Explore topic-wise MCQs in Engineering Mathematics.

This section includes 14 Mcqs, each offering curated multiple-choice questions to sharpen your Engineering Mathematics knowledge and support exam preparation. Choose a topic below to get started.

1.

Find the value of \(\int \frac{1}{16x^2+16x+10}dx\).

A. 1⁄8 sin-1(x + 1⁄2)
B. 1⁄8 tan-1(x + 1⁄2)
C. 1⁄8 sec-1(x + 1⁄2)
D. 1⁄4 cos-1(x + 1⁄2)
Answer» C. 1⁄8 sec-1(x + 1⁄2)
2.

Temperature of a rod is increased by moving x distance from origin and is given by equation T(x) = x2 + 2x, where x is the distance and T(x) is change of temperature w.r.t distance. If, at x = 0, temperature is 40 C, find temperature at x=10.

A. 473 C
B. 472 C
C. 474 C
D. 475 C
Answer» B. 472 C
3.

Find the value of ∫(x4 – 5x2 – 6x)4 4x3 – 10x – 6 dx.

A. \(\frac{(x^4-5x^2-6x)^4}{4}\)
B. \(\frac{(x^4-5x^2-6x)^5}{5}\)
C. \(\frac{(4x^3-10x-6)^5}{5}\)
D. \(\frac{(4x^3-10x-6)^4}{4}\)
Answer» C. \(\frac{(4x^3-10x-6)^5}{5}\)
4.

Find the area inside function \(\frac{(2x^3+5x^2-4)}{x^2}\) from x = 1 to a.a) a2⁄2 + 5a – 4ln(a)b) a2⁄2 + 5a – 4ln(a) – 11⁄2c) a2⁄2 + 4ln(

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» C. a2⁄2 + 4ln(a) – 11⁄2
5.

Find the area inside integral f(x)=\(\frac{sec^4⁡(x)}{\sqrt{tan⁡(x)}}\) from x = 0 to π.

A. π
B. 0
C. 1
D. 2
Answer» C. 1
6.

Find the area inside a function f(t) = \( \frac{t}{(t+3)(t+2)} dt\) from t = -1 to 0.

A. 4 ln⁡(3) – 5ln⁡(2)
B. 3 ln⁡(3)
C. 3 ln⁡(3) – 4ln⁡(2)
D. 3 ln⁡(3) – 5 ln⁡(2)
Answer» E.
7.

Find the area ln(x)⁄x from x = x = aeb to a.

A. b2⁄2
B. b⁄2
C. b
D. 1
Answer» B. b⁄2
8.

Find the area of a function f(x) = x2 + xCos(x) from x = 0 to a, where, a>0.a) a2⁄2 + aSin(a) + Cos(a) – 1b) a3⁄3 + aSin(a) + Cos(a)c) a3⁄3 + aSin(a) + Cos(a) – 1d) a3⁄3 + Cos(a) + Sin(

A. a2⁄2 + aSin(a) + Cos(a) – 1
B. a3⁄3 + aSin(a) + Cos(a)
C. a3⁄3 + aSin(a) + Cos(a) – 1
D. a3⁄3 + Cos(a) + Sin(a) – 1
Answer» D. a3⁄3 + Cos(a) + Sin(a) – 1
9.

Find the value of ∫x3 ex e2x e3x….enx dx.

A. \(\frac{2}{n(n+1)} e^{\frac{n(n+1)}{2}x} \left [x^3+3x^2 [\frac{2}{n(n+1)}]^1+6x[\frac{2}{n(n+1)}]^2 +6[\frac{2}{n(n+1)}]^3\right ]\)
B. \(\frac{2}{n(n+1)} e^{\frac{n(n+1)}{2}x} \left [x^3+3x^2 [\frac{2}{n(n+1)}]^1+6x[\frac{2}{n(n+1)}]^2 +6[\frac{2}{n(n+1)}]^3\right ]\)
C. \(\frac{2}{n(n+1)} e^{\frac{n(n+1)}{2}x} \left [x^3+3x^2 [\frac{2}{n(n+1)}]^1+6x[\frac{2}{n(n+1)}]^2 +6[\frac{2}{n(n+1)}]^3\right ]\)
D. \(\frac{2}{n(n+1)} e^{\frac{n(n+1)}{2}x} \left [x^3+3x^2 [\frac{2}{n(n+1)}]^1+6x[\frac{2}{n(n+1)}]^2 +6[\frac{2}{n(n+1)}]^3\right ]\)
Answer» B. \(\frac{2}{n(n+1)} e^{\frac{n(n+1)}{2}x} \left [x^3+3x^2 [\frac{2}{n(n+1)}]^1+6x[\frac{2}{n(n+1)}]^2 +6[\frac{2}{n(n+1)}]^3\right ]\)
10.

Find the value of ∫x7 Cos(x) dx.

A. x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x)
B. x7 Sin(x) – 7x6 Cos(x) + 42x5 Sin(x) – 210x4 Cos(x) + 840x3 Sin(x) – 2520x2 Cos(x) + 5040xSin(x) – 5040Cos(x)
C. x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x)
D. x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 10080Cos(x)
Answer» B. x7 Sin(x) – 7x6 Cos(x) + 42x5 Sin(x) – 210x4 Cos(x) + 840x3 Sin(x) – 2520x2 Cos(x) + 5040xSin(x) – 5040Cos(x)
11.

Value of ∫uv dx,where u and v are function of x.

A. \(\sum_{i=1}^n(-1)^i u_i v^{i+1}\)
B. \(\sum_{i=0}^nu_i v^{i+1}\)
C. \(\sum_{i=0}^n(-1)^i u_i v^{i+1}\)
D. \(\sum_{i=0}^n(-1)^i u_i v^{n-i}\)
Answer» D. \(\sum_{i=0}^n(-1)^i u_i v^{n-i}\)
12.

Find the value of ∫x3 Sin(x)dx.

A. x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x)
B. – x3 Cos(x) + 3x2 Sin(x) – 6Sin(x)
C. – x3 Cos(x) – 3x2 Sin(x) + 6xCos(x) – 6Sin(x)
D. – x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x)
Answer» E.
13.

Integration of (Sin(x) + Cos(x))ex is?

A. ex Cos(x)
B. ex Sin(x)
C. ex Tan(x)
D. ex (Sin(x) + Cos(x))
Answer» C. ex Tan(x)
14.

Find the value of ∫tan-1⁡(x)dx.

A. sec-1 (x) – 1⁄2 ln⁡(1 + x2)
B. xtan-1 (x) – 1⁄2 ln⁡(1 + x2)
C. xsec-1 (x) – 1⁄2 ln⁡(1 + x2)
D. tan-1 (x) – 1⁄2 ln⁡(1 + x2)
Answer» C. xsec-1 (x) – 1⁄2 ln⁡(1 + x2)