300+ REAL TIME Engineering Mathematics Objective Questions & Answers 2023

250+ TOP MCQs on Volume Integrals and Answers

Linear Algebra Multiple Choice Questions on “Volume Integrals”.

1. Find the value of (int_0^1 int_0^2 int_1^2 xy^2 z^3 dxdydz.).
a) 2
b) 3
c) 4
d) 5
Answer: d
Explanation: (int_0^1 int_0^2 int_1^2 xy^2 z^3 dxdydz = (int_0^1 xdx)(int_0^2 y^2 dy)(int_1^2 z^3 dz) )
(= frac{1}{2}*frac{8}{3}*(frac{16}{4}-frac{1}{4}) )
(= 5. )

2. Evaluate ∫∫∫z² dxdydz taken over the volume bounded by the surfaces x²+y²=a², x²+y²=z and z=0.
a) (pi frac{a^8}{12} )
b) (pi frac{a^8}{4} )
c) (pi frac{a^4}{4} )
d) (pi frac{a^8}{8} )
Answer: a
Explanation: z: 0 to (x^2+y^2 )
( y: pi frac{a^8}{12} ) to ( sqrt{a^2-x^2} )
x: -a to a
(displaystyleintintint z^2 dxdydz = int_{-a}^a int_{-sqrt{(a^2-x^2 )}}^{sqrt{a^2-x^2}}int_0^{x^2+y^2}z^2 dxdydz )
(displaystyle = int_{-a}^a int_{-sqrt{a^2-x^2}}^{sqrt{a^2-x^2}}frac{z^3}{3} dxdy ) from 0 to ( x^2+y^2. )
(displaystyle = int_{-a}^a int_{-sqrt{a^2-x^2}}^{sqrt{a^2-x^2}} frac{(x^2+y^2 )^3}{3} dxdy)
( = frac{1}{3} int_0^π int_{-a}^a r^7 drdθ )
( = π frac{a^8}{12}. )

3. Find the volume of the cylinder bounded by x²+y² = 4, y+z = 4 and z=0.
a) (16π-frac{32}{3} )
b) (32π-frac{32}{3} )
c) (16-32 frac{π}{3} )
d) (32-32 frac{π}{3} )
Answer: a
Explanation: (z: (4-y))
(x: -sqrt{4-y^2} ) to ( sqrt{4-y^2} )
y: -2 to 2
Volume = (displaystyleint_{-2}^2 int_{-sqrt{4-y^2}}^{sqrt{4-y^2}} (4-y)dxdy )
(= 2int_{-2}^2 (4y-y^2 )dxdy ) from 0 to ( sqrt{4-y^2} )
(= 4big{4(0 + π) – 4 + frac{4}{3}big} )
(= 16π-frac{32}{3}. )

4. Find the volume of sphere by triple integration.
a) (8a^3 frac{π}{3} )
b) (4a^3 frac{π}{3} )
c) (2a^3 frac{π}{3} )
d) (a^3 frac{π}{3} )
Answer: b
Explanation: The sphere is given by x²+y²+z²=a².
θ : 0 to 2 π
φ : 0 to π
r : 0 to a
Volume = ( int_0^a int_0^π int_0^{2π} r^2 sinθdθdrdϕ )
(= 4(int_0^a r^2 dr)(int_0^π sinθdθ)(int_0^{π/2}dϕ) )
(= 4a^3 frac{π}{3}.)

5. Using polar coordinates, find the volume of the cylinder with radius a and height h.
a) (πa^2h )
b) (frac{πa^2h}{3} )
c) (2 frac{πa^2h}{3} )
d) ( 4 frac{πa^2h}{3} )
Answer: a
Explanation: r: 0 to a
θ : 0 to 2 π
z: 0 to h
Therefore, volume = (int_0^a int_0^{2π}int_0^h rdθdrdz )
(= (int_0^a rdr)(int_0^{2π}dθ)(int_0^h dz) )
(= frac{a^2}{2} * 2π * h )
(= πa^2h. )

6. In multiple integrals, if the limits depends on variable, then the order of integration can be anything.
a) True
b) False
Answer: b
Explanation: In multiple integration, if the limits depend on variable then the order of integration can’t be anything. First the dependent integration should be done and then the independent integration.

7. To find volume _________________ can be used.
a) single integration
b) double integration
c) triple integration
d) double & triple integration
Answer: d
Explanation: To find volume, triple integration should be used and proper limits should be given for each variable.
But volume integration can also be done using double integration by using 1D equation of the 3D object as the function.

8. Evaluate (int_{-1}^1 int_0^z int_{x-z}^{x+z}(x+y+z)dxdydz.)
a) 0
b) 1
c) 2
d) 3
Answer: a
Explanation: (int_{-1}^1 int_0^z int_{x-z}^{x+z}(x+y+z)dxdydz=
int_{-1}^1 int_0^z (xy+frac{y^2}{2}+zy)dxdz ) from x-z to x+z
(= int_{-1}^1int_0^z(3x^2+z^2+2xz)dxdz= int_{-1}^1(x^3+z^2 x+x^2 z)dz ) from 0 to z
(= int_{-1}^1(z^3+z^3+z^3)dz= frac{z^4}{4} ) from -1 to 1
= 0.

9. Evaluate (int_0^1int_0^{sqrt{1-x^2}}int_0^{sqrt{1-x^2-y^2}}xyzdxdydz.)
a) (frac{1}{6} )
b) (frac{1}{12} )
c) (frac{1}{24} )
d) (frac{1}{48} )
Answer: c
Explanation: (int_0^1 int_0^{sqrt{1-x^2}}int_0^{sqrt{1-x^2-y^2}}xyzdxdydz=
int_0^1 int_0^{sqrt{1-x^2}} frac{xyz^2}{2} dxdy ) from 0 to ( sqrt{1-x^2-y^2} )
(= frac{1}{2} int_0^1 frac{xy^2}{2} – frac{x}{2} – frac{xy^4}{4} dx ) from 0 to (sqrt{1-x^2} )
(= frac{1}{2} int_0^1 frac{(x-x^3)}{2}-frac{(x^3-x^5)}{2}-frac{(x+x^5-2x^3)}{4} dx)
(= frac{1}{4}big{frac{1}{6} – frac{1}{2} + frac{1}{2}big} )
(= frac{1}{24}. )

10. What is the volume of a cube with side a?
a) (frac{a^3}{8} )
b) (a^2)
c) (a^3)
d) (frac{a2}{4} )
Answer: c
Explanation: (int_0^a int_0^a int_0^a dxdydz = (int_0^a dx)(int_0^a dy)(int_0^a dz)= a^3.)

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250+ TOP MCQs on DeMoivre’s Theorem and Answers

Complex Analysis Multiple Choice Questions on “DeMoivre’s Theorem”.

1. Find the value of (1+i)¹⁰⁰.
a) 2¹⁰⁰ (cos⁡100π+isin100π)
b) 2¹⁰⁰ (cos⁡25π+isin25π)
c) 2⁵⁰ (cos⁡100π+isin100π)
d) 2⁵⁰ (cos⁡25π+isin25π)
View Answer

Answer: d
Explanation: We know that,
1+i=(sqrt 2 (frac{1}{sqrt 2}+frac{i}{sqrt 2})=sqrt 2 (cos frac{pi}{4}+isin frac{pi}{4}))
((1+i)^{100}=(sqrt 2 left (cos frac{pi}{4}+isin frac{pi}{4}right ))^{100}=2^{50}(left(cos frac{pi}{4}+isin frac{pi}{4}right))^{100})
By Applying the DeMoivre’s Theorem
((1+i)^{100}=2^{50} left (cos 100frac{pi}{4}+isin 100frac{pi}{4} right ))
((1+i)^{100}=2^{50} (cos⁡25pi+isin25pi)).

2. Find the value of (1-i)¹⁰⁰.
a) 2¹⁰⁰ (cos⁡100π-isin100π)
b) 2¹⁰⁰ (cos⁡25π-isin25π)
c) 2⁵⁰ (cos⁡25π-isin25π)
d) 2⁵⁰ (cos⁡100π-isin100π)
View Answer

Answer: c
Explanation: We know that,
(1-i=sqrt 2 (frac{1}{sqrt 2}-frac{i}{sqrt 2})=sqrt 2 (cos frac{pi}{4}-isin frac{pi}{4}))
((1-i) ^{100}=(sqrt 2 (cos frac{pi}{4}-isin frac{pi}{4})) ^{100}=2^{50}((cos frac{pi}{4}-isin frac{pi}{4})) ^{100})
By Applying the DeMoivre’s Theorem
((1-i)^{100}=2^{50} (cos frac{100pi}{4}-isin frac{100pi}{4}))
((1-i)^{100}=2^{50} (cos⁡25pi-sin25pi)).

3. If (a=frac{1}{sqrt 2}+frac{i}{sqrt 2}), find the value of a⁵ + conjugate of a⁵=?
a) (cos⁡ frac{3pi}{4})
b) (2 sin ⁡frac{5pi}{4})
c) (2 cos frac{5pi}{4})
d) (cos frac{5pi}{4})
View Answer

Answer: c
Explanation: We know that,
(a=(frac{1}{sqrt 2}+frac{i}{sqrt 2})=(cos frac{pi}{4}+isin frac{pi}{4}))
Conjugate of a = ((frac{1}{sqrt 2}-frac{i}{sqrt 2})=(cos frac{pi}{4}-isin frac{pi}{4}))
(a^5) + (conj of a)⁵=((cos frac{pi}{4}+isin frac{pi}{4})^5+(cos frac{pi}{4}-isin frac{pi}{4})^5)
By Applying the DeMoivre’s Theorem
a⁵ + (conj of a)⁵ = (cos⁡ frac{5pi}{4} +isin frac{5pi}{4}+cos⁡ frac{5pi}{4} -isin frac{5pi}{4})
a⁵ + (conj of a)⁵ = (2 cos ⁡frac{5pi}{4}).

4. Evaluate (frac{(1+sqrt 3 i) ^{16}}{(sqrt 3-i) ^{17}}).
a) (frac{1}{2}×(cos frac{pi}{6}+isin frac{pi}{6}))
b) (frac{1}{3}×(cos frac{pi}{6}+isin frac{pi}{6}))
c) (frac{1}{2}×(cos frac{pi}{3}+isin frac{pi}{3}))
d) (frac{1}{4}×(cos frac{pi}{6}+isin frac{pi}{6}))
View Answer

Answer: a
Explanation: We know that,
(1+sqrt 3 i=2(cos frac{pi}{3}+isin frac{pi}{3}))
(sqrt 3-i=2(cos frac{pi}{6}-isin frac{pi}{6}))
(frac{(1+sqrt 3 i) ^{16}}{(sqrt 3-i) {^17}}=frac{2^{16}(cos frac{pi}{3}+isin frac{pi}{3}) ^{16}}{2^{17}(cos frac{pi}{6}-isin frac{pi}{6}) ^{17}})
By Demoivre’s Theorem
= (frac{1}{2}×(cos frac{16pi}{3}+isin frac{16pi}{3})(cos frac{17pi}{6}+isin frac{17pi}{6}))
= (frac{1}{2}×(cos(frac{⁡49pi}{6})+isin(frac{49pi}{6})))
= (frac{1}{2}×(cosfrac{pi}{6}+isinfrac{pi}{6})).

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250+ TOP MCQs on Taylor Mclaurin Series and Answers

Engineering Mathematics Quiz focuses on “Taylor Mclaurin Series – 4”.

1. The expansion of f(x), about x = a is
a) (f(a)+frac{h}{1!} f’ (a)+frac{h^2}{2!} f” (a)….+frac{h^n}{n!} f^n (a))
b) (f(a)+frac{h}{1!} f’ (a)+frac{h^2}{2!} f” (a)….)
c) (hf(a)+frac{h^2}{1!} f’ (a)+frac{h^3}{2!} f” (a)…+frac{h^n}{n!} f^n (a))
d) (hf(a)+frac{h^2}{1!} f’ (a)+frac{h^3}{2!} f” (a)…..)
Answer: a
Explanation: By taylor expansion,
f(a+h) = f(a) + ^h⁄_1! f’ (a) + ^h²⁄_2! f^” (a)…….

2. Find the expansion of e^x in terms of x + m, m > 0.
a) (e^m [1+(x+m)+frac{(x+m)^2}{2!}+frac{(x+m)^3}{3!}+….])
b) (e^{-m} [1+(x-m)+frac{(x-m)^2}{2!}+frac{(x-m)^3}{3!}+….])
c) (e^m [1+(x-m)+frac{(x-m)^2}{2!}+frac{(x-m)^3}{3!}+….])
d) (e^{-m} [1+(x+m)+frac{(x+m)^2}{2!}+frac{(x+m)^3}{3!}+….])
Answer: d
Explanation: Let, h = x + m = > f(x) = f(h-m) = e^(h-m)
By taylor theorem, putting a = -m, we get,
f(a+h) = (f(a)+frac{h}{1!} f’ (a)+frac{h^2}{2!} f” (a)…)
f(h-m) = (f(-m)+h/1! f'(-m)+frac{h^2}{2!} f” (-m)….(1))
now, f(-m) = f’(-m) = f’’(-m)=e^-m
hence,
f(x)=e^x=f(h-m)=(e^{-m} [1+(x+m)+frac{(x+m)^2}{2!}+frac{(x+m)^3}{3!}+….])

3. Expand ln(x) in the power of (x-m).
a) ln⁡(m)+(frac{h}{m}-frac{1}{2!} (h/m)^2+frac{2}{3!} (h/m)^3-……)
b) ln⁡(m)-(frac{h}{m}-frac{1}{2!} (h/m)^2-frac{2}{3!} (h/m)^3-……)
c) ln⁡(m)-(frac{1}{2!} (h/m)^2+frac{2}{3!} (h/m)^4-……)
d) ln⁡(m)+(frac{h}{m}+frac{2}{3!} (h/m)^3-……)
Answer: a
Explanation: where, h = x-m
Let, h = x – m => f(x) = f(h+m) = e^(h+m)
By taylor theorem, putting a = m , we get,
f(a+h) = (f(a)+frac{h}{1!} f’ (a)+frac{h^2}{2! }f” (a)…)
f(h-m) = (f(m)+frac{h}{1!} f’ (m)+frac{h^2}{2!} f” (m))…….(1)
now,f(m) = ln(m), f’(m)=1/m, f” (m)=-1/m², f”’ (m)=2/m³,……
hence,
f(x)=ln(x)=f(h+m)=(ln⁡(m)+frac{h}{m}-frac{h^2}{2!}frac{1}{m^2}+frac{h^3}{3!} frac{2}{m^3}-……)
f(x)=ln(x)=ln⁡(m)+(frac{h}{m}-frac{1}{2!} (h/m)^2+frac{2}{3!} (h/m)^3-……)
where, h = x-m

4. Find the value of √10
a) 3.1633
b) 3.1623
c) 3.1632
d) 3.1645
Answer: b
Explanation: Now f(x)=√x,
Hence, f’ (x)=(frac{1}{2} x^{-1/2})
f” (x)=(-frac{1}{4} x^{-3/2})
f”’ (x)=(frac{3}{8} x^{-5/2})
Hence,
f(x+h)=(sqrt{x+h}=sqrt{x}+frac{h}{2} x^{-1/2}-frac{h^2}{8} x^{-3/2}+frac{h^3}{16} x^{-5/2}+…. )
(By Taylor’s expansion)
Putting,
h=1, and x=9 we get,
f(10)=√10=3+1/6-1/216+1/3888+….=3.1623

5. Expand f(x) = ¹⁄_x about x = 1.
a) 1 – (x-1) + (x-1)² – (x-1)³ + ….
b) 1 + (x-1) + (x-1)² + (x-1)³ + ….
c) 1 + (x-1) – (x-1)² + (x-1)³ + ….
d) 1 – (x+1) + (x+1)² – (x+1)³ + ….
Answer: a
Explanation: Given f(x) = ¹⁄_x
Let, x – 1 = h
Hence, x = 1 + h
Hence, f(x) = f(1 + h) = f(1) + ^h⁄_1! f’ (1) + ^h²⁄_2! f^” (1) +^h³⁄_3! f^”’ (1)+…
Now, f(1) = 1, f'(1) = -1, f”(1) = 2 ,f”'(1) = -6,…….
Hence, f(1 + h) = 1 – h + h² – h³+….
hence, 1 – (x-1) + (x-1)² – (x-1)³ +….

6. Find the expansion of f(x) = ^{e^x} ⁄_1+e^x, given ∫f(x)dx = ln⁡(2), for x = 0
a) (frac{1}{2}-frac{x}{4}-frac{x^3}{48}-…)
b) (frac{1}{2}+frac{x}{4}-frac{x^3}{48}+….)
c) (frac{1}{2}+frac{x}{4}+frac{x^3}{48}+….)
d) (frac{1}{2}+frac{x}{4}-frac{x^3}{48}+….)
Answer: b
Explanation:
Given,f(x)=(frac{e^x}{1+e^x})
Differentiating it we get
(f^1 (x)=ln⁡(1+e^x)+C), now putting x=0 we get,c=0
Hence,
(f^1 (x)=ln⁡(1+e^x))
Now,(f^1 (x)=ln⁡(1+e^x)=ln⁡(2)+frac{x}{2}+frac{x^2}{8}-frac{x^4}{192}+..)(By,Mclaurin’s expansion)
Hence, Differentiating it we get,
f(x)=(frac{1}{2}+frac{x}{4}-frac{x^3}{48}+….).

7. Find the value of e^{^π⁄_4√2}
a) 1.74
b) 1.84
c) 1.94
d) 1.64
Answer: a
Explanation: Let, f(x) = e^xSin(x), f(0) = 1
Now, the expansion of xSin(x) is (x^2-frac{x^3}{3!}+frac{x^6}{5!}+….)
Hence, (e^xSin(x)=e^y=1+y+frac{y^2}{2!}+frac{y^3}{3!}+….)
Hence,
(e^{xSin(x)}=1+(x^2-frac{x^4}{3!}+frac{x^6}{6!}+…..)+frac{(x^2-frac{x^4}{3!}+frac{x^6}{6!}+…..)^2}{2!})
(+frac{(x^2-frac{x^4}{3!}+frac{x^6}{6!}+…..)^3}{6}+….)
(e^{xSin(x)}=1+x^2-frac{x^4}{3!}+frac{x^6}{5!}+frac{x^4}{2}-frac{x^6}{6}+frac{x^6}{6}+….) (we neglect all other other terms by considering the options given)
Hence, (e^{xSin(x)}=1+x^2+frac{x^4}{3}+frac{x^6}{120}+……)
Putting, x = π/4,
We get,
f(π/4)=e^{π/4 Sin(π/4)}=e^π/(4√2)=1+(π/4)²+1/3 (π/4)⁴+….=1+.6168+.1268=1.74

8. Find the value of ln(sin(31^o)) if ln(2) = 0.69315
a) -0.653
b) -0.663
c) -0.764
d) -0.662
Answer: b
Explanation: Let, f(x) = ln⁡(sin⁡(x+h))
Then, f(x) = ln⁡(sin⁡(x)), if h=0
f’ (x)=cot⁡(x), f” (x)=-cosec² (x), f”’ (x)=2cosec² (x)cot⁡(x)
Hence, by Taylor’s theorem,
f(x+h)=f(x)+hf'(x)+(frac{h^2}{2!}) f” (x)+(frac{h^3}{3!}) f”’ (x)+⋯..
Hence, ln⁡(sin⁡(x+h))=ln⁡(sin⁡(x))+h cot⁡(x)-(frac{h^2}{2!}) cosec² (x)+(frac{h^3}{3!}) (2cosec² (x) cot⁡(x))+⋯.
Now let, x=30^o, h=1^o;
ln(sin(31^o)) = (ln(sin(30))+frac{π}{180} cot⁡(frac{π}{6})-frac{1}{2!} (frac{π}{180})^2 cosec^2 (frac{π}{6})+⋯)
ln(sin(31^o)) = -0.6935+.030231-.000304+0
ln(sin(31^o)) = -0.663

9. The expansion of f(x,y), is
a) f(0,0)+([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+frac{1}{2!} [x^2 frac{∂^2 f}{∂x^2}-2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}]+….)
b) f(0,0)+([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+frac{1}{2!} [x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}]+…)
c) f(0,0)+([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+frac{1}{2!} [x^2 frac{∂^2 f}{∂x^2}-2xy frac{∂^2 f}{∂x∂y}-y^2 frac{∂^2 f}{∂y^2}]+…)
d) f(0,0)-([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+frac{1}{2!} [x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}]-…)
Answer: b
Explanation: By taylor expansion,
f(x,y) = f(0,0)+([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+frac{1}{2!} [x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}]+…)

10. The expansion of f(x, y)=e^{x Sin(y)}, is
a) x + xy + ….
b) y + y² x + ….
c) x + x² y + ….
d) y + xy + ……..
Answer: d
Explanation: Now, f(x, y)=e^{x Sin(y)}, f(0,0) = 0
Therefore,
f_x (x,y) = e^x Sin(y), hence f_x (0,0) = 0

f_y (x,y) = e^x Cos(y), hence f_y (0,0) = 1

f_xx (x,y) = e^x Sin(y), hence f_xx (0,0) = 0

f_yy (x,y) = -e^x Sin(y), hence f_yy (0,0) = 0

f_xy (x,y) = e^x Cos(y), hence f_xy (0,0) = 1
By taylor expansion,
f(x,y) = f(0,0)+([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+frac{1}{2!} [x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}]+…)
f(x,y) = 0 + 0 + y + (frac{1}{2!}) [0 + 2xy + 0] +…..
f(x,y) = y + xy + ……..

11. The expansion of f(x, y) = e^x ln(1 + y), is
a) f(x,y) = y + xy – ^y²⁄₂ +…….
b) f(x,y) = y – xy + ^y²⁄₂ -…….
c) f(x,y) = y + x – ^y²⁄₂ +……..
d) f(x,y) = x + y – ^x²⁄₂ +……..
Answer: a
Explanation: Now, f(x, y) = e^x ln(1 + y) , f(0,0) = 0
Therefore,
(f_x (x,y)=e^x ln(1+y)), hence f_x (0,0) = 0
(f_y (x,y)=frac{e^x}{(1+y)}), hence f_y (0,0) = 1
(f_{xx} (x,y)=e^x ln(1+y)), hence f_xx (0,0) = 0
(f_{yy} (x,y)=-frac{e^x}{(1+y)^2}), hence f_yy (0,0) = -1
(f_{xy} (x,y)=frac{e^x}{(1+y)}), hence f_xy (0,0) = 1
By taylor expansion,
f(x,y) = f(0,0)+([x frac{∂f}{∂x}+y frac{∂f}{∂y}]+1/2! [x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}]+…)
f(x,y) = y + xy – ^y²⁄₂ +…….

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250+ TOP MCQs on Total Derivative and Answers

Differential and Integral Calculus Multiple Choice Questions on “Total Derivative”.

1. The total derivative is the same as the derivative of the function.
a) True
b) False
View Answer

Answer: a
Explanation: In mathematics, the total derivative of a function at a point is the best linear approximation near this point of the function with respect to its arguments.

2. What is the derivative of (7sqrt[3]{x}-frac{3}{x^4}+5x) with respect x?
a) (7x^{frac{-1}{3}}-3x^{-5}+5)
b) (7x^{frac{-2}{3}}-3x^{-5}+5)
c) (7x^{frac{-2}{3}}-3x^{-3}+5)
d) (7x^{frac{-1}{3}}-3x^{-3}+5)
View Answer

Answer: b
Explanation: Given: y= (7sqrt[3]{x}-frac{3}{x^4}+5x)
(frac{dy}{dx}=frac{d(7x^{frac{-1}{3}}-3x^{-4}+5x)}{dx})
(frac{dy}{dx}=7x^{frac{-1}{3}-1}-3x^{-4-1}+5x^{1-1})
(frac{dy}{dx}=7x^{frac{-2}{3}}-3x^{-5}+5)

3. Find the range in which the function f(x) = 8 + 40x³ – 5x⁴ – 4x⁵ is increasing.
a) 2<z<0, 0<z<3
b) 1<z<0, 0<z<2
c) 3<z<0, 0<z<2
d) 3<z<0, 0<z<4
View Answer

Answer: c
Explanation: Given: f(x)=8 + 40x³ – 5x⁴ – 4x⁵
f'(x) = 120x² – 20x³ – 20x⁴
f'(x) = -20x² (x+x² – 6)
f'(x) = -20x² (x+3)(x-2)
Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve.
f'(x) = -20x² (x+3)(x-2)=0
From this it is pretty easy to see that the derivative will be zero, and hence the function will not be moving, at,
x=0,-3,-2
Because the derivative is continuous, we know that the only place it can change sign is where the derivative is zero. So, a quick number line will give us the sign of the derivative for the various intervals.

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From this we get,
Increasing: 3

4. What is the maximum area of the rectangle with perimeter 500 mm?
a) 15,625 mm²
b) 15,025 mm²
c) 15,600 mm²
d) 10,625 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)
500=2(x+y)
x+y=250
y=250-x
We can now substitute the value of y in (1)
A=x*(250-x)
A=250x-x²
To find maximum value we need derivative of A,
(frac{dA}{dx}=250-2x)
To find maximum value, (frac{dA}{dx}=0)
250-2x=0
2x=250
x=125 mm
Therefore, when the value of x=125 mm and the value of y=250-125=125 mm, the area of the rectangle is maximum, i.e., A=125*125=15,625 mm²

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 = -1
d) k × i = -j, i × k = j
View Answer

Answer: b
Explanation: The properties of vector or cross product, for the orthogonal vectors, i, j, and k are,
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. Which of the following trigonometric function derivatives is correct?
a) (frac{d(sinx)}{dx}=-cosx)
b) (frac{d(secx)}{dx}=tanx)
c) (frac{d(tanx)}{dx}=sec^2 x)
d) (frac{d(cosx)}{dx}=sinx)
View Answer

Answer: c
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)

7. The division rule of differentiation for two functions is given by, ((frac{f(x)}{g(x)})’= frac{f'(x)-g’ (x)}{(g(x))^2}. )
a) True
b) False
View Answer

Answer: b
Explanation: The division rule of differentiation for two functions is given by,
((frac{f(x)}{g(x)})’= frac{g(x)f'(x)- g'(x)f(x)}{(g(x))^2} )

8. What is the derivative of z=3x*logx+5x⁶ e^x with respect to x?
a) 3+30x⁵ e^x
b) 3+5x⁶ e^x+30x⁵ e^x
c) 3+5x⁶ e^x
d) 3+3logx+5x⁶ e^x+30x⁵ e^x
View Answer

Answer: d
Explanation: Given: z=3x*logx+5x⁶ e^x
(frac{dz}{dx}=3x(frac{1}{x}))+3logx+5x⁶ e^x+30x⁵ e^x
(frac{dz}{dx})=3+3logx+5x⁶ e^x+30x⁵ e^x

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 2%?
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a) 0.06366%
b) 0.006366%
c) 0.6366%
d) 6.366%
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 2%,
(dR=frac{dV}{4πR^2}=frac{0.02}{4π(5)^2}=0.00006366)
Error in radius = 0.006366%

10. Which of the following is correct?
a) (frac{d}{dx} (sin^{-1}(⁡x))= frac{1}{sqrt{1-x}})
b) (frac{d}{dx} (sec^{-1}(⁡x))= frac{1}{xsqrt{x^2-1}})
c) (frac{d}{dx} (tan^{-1}(⁡x))= frac{1}{sqrt{x^2+1}})
d) (frac{d}{dx} (sin^{-1)}(⁡x))= frac{1}{x+1} )
View Answer

Answer: a
Explanation: Rules for derivatives of inverse trigonometric functions are:

(frac{d}{dx} (sin^{-1}⁡(⁡x))= frac{1}{sqrt{1-x^2}})
(frac{d}{dx} (sec^{-1}⁡(⁡x))= frac{1}{xsqrt{x^2-1}})
(frac{d}{dx} (tan^{-1}(⁡x))= frac{1}{1+x^2})

Global Education & Learning Series – Differential and Integral Calculus.

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250+ TOP MCQs on Quadrature and Answers

Differential and Integral Calculus Interview Questions and Answers focuses on “Quadrature”.

1. What is meant by quadrature process in mathematics?
a) Finding area of plane curves
b) Finding volume of plane curves
c) Finding length of plane curves
d) Finding slope of plane curves
Answer: a
Explanation: The process of finding area of plane curves is often called quadrature. It is an important application of integral calculus.

2. What is the formula used to find the area surrounded by the curves in the following diagram?

a) (int_a^b y ,dx)
b) (int_a^b -y ,dx)
c) (int_a^b x ,dy)
d) (int_a^b -x ,dy)
Answer: a
Explanation: The area is present above the x-axis. Area above the x-axis is positive. The area is bounded by x-axis, curve y = f(x), straight lines x=a and x=b. Hence, area is found by integrating the curve with the lines as limits.

3. What is the formula used to find the area surrounded by the curves in the following diagram?

a) (int_a^b y ,dx)
b) (int_a^b -y ,dx)
c) (int_a^b x ,dy)
d) (int_a^b -x ,dy)
Answer: b
Explanation: The area is present below the x-axis. Area below the x-axis is negative. The area is bounded by x-axis, curve y = f(x), straight lines x=a and x=b. Hence, area is found by integrating the curve with the lines as limits.

4. What is the formula used to find the area surrounded by the curves in the following diagram?

a) (int_c^d y ,dx)
b) (int_c^d -y ,dx)
c) (int_c^d x ,dy)
d) (int_c^d -x ,dy)
Answer: c
Explanation: The area is present right of y-axis. Area right to y-axis is positive. The area is bounded by the y-axis, curve x = f(y), straight lines y=c and y=d. Hence, area is found by integrating the curve with the lines as limits.

5. What is the formula used to find the area surrounded by the curves in the following diagram?

a) (int_c^d y ,dx)
b) (int_c^d -y ,dx)
c) (int_c^d x ,dy)
d) (int_c^d -x ,dy)
Answer: d
Explanation: The area is present left of y-axis. Area left to y-axis is negative. The area is bounded by y-axis, curve x = f(y), straight lines y=c and y=d. Hence, area is found by integrating the curve with the lines as limits.

6. Find the area bounded in the following diagram.

a) 6
b) 12
c) 8
d) 10
Answer: b
Explanation: Area is bounded by y = (frac{3}{2}) (x + 2), lines x = 1 and x = 3.
Area = (int_1^3 y dx = frac{3}{2} int_1^3 (x+2),dx )
( = frac{3}{2} bigg[frac{x^2}{2} + 2xbigg]_1^3)
( = frac{3}{2} [frac{1}{2} (9 − 1) + 2(3 − 1)] = frac{3}{2} [4 + 4])
= 12 sq.units.

7. What is the area bounded by the curve y = x² – 5x + 4, x = 2, x = 3, x-axis in the following diagram?

a) 13
b) 6
c) (frac{13}{6})
d) (frac{6}{13})
Answer: c
Explanation: The area lies below the x-axis.
Area = ∫ -y dx
= (int_2^3 -(x^2-5x+4) dx)
= (displaystylebigg[frac{x^3}{3} – 5 frac{x^2}{2} + 4xbigg]_2^3)
= (-[(9 – (frac{45}{2}) +12) – ((frac{8}{3}) – (frac{20}{2}) + 8)] )
= (– left(-frac{13}{6}right))
= (frac{13}{6} ) sq.units.

Global Education & Learning Series – Differential and Integral Calculus.

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250+ TOP MCQs on Reducible to Homogenous Form and Answers

Ordinary Differential Equations Interview Questions and Answers focuses on “Reducible to Homogenous Form”.

1. Solution of the differential equation (frac{dy}{dx} = frac{3x-6y+7}{x-2y+4}) is _____
a) log(x-2y+4)²=c
b) 6x-2y+log(x-2y+2)²=c
c) x-2y+log(x-2y+6)²=c
d) -5x+log(x-2y+3)²=c
Answer: b
Explanation: (frac{dy}{dx} = frac{3x-6y+7}{x-2y+4} rightarrow frac{dy}{dx} = frac{3(x-2y)+7}{(x-2y)+4})…..here coefficient of x and y in the numerator & denominator are proportional hence substituting (x-2y = t rightarrow 1 – 2frac{dy}{dx} = frac{dt}{dx})
(1- frac{dt}{dx} = 2left(frac{3t+7}{t+4}right) rightarrow frac{dt}{dx} = frac{t+4-6t-14}{t+4} =frac{-5(t+2)}{t+4})
separating the variables and hence integrating
( int frac{t+4}{t+2} ,dt = int -5 ,dx rightarrow int left(1 + frac{2}{t+2}right) ,dt = -5x + c)
t + 2 log(t+2) = -5x + c –> x – 2y + 2 log(x-2y+2) + 5x = c
6x-2y+log(x-2y+2)² = c is the solution.

2. Solution of the differential equation (3y-7x+7)dx+(7y-3x+3)dy=0 is ______
a) p=(y+x)⁵ (y-x)²
b) p=(y+x+2)⁵ (y-x+2)²
c) p=(y+x)² (y-x)⁵
d) p=(y+x-1)⁵ (y-x+1)²
Answer: d
Explanation: (3y-7x+7)dx + (7y-3x+3)dy = 0 –> (frac{dy}{dx} = frac{7x-3y-7}{-3x+7y+3})
substituting x=X+h, y=Y+k where (h,k) will satisfy the equation
– 3y+7x-7=0 & 7y-3x+3=0…..(1)
after substitution we get (frac{dY}{dX} = frac{(7X-3Y)+(7h- 3k-7)}{(-3X+7Y)+(-3h+7k+3)})…..(2)
from (1) we can write 7h- 3k-7=0 & -3h+7k+3=0 solving for
h & k we get h=1 & k=0 (2) can be written as (frac{dY}{dX} = frac{7X-3Y}{-3X+7Y})
(frac{dY}{dX} = frac{7X-3Y}{-3X+7Y})…..it is a homogenous equation hence substituting Y=VX
(V + Xfrac{dV}{dX} = frac{7-3V}{-3+7V})……separating the variables we get & integrating
(Xfrac{dV}{dX} = frac{7-3V}{-3+7V} – V = frac{7-7V^2}{7V-3})
( int frac{7V-3}{1-V^2} ,dV = 7int frac{1}{X} ,dX)
( int frac{7V}{1-V^2} ,dV – int frac{3}{1-V^2} ,dV) = 7log X + 7log c……7log c=log k
substituting 1-V²=t –> -2V dV=dt
( int frac{7}{-2t} ,dt + frac{3}{2} log(frac{V-1}{V+1})- 7 log X = log k)
(frac{7}{-2t} log(1-V^2) + frac{3}{2} log(frac{V-1}{V+1}) – 7 log X = log k)
(log(1-V^2)^{frac{7}{-2t}} + log(frac{V-1}{V+1})^{frac{3}{2}} + log(X^{-7}) = log k)
((1-V^2)^{frac{7}{-2t}} (frac{V-1}{V+1})^{frac{3}{2}} X^{-7} = k)
(1+V)⁵ (V-1)² X⁷=p where 1/k = p
p=(y+x-1)⁵ (y-x+1)² is the solution…after substituting back V=Y/X & Y=y, X=x-1.

3. Solution of the differential equation (x-y)dy=(x+y+1)dx is _____
a) ( sqrt{x^2+y^2} = e^{c tan^{-1}⁡left(frac{y+0.5}{x+0.5}right)})
b) ((x^2+y^2)^1.5 = c cot^{-1} ⁡left(frac{y+0.5}{x+0.5}right))
c) ((x^2+y^2)^2 = e^{c cot⁡⁡left(frac{y+0.5}{x+0.5}right)})
d) ((x^2+y^2)^1 = c tan⁡left(frac{y+0.5}{x+0.5}right))
Answer: a
Explanation: (frac{dy}{dx} = frac{x+y+1}{x-y}), by substituting x=X+h, y=Y+k
w.k.t (h,k) satisfies the equations as h+k+1=0 & h-k=0 –> h=k=-0.5
and hence the differential equation changes to the form (frac{dy}{dx}=frac{X+Y}{X-Y} ) is a homogenous equation thus put (v = frac{Y}{X} rightarrow ,v + X frac{dv}{dX} = frac{1+v}{1-v})
(X frac{dv}{dX} = frac{1+v^2}{1-v} rightarrow int frac{1-v}{1+v^2} ,dv = ∫frac{1}{X} ,dx = int(frac{1}{1+v^2} – frac{v}{1+v^2}) ,dv = log X + c)
tan^-1⁡v-0.5log (1+v²) = log X+c –> ( tan^{-1} frac{⁡Y}{X} – log⁡(sqrt{X^2+Y^2}) = c)
(sqrt{X^2+Y^2} = e^{c tan^{-1}⁡left(frac{y+0.5}{x+0.5}right)})……from the equation X=x+0.5 & Y=y+0.5.

4. Solution of the differential equation (frac{dy}{dx} = frac{x+2y-3}{2x+y-3}) is _____
a) (x+y)=c(x-y)²
b) (x+y-2)=c(x-y)³
c) (2x+y)=c(x+2y)²
d) (2x+y-3)=c(x-2y-3)³
Answer: b
Explanation: (frac{dy}{dx} = frac{x+2y-3}{2x+y-3}) –>x=X+h, y=Y+k
h+2k-3=0 & 2h+k-3=0 solving we get h=1=k
new differential equation is (frac{dY}{dX} = frac{X+2Y}{2X+Y}), put Y=vX
(v + Xfrac{dv}{dX} = frac{1+2v}{2+v} rightarrow X frac{dv}{dX} = frac{1-v^2}{2+v})
or
( int(frac{2+v}{1-v^2}) dv = int frac{1}{X} ,dX rightarrow int(frac{2}{1-v^2} + frac{v}{1-v^2}) ,dv = log X + log c)
(logleft(frac{1+v}{1-v}right) – 0.5 log (1-v^2) = log X + log c)
(log left(frac{X+Y}{X-Y}right) – 0.5 log (X^2-Y^2) + log X = log X + log c)
(log left(frac{X+Y}{X-Y}.frac{1}{sqrt{X^2-Y^2}}right) = log c)
(sqrt{X+Y}) = (X-Y)^1.5 c or (X+Y)=(X-Y)^3c
(x+y-2)=c(x-y)³ is the solution.

5. Solution of the differential equation (frac{dy}{dx} = frac{y-x+1}{y+x+5}) is ______
a) (π-tan{-1}⁡ frac{y+3}{x+2} – logsqrt{(x+2)^2+(y+3)^2}=c)
b) (-tan{-1}⁡⁡ frac{y+3}{x+2} – logsqrt{(x+2)^2+(y+3)^2}=c)
c) (tan{-1}⁡⁡ frac{y+3}{x+2} – logsqrt{x+y+5}=c)
d) (-cot{-1}⁡⁡ frac{y+3}{x+2} – logsqrt{x+y+5}=c)
Answer: b
Explanation: (frac{dy}{dx} = frac{y-x+1}{y+x+5}) –> x=X+h, y=Y+k
h+k+5=0 & k-h+1=0 solving we get h=-2, k=-3
(frac{dY}{dX} = frac{Y-X}{Y+X}), put Y=vX we get an new equation
(V + Xfrac{dV}{dX} = frac{v-1}{v+1} rightarrow Xfrac{dV}{dX} = frac{-(1+v^2)}{1+v})
or
(-int frac{1+v}{1+v^2} dv = int frac{1}{X} dX = int frac{1}{1+v^2} + frac{v}{1+v^2} ,dv)
-tan^-1⁡v-0.5 log(1+v²)=log x +c
(-tan^{-1} frac{⁡Y}{X} – logsqrt{X^2+Y^2} = c)
(-tan{-1}⁡⁡ frac{y+3}{x+2} – logsqrt{(x+2)^2+(y+3)^2}=c) is the solution.

Global Education & Learning Series – Ordinary Differential Equations.

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