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.

JEE Mathematics Solved Question Paper 2014

Multiple Choice Questions

21.

The area of the portion of the circle x² + y² = 64 which is exterior to the parabola y² = 12x, is

$(8 π - \sqrt{3}) sq units$
$\frac{16}{3} (8 - \sqrt{3}) sq units$
$\frac{16}{3} (8 π - \sqrt{3}) sq units$
None of the above

22.

$\lim_{n \to \infty} \frac{1}{n} (\frac{1}{n + 1} + \frac{2}{n + 2} + . . . + \frac{3 n}{4 n})$ is equal to

log(4)
- log(4)
1 - log(4)
None of these

23.
The value of the integral $\int_{0}^{\frac{π}{2}} \frac{\sin^{2} (x)}{\sin (x) + \cos (x)} dx$ is equal to

$\sqrt{2} (\log (\sqrt{2}))$

$\sqrt{2} (\sqrt{2} + 1)$

$\log (\sqrt{2} + 1)$

None of the above

None of the above

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} (x)}{\sin (x) + \cos (x)} dx . . . (i) Now, I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} (\frac{π}{2} - x)}{\sin (\frac{π}{2} - x) + \cos (\frac{π}{2} - x)} dx \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} (x)}{\sin (x) + \cos (x)} dx . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} (x) + \cos^{2} (x)}{\sin (x) + \cos (x)} dx = \int_{0}^{\frac{π}{2}} \frac{1}{\sin (x) + \cos (x)} dx = \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{1}{\cos (x - \frac{π}{4})} dx = \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \sec (x - \frac{π}{4}) dx = \frac{1}{\sqrt{2}} {[\log \{\sec (x - \frac{π}{4}) + \tan (x - \frac{\log () π}{4})\}]}_{0}^{\frac{π}{2}} = \frac{1}{\sqrt{2}} [\log (\sqrt{2} + 1) - \log (\sqrt{2} - 1)] = \frac{1}{\sqrt{2}} (\log (\frac{\sqrt{2} + 1}{\sqrt{2} - 1})) = \frac{1}{\sqrt{2}} \log {(\sqrt{2} + 1)}^{2} = \sqrt{2} \log (\sqrt{2} + 1)$