Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Integrals

Multiple Choice Questions

401.

If $\int_{0}^{\infty} \frac{\log (1 + x^{2})}{1 + x^{2}} d x = k \int_{0}^{1} \frac{\log (1 + x)}{1 + x^{2}} dx$ , then k is equal to

4
8
$π$
2 $π$

402.

$\int_{0}^{x} |\sin (t)| dt, where x \in [2 nπ, (2 n + 1) π], n \in N$ , is equal to

4n - 1 - cos(x)
4n - sin(x)
4n - cos(x)
4n + 1 - cos(x)

403.

Let I₁ = $\int_{0}^{1} \frac{e^{x} dx}{1 + x}$ and I₂ = $\int_{0}^{1} \frac{x^{2} dx}{e^{x^{3}} (2 - x^{3})}$ Then, $\frac{I_{1}}{I_{2}}$ is equal to

$\frac{1}{3 e}$
3e
$\frac{e}{3}$
$\frac{3}{e}$

404.

$\int_{\frac{5}{2}}^{5} \frac{\sqrt{{(25 - x^{2})}^{3}}}{x^{4}} dx$ is equal to

$\frac{π}{3}$
$\frac{2 π}{3}$
$\frac{π}{6}$
$\frac{5 π}{6}$

405.

$\int \frac{dx}{\cos (x) + \sqrt{3} \sin (x)}$ equals

$\frac{1}{2} \log (\tan (\frac{x}{2} + \frac{π}{12})) + C$
$\frac{1}{3} \log (\tan (\frac{x}{2} - \frac{π}{12})) + C$
$\log (\tan (\frac{x}{2} + \frac{π}{6})) + C$
$\frac{1}{2} \log (\tan (\frac{x}{2} - \frac{π}{6})) + C$

406.

$\int_{0}^{\frac{π}{2}} \sin (2 x) . \log (\tan (x)) dx$

407.

If $∆ x = |\begin{matrix} 1 & \cos (x) & 1 - \cos (x) \\ 1 + \sin (x) & \cos (x) & 1 + \sin (x) + \cos (x) \\ \sin (x) & \sin (x) & 1 \end{matrix}|$ , then $\int_{0}^{\frac{π}{4}} ∆ (x) dx$ is equal to

$\frac{1}{4}$
$\frac{1}{2}$
0
$- \frac{1}{4}$

408.

$\int_{0}^{2 π} (\sin (x) + |\sin (x)|) dx$ is equal to

409.

The value of $\int_{0}^{\sqrt{2}} [x^{2}] dx$ , where [.] is the greatest integer function, is

2 - $\sqrt{2}$
2 + $\sqrt{2}$
$\sqrt{2}$ - 1
$\sqrt{2}$ - 2

410.
If l (m,n) = $\int_{0}^{1} t^{m} {(1 + t)}^{n} dt$ , then the expression for l (m, n) in terms of l (m + 1, n + 1) is

$\frac{2^{n}}{m + 1} - \frac{n}{m + 1} . l (m + 1, n - 1)$

$\frac{n}{m + 1} . l (m + 1, n - 1)$

$\frac{2 n}{m + 1} + \frac{n}{m + 1} . l (m + 1, n - 1)$

$\frac{m}{n + 1} . l (m + 1, n - 1)$

$\frac{2^{n}}{m + 1} - \frac{n}{m + 1} . l (m + 1, n - 1)$

$We have, l (m, n) = l = \int_{0}^{1} t^{m} {(1 + t)}^{n} dt \Rightarrow l (m, n) = {[{(1 + t)}^{n} . \frac{t^{m + 1}}{m + 1}]}_{0}^{1} - \frac{n}{m + 1} \int_{0}^{1} {(1 + t)}^{n - 1} . t^{m + 1} dt$

= $\frac{2^{n}}{m + 1} - \frac{n}{m + 1} . l (m + 1, n - 1)$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

410. If l (m,n) = ∫01tm1 + tndt, then the expression for l (m, n) in terms of l (m + 1, n + 1) is 2nm + 1 - nm + 1 . l m + 1, n - 1 nm + 1 . l m + 1, n - 1 2nm + 1 + nm + 1 . l m + 1, n - 1 mn + 1 . l m + 1, n - 1