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.

Limits and Derivatives

Multiple Choice Questions

51.

$\lim_{x \to α} {(\tan (x) \cot (x))}^{\frac{1}{x - α}}$ is equal to

$2 \csc (2 α)$
$\frac{1}{2} \sin (2 α)$
$- 2 \csc (2 α)$
None of these

52.

The value of $\lim_{x \to \infty} \frac{{(2^{x^{n}})}^{{\frac{1}{e}}^{x}} - {(3^{x^{n}})}^{{\frac{1}{e}}^{x}}}{x^{n}}$ (where, x $\in$ N) is

0
$n \log (n (\frac{2}{3}))$
$\log ((\frac{2}{3}))$
not defined

53.

$\lim_{x \to \frac{π}{4}} \frac{\tan (x) - 1}{\cos (2 x)}$ is equal to

54.
Which of the following inequality is true for x > 0?

$\log (1 + x) < \frac{x}{1 + x} < x$

$\frac{x}{1 + x} < x < \log (1 + x)$

$x < \log (1 + x) < \frac{x}{1 + x}$

$\frac{x}{1 + x} < \log (1 + x) < x$

$\frac{x}{1 + x} < \log (1 + x) < x$

Let f(x) = log(1 + x) - $\frac{x}{1 + x}$

$∴ f' (x) = \frac{1}{1 + x} - \frac{(1 + x) . 1 - x . 1}{{(1 + x)}^{2}} = \frac{1}{1 + x} - \frac{1}{{(1 + x)}^{2}} = \frac{1 + x - 1}{{(1 + x)}^{2}} = \frac{x}{{(1 + x)}^{2}}$

which is positive. [ $∵$ x > 0]

$∴$ f(x) is monotonic increasing, when x > 0

$\Rightarrow f (x) > f (0) Now, f (0) = \log (1) - 0 = 0 ∴ f (x) > 0 \Rightarrow \log (1 + x) - \frac{x}{1 + x} > 0 \Rightarrow \frac{x}{1 + x} < \log (1 + x) . . . (i) Also, for x > 0, x^{2} > 0 \Rightarrow x^{2} + x > x \Rightarrow x (x + 1) > x \Rightarrow x > \frac{x}{x + 1} . . . (ii) From Eqs . (i) and (ii), we get \frac{x}{1 + x} < \log (1 + x) < x$