} = ae^t sqrt{(cos t + sin t)^2+(-sin t+cos⁡t)^2})
(frac{ds}{dt} = ae^t sqrt{1+2 sin t cos + 1 – 2 sin⁡ t cos⁡t} = sqrt{2} ae^t…(cos^{2} t + sin{2} t = 1).)

4. For the curve in polar form (sqrt{frac{r}{a}} = sec⁡(frac{θ}{2}) ,the, ,value, ,of, frac{ds}{dθ}) is _____
a) r sec θ
b) r sec ((frac{θ}{2}))
c) r sec (2θ)
d) r cosec ((frac{θ}{2}))
View Answer

Answer: c
Explanation: Squaring the given curve on both side i.e (r=a sec ^2 (frac{θ}{2}))…(1)
(frac{dr}{dθ} = a.2 sec⁡(frac{θ}{2}) sec⁡(frac{θ}{2}) tan (frac{θ}{2}).1/2 = a sec^2 (frac{θ}{2}) tan(frac{θ}{2}) )
from (1)
(frac{dr}{dθ} = r tan (frac{θ}{2}) ,the, ,equation, ,for, frac{ds}{dθ} ,is, = sqrt{(r^2+(frac{dr}{dθ})^2)} = sqrt{r^2(1+ tan^2 (frac{θ}{2}))} )
(frac{ds}{dθ} = r sec (frac{θ}{2}).)

5. If r=b eθ cot⁡a, where a, b are constants then ‘s’ is represented by which one of the following equation?
a) s=r sec(a)+c
b) s=r cos(a)+c
c) s=r+c
d) s=r tan(a)+c
View Answer

Answer: a
Explanation: r=b eθ cot⁡a
(frac{dr}{dθ} = b e^{θ cot⁡a}(cot a)=r cot(a) )
w.k.t (frac{ds}{dr} = sqrt{1+r^2 (frac{dθ}{dr})^2} ,where, frac{dθ}{dr}=frac{1}{r} ,tan⁡a)
(frac{ds}{dr} = sqrt{1+tan^2 a} = sec ,a rightarrow s = int sec ,a ,dr , = r sec(a)+c,) where c is constant of integration.

250+ TOP MCQs on Errors and Approximations and Answers

Engineering Mathematics Multiple Choice Questions on “Errors and Approximations”.

1. Error is the Uncertainty in measurement.
a) True
b) False
Answer: a
Explanation: In the term of mathematics, “Error tells the person how much correct or certain its measurement is.”

2. Relative error in x is?
a) δx
b) δx⁄x
c) δx⁄x * 100
d) 0
Answer: b
Explanation: Option ‘δx’ is called absolute error.
Option ‘δx⁄x’ is called relative error.
Option ‘δx⁄x * 100’ is called Percentage error.

3. Find the percentage change power in the circuit if error in value of resistor is 1% and that of voltage source is .99%.
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: Power is given by P = V2⁄R
Taking log on both sides,
log(P) = 2log(V) – log(r)
Differentiating it ,
δp⁄p = 2δV⁄V – δr⁄r
Multiplying by 100 we get,
%P = 2%V – %r = (2*.99) – 1 = 0.98%.

4. Magnitude of error can be negative or positive.
a) True
b) False
Answer: b
Explanation: Magnitude of error can not be negative.Negative or positive sign only shows the increase or decrease in the quatity.

5. Given the kinetic energy of body is T = 1⁄2 mv2. If the mass of body changes from 100 kg to 100 kg and 500 gm and velocity of a body changes from 1600 mt/sec to 1590 mt/sec. Then find the approximate change in T.
a) 960000 J decrease in value
b) 960000 J increase in value
c) 450000 J decrease in value
d) 450000 J increase in value
Answer: a
Explanation: Given T = 1⁄2 mv2
Now taking log and differentiating,
δT = 0.5[v2 δm + 2mvδv]
Now, v = 1600 mt/sec, m = 100kg, δv = -10, δm = 0.5
Then,
δT = -960000 J => decrese in value of T by 960000 J.

6. The speed of a boat is given by, v = k(1⁄t – at), where k is the constant and l us the distance travel by boat in time t and a is the acceleration of water. If there is an change in ‘l’ from 2cm to 1cm in time 2sec to 1sec. If the acceleration of water changes from 0.95 mt/sec2 to 2 mt/sec2 find the motion of boat.
a) -2
b) 2
c) 0.5
d) -0.5
Answer: a
Explanation: Given, v = k(1⁄t – at)
Differentiating it we get
(δv=kleft [frac{(tδl-lδt)}{t^2} – aδt – tδa right ])
Hence,
(frac{δv}{v}=frac{kleft [frac{(tδl-lδt)}{t^2} – aδt – tδa right ]}{v})
(frac{δv}{v}=frac{left [frac{(tδl-lδt)}{t^2} – aδt – tδa right ]}{(frac{l}{t}-at)})
Put, l = 2cm, t = 2sec, a = 0.95 mt/sec2
and δl = 1cm, δt = 1 sec⁡and δa = -1.05 mt/sec2
we get,
δv⁄v = 2.

7. The relative error in the volume of figure having hemispherical ends and a body of right circular cylinder is, if error in radius(r) is 0 and in height(h) is 1.
a) 1/(h + 4⁄3 r)
b) 1/(h + 2⁄3 r)
c) h/(h + 4⁄3 r)
d) r/(h + 4⁄3 r)
Answer: a
Explanation: Given V = πr2 h + 4⁄3 πr3
Now since error in radius is zero , it should be treated as constant, Hence,
(δV=frac{πr^2 δh}{πr^2 (h+frac{4}{3} r)}=frac{1}{(h+frac{4}{3} r)})

8. If n resistors of unequal resistances are connected in parallel,and the percenrage error in all resistors are k then,total error in parallel combination is?
a) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
b) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=1-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
c) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=1+frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
d) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
Answer: b
Explanation: Given 1⁄r = 1⁄a + 1⁄b + 1⁄c +⋯.. + 1⁄n
Differentiating all,
– 1⁄r2 dr = – 1⁄a2 da – 1⁄b2 db – ….- 1⁄n2 dn
Now,
1⁄r2 dr = + 1⁄a2 da + 1⁄b2 db + ….+ 1⁄n2 dn
Multiplying by 100 and putting da = db = ……. = dn =k.

9. The approximate value of function f(x + δx, y + δy) is?
a) f + ∂f⁄∂x dx + ∂f⁄∂y dy
b) ∂f⁄∂x dx + ∂f⁄∂y dy
c) f – ∂f⁄∂x dx + ∂f⁄∂y dy
d) ∂f⁄∂x dx – ∂f⁄∂y dy
Answer: a
Explanation: f(x + δx, y + δy) = f(x,y) + df = f + ∂f⁄∂x dx + ∂f⁄∂y dy.

10. At a distance of 120ft from the foot of a tower the elevation of its top is 60o. If the possible error in measuring the distance and elevation are 1 inch and 1 minute, find approximate error in calculated height.
a) 0.342
b) 0.284
c) 0.154
d) 0.986
Answer: b
Explanation: Tan(z) = h⁄x
h = x Tan(z)
Taking log and then differentiate we get,
∂h⁄h = ∂x⁄x + 1⁄Tan(z) Sec2 (z)δz
Now h = 120 tan(60o) = 120√3
Putting, δx = 1⁄12 ft, δz = π⁄(60*180)
Putting the values we get,
δh = 0.284.

11. Find the approximate value of (1.04)3.01.
a) 1.14
b) 1.13
c) 1.11
d) 1.12
Answer: d
Explanation: Let, f(x,y) = xy
Now,
∂f⁄∂x = yx(y-1) and ∂f⁄∂y = xy log⁡(x)
Putting, x = 1, y = 3, δx = 0.04, δy = 0.01
Now, df = δx ∂f⁄∂x + δy∂f⁄∂y = 0.12
Hence, f(x + δx,y + δy) = (1.04)3.01 = 1.12.

12. 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 = ∂f⁄∂x dx + ∂f⁄∂y dy + ∂f⁄∂z dz = ((xdx + ydy + zdz))/f = (-0.02 + 0.02 – 0.12)/3 = -0.04
Hence,
[0.982+2.012+1.942 ](1⁄2) = f(1, 2, 2) + df = 3-0.04 = 2.96.

13. Find the approximate value of log⁡(11.01-log⁡(10.1)), Given log(10) = 2.30 and and log(8.69) = 2.16, all the log are in base ‘e’.
a) 2.1654
b) 2.1632
c) 2.1645
d) 2.1623
Answer: d
Explanation: Let, f(x,y) = log(x-log(y))
Now by differentiating,
(frac{∂f}{∂x}=frac{1}{x-log⁡(y)}) and (frac{∂f}{∂y}=frac{1}{y(x-log⁡(y))})
Now, putting, x = 11, y = 10, δx=.01 and δy=.1
We get,
(frac{∂f}{∂x}∂f/∂x)=1/8.69 and (frac{∂f}{∂y}∂f/∂y)=1/86.9
Hence, df = 0.0023
Hence, f(x + δx, y + δy) = log⁡(11.01 – log⁡(10.1))= 2.16 + df = 2.1623.

14. Find approximate value of e10.19.09, given e90 = 1.22 * 1039.
a) 2.41 * 1039
b) 2.42 * 1039
c) 2.43 * 1039
d) 2.44 * 1039
Answer: b
Explanation: Let, f(x,y) = exy = exy
Now by differentiating,
∂f⁄∂x = yexy and ∂f⁄∂y = xexy
Now, putting, x = 10, y = 9, δx = .01 and δy = .09
We get,
∂f⁄∂x = 1.09* 1040 and ∂f⁄∂y = 1.22* 1040
Hence, df = 1.27* 1039
Hence, f(x + δx, y + δy) = e10.19.09 = 2.42 * 1039.

250+ TOP MCQs on Double Integrals and Answers

Engineering Mathematics Multiple Choice Questions on “Double Integrals”.

1. Find the value of ∫∫xyex + y dxdy.
a) yey (xex-ex)
b) (yey-ey)(xex-ex)
c) (yey-ey)xex
d) (yey-ey)(xex+ex)
Answer: b
Add constant automatically
Explanation: Given, ∫∫xyex + y dxdy
∫∫xyex ey dxdy= ∫yey dy∫xex dx=(yey-ey)(xex-ex).

2. Find the value of ∫∫ x⁄x2 + y2 dxdy.
a) [ytan(-1) (y)- 1⁄2 ln⁡(1+y2)]
b) x [ytan(-1) (y)- 1⁄2 ln⁡(1+y2)]
c) y [xtan(-1) (x)- 1⁄2 ln⁡(1+x2)]
d) x [ytan(-1) (y)- 1⁄2 ln⁡(1+y2)]
Answer: d
Explanation: Add constant automatically
Given, (intint frac{x}{x^2+y^2} ,dxdy)
(int x int frac{1}{x^2+y^2} ,dydx=int x frac{1}{x} tan^{-1}⁡(frac{y}{x}) ,dy=int tan^{-1}⁡(frac{y}{x}) ,dy)
(int tan^{-1}⁡(frac{y}{x}),dy = xint tan^{-1}⁡(t),dt)
Putting, x = tan(z),
We get, dz = sec2⁡(z)dz,

x∫ zsec2 (z)dz

By integration by parts,

x ∫ zsec2 (z)dz=x[ztan(z)-log⁡(sec⁡(z))]= x[ytan(-1) (y)- 1⁄2 ln⁡(1+y2)].

3. Find the ∫∫x3 y3 sin⁡(x)sin⁡(y) dxdy.
a) (x3 Cos(x) + 3x2 Sin(x) + 6xCos(x)-6Sin(x))(y3 Cos(y) + 3[y2 Sin(y) – 2[-yCos(y) + Sin(y)]])
b) (-x3 Cos(x) – 3x2 Sin(x) – 6xCos(x)-6Sin(x))(-y3 Cos(y) + 3[y2 Sin(y) – 2[-yCos(y) + Sin(y)]])
c) (-x3 Cos(x) + 3x2 Sin(x) + 6xCos(x)-6Sin(x))(-y3 Cos(y) + 3y2 Sin(y) + 6yCos(y) – 6Sin(y))
d) (–x3 Cos(x) + 6xCos(x) – 6Sin(x))(-y3 Cos(y))
Answer: c
Explanation: Add constant automatically
∫x3 Sin(x)dx = -x3 Cos(x) + 3∫x2 Cos(x)dx
∫x2 Cos(x)dx = x2 Sin(x) – 2∫xSin(x)dx
∫xSin(x)dx = -xCos(x) + ∫Cos(x)dx = -xCos(x) + Sin(x)
=> ∫x3 Sin(x)dx = -x3 Cos(x) + 3[x2 Sin(x) – 2[-xCos(x) + Sin(x)]]
=> ∫x3 Sin(x)dx = -x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x)
and, ∫y3 Sin(y)dy = -y3 Cos(x) + 3∫y2 Cos(y)dy
∫y2 Cos(y)dy = y2 Sin(y) – 2∫ySin(y)dy
∫ySin(y)dy = -yCos(y) + ∫Cos(y)dy = -yCos(y) + Sin(y)
=> ∫y3 Sin(y)dy = -y3 Cos(y) + 3[y2 Sin(y) – 2[-yCos(y) + Sin(y)]]
=> ∫y3 Sin(y)dy = -y3 Cos(y) + 3y2 Sin(y) + 6yCos(y) – 6Sin(y)
Hence, ∫∫x3 y3 sin⁡(x) sin⁡(y) dxdy = (∫x3 Sin(x)dx)(∫y3 Sin(y)dy) = (-x3 Cos(x) + 3x2 Sin(x)+6xCos(x) – 6Sin(x))(-y3 Cos(y) + 3y2 Sin(y) + 6yCos(y) – 6Sin(y)).

4. Find the integration of (intint_0^{sqrt{2ax-x^2}}x ,dxdx).
a) ax2⁄2 – x5⁄30
b) ax2⁄2 – x3⁄6
c) ax2⁄2
d) ax4⁄8 – x3⁄6
Answer: b
Explanation: Add constant automatically
Given, f(x)=(intint_0^{sqrt{2ax-x^2}}x ,dxdx = int [frac{x^2}{2}]_0^{sqrt{2ax-x^2}} ,dxdx = int frac{2ax-x^2}{2} ,dx=frac{ax^2}{2}-frac{x^3}{6})

5. Find the value of ∫∫xy7 Cos(x)Cos(y) dxdy.
a) (7y6 Cos(y) + 42y5 Sin(y) + 210y4 Cos(y) + 840y3 Sin(y) + 2520y2 Cos(y) + 5040ySin(y) + 5040Cos(y))(7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x))
b) (y7 Sin(y) + 7y6 Cos(y) + 42y5 Sin(y) + 210y4 Cos(y) + 840y3 Sin(y) + 2520y2 Cos(y) + 5040ySin(y) + 5040Cos(y))(x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x))
c) (y7 Sin(y) + 42y5 Sin(y) + 210y4 Cos(y) + 840y3 Sin(y) + 2520y2 Cos(y) + 5040ySin(y) + 5040Cos(y))(x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x))
d) (y7 Sin(y) + 7y6 Cos(y) + 42y5 Sin(y) + 210y4 Cos(y) + 840y3 Sin(y) + 2520y2 Cos(y) + 5040ySin(y) + 5040Cos(y))(x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x))
Answer: d
Explanation: Add constant automatically
By, f(x)=(int uvdx=sum_{i=0}^n (-1)^i u_i v^{i+1}),
Let, u = x7 and v=Cos(x),
∫x7 Cos(x) dx=x7 Sin(x)+7x6 Cos(x)+42x5 Sin(x)+210x4 Cos(x)+840x3 Sin(x)+2520x2 Cos(x)+5040xSin(x)+5040Cos(x)
Similarly,
∫y7 Cos(y) dy=y7 Sin(y)+7y6 Cos(y)+42y5 Sin(y)+210y4 Cos(y)+840y3 Sin(y)+2520y2 Cos(y)+5040ySin(y)+5040Cos(y)
Now,
∫∫xy7 Cos(x)Cos(y) dxdy=∫y7 Cos(y) dy∫x7 Cos(x) dx=(y7 Sin(y)+7y6 Cos(y)+42y5 Sin(y)+210y4 Cos(y)+840y3 Sin(y)+2520y2 Cos(y)+5040ySin(y)+5040Cos(y))(x7 Sin(x)+7x6 Cos(x)+42x5 Sin(x)+210x4 Cos(x)+840x3 Sin(x)+2520x2 Cos(x)+5040xSin(x)+5040Cos(x)).

6. Find the integration of ∫∫0x x2 + y2 dxdy.
a) x4⁄6
b) y
c) 2x3⁄3y
d) 1
Answer: c
Explanation: Add constant automatically
Given, (f(x)=∫_0^x (x^2+y^2) ,dxdy = ∫ (frac{x^3}{3}+frac{x^3}{3}),dxdy = frac{2x^3}{3} y).

7. Find the value of (intint_0^y frac{2xy^5}{sqrt{1+x^2 y^2-y^4}} dxdy).
a) (2[frac{y^4}{4}- frac{2}{3} (1-y^4)^{frac{3}{2}}])
b) (2[frac{y^4}{4}- (1-y^4)^{frac{3}{2}}])
c) (2[frac{y^4}{4}-frac{2}{3} (1-y^4)^{frac{3}{2}}])
d) (2[frac{y^3}{3}-frac{2}{3} (1-y^4)^{frac{3}{2}}])
Answer: c
Explanation:
Given, f(x)=(intint_0^y frac{2xy^5}{sqrt{1+x^2 y^2-y^4}} dxdy)
=(intint_0^y frac{1}{y} frac{2xy^5}{sqrt{(frac{1-y^4}{y^2})+x^2}} dxdy=int 2y^4 left |(frac{1-y^4}{y^2})+x^2right |_0^y dy)
(=2int [y^3-sqrt{1-y^4}y^3]dy=2[frac{y^4}{4}-frac{2}{3} (1-y^4)^{3/2}])

8. Find the value of (intint_0^{1-y} xysqrt{1-x-y} ,dxdy).
a) 16⁄946
b) 8⁄945
c) 16⁄936
d) 16⁄945
Answer: d
Explanation:
Given, f(x)=(int_0^1 int_0^{1-y} xysqrt{1-x-y} dxdy)
putting,t=x/(1-y)=>x=t(1-y)=>dx=(1-y)dt
(int_0^1 int_0^1 t(1-y)ysqrt{1-t(1-y)-y} (1-y)dtdy)
=(int_0^1 int_0^1 y(1-y)^{5/2} t(1-t)^{1/2} dtdy)
=(int_0^1y(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} dyint_0^1 t^{2-1} (1-t)^{3/2-1} dt=beta(2,frac{7}{2})beta(2,frac{3}{2})=frac{16}{945})

9. Find the area inside function (2x3 + 5 x2 – 4)⁄x2 from x = 1 to a.
a) a2⁄2 + 5a – 4ln(a)
b) a2⁄2 + 5a – 4ln(a) – 11⁄2
c) a2⁄2 + 4ln(a) – 11⁄2
d) a2⁄2 + 5a – 11⁄2
Answer: b
Explanation: Add constant automatically
Given,
(f(x) = frac{(2x^3+5x^2-4)}{x^2})

Integrating it we get, F(x) = x2⁄2 + 5x – 4ln(x)

Hence, area under, x = 1 to a, is

F(a) – F(1) = a2⁄2 + 5a – 4ln(a) – 1⁄2 – 5 = a2⁄2 + 5a – 4ln(a) – 11⁄2.

10. Find the value of (intintfrac{1}{16x^2+16x+10} ,dx).
a) (frac{1}{8} (x+frac{1}{2})Sin^{-1} (x+frac{1}{2})-frac{1}{2} ln⁡(1+(x+frac{1}{2})^2))
b) (frac{1}{8} (x+frac{1}{2})tan^{-1} (x+frac{1}{2})-frac{1}{16} ln⁡(1+(x+frac{1}{2})^2))
c) ((x+1/2) cos^{-1} (x+frac{1}{2})-frac{1}{16} ln⁡(1+(x+frac{1}{2})^2))
d) (frac{1}{8} (x+frac{1}{2}) sec^{-1} (x+frac{1}{2})-frac{1}{16} ln⁡(1+(x+frac{1}{2})^2))
Answer: b
Explanation: Add constant automatically
Given,(int frac{1}{16x^2+16x+10} dx=frac{1}{2}int frac{1}{4x^2+4x+5} dx)
=(int frac{1}{8(x^2+x+5/4+1/4+1/4)} dx)
=(int frac{1}{8[(x+1/2)^2+1^2]}dx=frac{1}{8} tan^{-1}⁡(x+1/2))
Hence, (frac{1}{8} int tan^{-1}⁡(x+frac{1}{2})dx)
Now, Putting, x+1/2 = tan(y),
We get, dx = sec2⁡(y)dy,
=1/8 (int ysec^2 (y)dy)
By integration by parts,
ytan(y)-log⁡(sec⁡(y))=(frac{1}{8} (x+frac{1}{2})tan^{-1} (x+frac{1}{2})-frac{1}{16} ln⁡(1+(x+frac{1}{2})^2))

250+ TOP MCQs on Geometrical Applications and Answers

Ordinary Differential Equations Multiple Choice Questions on “Geometrical Applications”.

1. Who invented the rectangular coordinate system?
a) Frans van Schooten
b) René Descartes
c) Isaac Newton
d) Gottfried Wilhelm Leibniz
Answer: b
Explanation: The rectangular coordinate system was invented by French mathematician and philosopher René Descartes, who published this idea in 1637.

2. What is the formula for the intercept of tangent line on x-axis?
a) (x-frac{y}{y’})
b) (y-frac{x}{y’})
c) y – xy’
d) y – x’y
Answer: a
Explanation: Intercept of a tangent line on the:
x-axis: (x-frac{y}{y’})
y-axis: y – xy’

3. The intercept of a normal line on y axis is given by, (y+frac{x}{y’}.)
a) True
b) False
Answer: a
Explanation: Intercept of a normal line on the:
x-axis: x + yy’
y-axis: (y+frac{x}{y’})

4. What is the time taken by the particle to cover a distance of 98 meters, if the particle is moving with a velocity given by, (frac{dx}{dt}= frac{1}{x} ) (x is the distance)?
a) 48 s
b) 50 s
c) 98 s
d) 49 s
Answer: d
Explanation: Given: (frac{dx}{dt}= frac{1}{x} )
dt=xdx
Integrating, (∫dt=∫_0^{98}xdx)
(t= (frac{x^2}{2})_0^{98}=frac{98}{2}=49 s)

5. Which of the following relations hold true?
a) i × i = j × j = k × k = 1
b) i × j = -k, j × i = k
c) i × i = j × j = k × k = 0
d) k × i = -j, i × k = j
Answer: c
Explanation: One of the properties of vector or cross product is, for the orthogonal vectors, i, j, and k, we have the relation,
i × i = j × j = k × k = 0,
i × j = k, j × i = -k,
j × k = i, k × j = -i,
k × i = j, i × k = -j

6. In recurrence relation, each further term of a sequence or array is defined as a function of its succeeding terms.
a) True
b) False
Answer: b
Explanation: An equation that gives a sequence such that each next term of the sequence or array is defined as a function of its preceding terms, is called a recurrence relation.

7. What is the maximum area of the rectangle with perimeter 650 mm?
a) 26,406.25 mm2
b) 26,406 mm2
c) 24,000 mm2
d) 24,075 mm2
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)
650=2(x+y)
x+y=325
y=325-x
We can now substitute the value of y in (1)
A=x*(325-x)
A=325x-x2
To find maximum value we need derivative of A,
(frac{dA}{dx}=325-2x)
To find maximum value, (frac{dA}{dx}=0)
325-2x=0
2x=325
x=162.5 mm
Therefore, when the value of x=162.5 mm and the value of y=325-162.5=162.5 mm, the area of the rectangle is maximum, i.e., A=162.5*162.5=26,406.25 mm2

8. What is the derivative of z=3x*logx+5x6 with respect to x?
a) 3+30x5 ex
b) 3+5x6 ex+30x5 ex
c) 3+5x6 ex
d) 3+3logx+30x5 ex
Answer: d
Explanation: Given: z = 3x*logx+5x6 ex
(frac{dz}{dx}=3x(frac{1}{x}))+3logx+30x5 ex
(frac{dz}{dx}) = 3+3logx+30x5 ex

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 10%?

a) 0.3183%
b) 0.03183%
c) 31.83%
d)3.183%
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 10%,
(dR=frac{dV}{4πR^2}=frac{0.1}{4π(5)^2}=0.0003183)
Error in radius = 0.03183%

10. What is the degree of the differential equation, x3-6x3y3+2xy=0?
a) 3
b) 5
c) 6
d) 8
Answer: c
Explanation: The degree of an equation that has not more than one variable in each term is the exponent of the highest power to which that variable is raised in the equation. But when more than one variable appears in a term, it is necessary to add the exponents of the variables within a term to get the degree of the equation. Hence, the degree of the equation, x3-6x3y3+2xy=0, is 3+3 = 6.

Global Education & Learning Series – Ordinary Differential Equations.

To practice all areas of Ordinary Differential Equations,

250+ TOP MCQs on General Properties of Inverse Laplace Transform and Answers

Ordinary Differential Equations Multiple Choice Questions on “General Properties of Inverse Laplace Transform”.

1. Find the (L^{-1} (frac{s+3}{4s^2+9})).
a) (frac{1}{4} cos⁡(frac{3t}{2})+frac{1}{2} cos⁡(frac{3t}{2}))
b) (frac{1}{4} cos⁡(frac{3t}{4})+frac{1}{2} sin⁡(frac{3t}{2}))
c) (frac{1}{2} cos⁡(frac{3t}{2})+frac{1}{2} sin⁡(frac{3t}{2}))
d) (frac{1}{4} cos⁡(frac{3t}{2})+frac{1}{2} sin⁡(frac{3t}{2}))
Answer: d
Explanation: In the given question
=(frac{1}{4} L^{-1}left (frac{s+3}{s^2+frac{9}{4}}right ))
=(frac{1}{4} Big{L^{-1}left (frac{s}{s^2+frac{9}{4}}right)+L^{-1}left (frac{3}{s^2+frac{9}{4}}right)Big})
=(frac{1}{4} Big{cos⁡(frac{3t}{2})+2 sin⁡(frac{3t}{2})Big})
=(frac{1}{4} cos⁡(frac{3t}{2})+frac{1}{2} sin⁡(frac{3t}{2})).

2. Find the (L^{-1} (frac{1}{(s+2)^4})).
a) (e^{-2t}×3)
b) ⁡(e^{-2t}×frac{t^3}{3})
c) (e^{-2t}×frac{t^3}{6})
d) (e^{-2t}×frac{t^2}{6})
Answer: c
Explanation: In the given question,
(L^{-1} (frac{1}{(s+2)^4})=e^{-2t} L^{-1} frac{1}{s^4}) —————– By the first shifting property
=(e^{-2t}×frac{t^3}{3!})
=(e^{-2t}×frac{t^3}{6}).

3. Find the (L^{-1} (frac{s}{(s-1)^7})).
a) (e^{-t} left (frac{t^6}{5!}+frac{t^5}{6!}right ))
b) (e^t left (frac{t^6}{5!}+frac{t^5}{6!}right ))
c) (e^t left (frac{t^6}{6!}+frac{t^5}{5!}right ))
d) (e^{-t} left (frac{t^6}{6!}+frac{t^5}{5!}right ))
Answer: c
Explanation: In the given question,
=(L^{-1} left (frac{s-1+1}{(s-1)^7}right))
=(e^t L^{-1} left (frac{s+1}{s^7}right))
=(e^t L^{-1} left (frac{1}{s^7}+frac{1}{s^6}right))
=(e^t left (frac{t^6}{6!}+frac{t^5}{5!}right))

4. Find the (L^{-1} (frac{s}{2s+9+s^2})).
a) (e^{-t} {cos⁡(2sqrt{2t})-sin⁡(sqrt{2t})})
b) (e^{-t} {cos⁡(2sqrt{2t})-sin⁡(2sqrt{2t})})
c) (e^{-t} {cos⁡(2sqrt{2t})-cos(sqrt{2t})})
d) (e^{-2t} {cos⁡(2sqrt{2t})-sin⁡(2sqrt{2t})})
Answer: b
Explanation: In the given question,
(L^{-1} left (frac{s}{2s+9+s^2}right )=L^{-1} left (frac{s}{(s+1)^2}+8)right ))
=(L^{-1} left (frac{(s+1)-1}{(s+1)^2+8}right ))
=(e^{-t} L^{-1} left (frac{(s-1)}{s^2+8}right )) ———————–By First Shifting Property
=(e^{-t} L^{-1} left (frac{s}{s^2+8}right )-e^{-t} L^{-1} left (frac{1}{s^2+8}right ))
=(e^{-t} {cos⁡(2sqrt{2t})-sin⁡(2sqrt{2t})}).

5. Find the (L^{-1} left (frac{(s+1)}{(s+2)(s+3)}right )).
a) 2e-3t-e-2t
b) 3e-3t-e-2t
c) 2e-3t-3e-2t
d) 2e-2t-e-t
Answer: a
Explanation: In the given question,
(L^{-1} left (frac{(s+1)}{(s+2)(s+3)}right )=L^{-1} left (frac{2(s+2)-(s+3)}{(s+2)(s+3)}right ))
=(L^{-1} left (frac{2}{(s+3)}right )+L^{-1} left (frac{1}{(s+2)}right ))
=2e-3t-e-2t.

6. Find the (L^{-1} left (frac{(3s+9)}{(s+1)(s-1)(s-2)}right )).
a) e-t+6et+5e2t
b) e-t-et+5e2t
c) e-3t-6et+5e2t
d) e-t-6et+5e2t
Answer: d
Explanation: In the given question,
(L^{-1} left (frac{(3s+9)}{(s+1)(s-1)(s-2)}right ))
=(L^{-1} left (frac{1}{(s+1)}right )-6L^{-1} left (frac{-6}{(s-1)}right )+5L^{-1} left (frac{-6}{(s-2)}right ) )————-Using properties of Partial Fractions
=e-t-6et+5e2t.

7. Find the (L^{-1} (frac{1}{(s^2+4)(s^2+9)})).
a) (frac{1}{5} left (frac{sin⁡(2t)}{2}-frac{sin⁡(t)}{3}right ))
b) (frac{1}{5} left (frac{sin⁡(2t)}{2}+frac{sin⁡(3t)}{3}right ))
c) (frac{1}{5} left (frac{sin⁡(t)}{2}-frac{sin⁡(3t)}{3}right ))
d) (frac{1}{5} left (frac{sin⁡(2t)}{2}-frac{sin⁡(3t)}{3}right ))
Answer: d
Explanation: In the given question,
(L^{-1} left (frac{1}{(s^2+4)(s^2+9)}right))
=(frac{1}{5} L^{-1} left (frac{5}{(s^2+4)(s^2+9)}right))
=(frac{1}{5} L^{-1} left (frac{(s^2+9)-(s^2+4)}{(s^2+4)(s^2+9)}right))
=(frac{1}{5} L^{-1} left (frac{1}{(s^2+4)}right )-frac{1}{5} L^{-1} left (frac{1}{(s^2+9)}right))
=(frac{1}{5} left (frac{sin⁡(2t)}{2}-frac{sin⁡(3t)}{3}right)).

8. Find the (L^{-1} left (frac{s}{(s^2+1)(s^2+2)(s^2+3)}right )).
a) (frac{1}{2} cos⁡(t)-cos⁡(sqrt3t)-frac{1}{2} cos⁡(sqrt3t))
b) (frac{1}{2} cos⁡(t)+cos⁡(sqrt2t)-frac{1}{2} cos⁡(sqrt3t))
c) (frac{1}{2} cos⁡(t)-cos⁡(sqrt2t)-frac{1}{2} cos⁡(sqrt3t))
d) (frac{1}{2} cos⁡(t)+cos⁡(sqrt2t)+frac{1}{2} cos⁡(sqrt3t))
Answer: c
Explanation: In the given question,
(L^{-1} left (frac{s}{(s^2+1)(s^2+2)(s^2+3)}right ))
=(L^{-1} left (frac{frac{1}{2}}{(s^2+1)}+frac{(-1)}{(s^2+2)}+frac{frac{(-1)}{2}}{(s^2+3)}right )) ——————-By method of Partial fractions
=(frac{1}{2} cos⁡(t)-cos⁡(sqrt2t)-frac{1}{2} cos⁡(sqrt3t)).

9. Find the (L^{-1} left (frac{s+1}{(s-1)(s+2)^2}right )).
a) (frac{2}{7} e^t-frac{2}{9} e^{-2t}+frac{1}{3} e^{-2t}×t)
b) (frac{2}{9} e^t-frac{2}{9} e^{-2t}+frac{1}{3} e^{-2t}×t)
c) (frac{2}{9} e^t-frac{2}{9} e^{-3t}+frac{1}{3} e^{-2t}×t)
d) (frac{2}{9} e^t-frac{2}{9} e^{-2t}+frac{1}{3} e^{-2t})
Answer: b
Explanation: In the given question,
(L^{-1} left (frac{s+1}{(s-1)(s+2)^2}right ))
Using properties of partial fractions-
s+1=A(s+2)2+B(s-1)(s+2)+C(s-1)
At s=1, A=(frac{2}{9})
At s=2, C=(frac{1}{3})
At s=0, B=(frac{-2}{9})
Re substituting all these values in the original fraction,
=(L^{-1} left (frac{2}{9(s-1)} + frac{-2}{9(s+2)} + frac{1}{3(s+2)^2}right))
=(frac{2}{9} e^t-frac{2}{9} e^{-2t}+frac{1}{3} e^{-2t}×t).

10. The (L^{-1} left (frac{3s+8}{s^2+4s+25}right )) is (e^{-st} (3cos⁡(sqrt{21}t+frac{2sin⁡(sqrt{21}t)}{sqrt{21}})). What is the value of s?
a) 0
b) 1
c) 2
d) 3
Answer: c
Explanation: In the given question,
(L^{-1} left (frac{3s+8}{s^2+4s+25}right )=L^{-1} left (frac{3(s+2)+2}{(s+2)^2+21}right ))
By the first shifting property
=(e^{-2t} L^{-1} left (frac{3s+2}{s^2+21}right ))
=(e^{-2t} L^{-1} left (frac{3s}{s^2+21}right )+e^{-2t} L^{-1} left (frac{2}{s^2+21}right ))
=(e^{-2t} (3cos⁡(sqrt{21}t+frac{2sin⁡(sqrt{21}t)}{sqrt{21}})).

Global Education & Learning Series – Ordinary Differential Equations.

To practice all areas of Ordinary Differential Equations,