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Limits and Derivatives

Multiple Choice Questions

181.

The equation $\sin (x) + 2 \sin (2 x) + 3 \sin (3 x) = \frac{8}{π}$ has atleast one root in

$[π, \frac{3 π}{2}]$
$[0, \frac{π}{2}]$
$[\frac{π}{2}, π]$
$[\frac{π}{2}, \frac{3 π}{4}]$

182.

If xy = e^{y - x}, then $\frac{dy}{dx}$ =

$\frac{y (x - 1)}{x (y + 1)}$
$\frac{y (x + 1)}{x (y - 1)}$
$\frac{x + 1}{y - 1}$
$\frac{x - 1}{y + 1}$

183.

If y = $\cos^{- 1} (\frac{\sin (x) + \cos (x)}{\sqrt{2}}), - \frac{π}{4} < x < \frac{π}{4}$ , then $\frac{dy}{dx}$ =

- 1
1
0
None of these

184.

$\lim_{x \to \infty} {(\frac{x + a}{x + b})}^{a + b}$ is equal to

1
1^{b - a}
e^{a - b}
e^b

185.

$\lim_{x \to 0} (\frac{x . 10^{x} - x}{1 - \cos (x)})$ is equal to

$\log (10)$
$2 \log (10)$
$3 \log (10)$
$4 \log (10)$

186.

If y_k is the kth derivative of y with respect to 'x', and y = cos(sin(x)), then $y_{1} \sin (x) + y_{2} \cos (x)$ is equal to

${ysin}^{3} (x)$
$- {ysin}^{3} (x)$
${ycos}^{3} (x)$
$- {ycos}^{3} (x)$

187.

$\lim_{x \to 0} \frac{\sin (x) \sin^{- 1} (x)}{x^{2}}$ is equal to

0
1
- 1
$\infty$

188.
$\lim_{x \to 0} \frac{4^{x} - 9^{x}}{x (4^{x} + 9^{x})} is equal to$

$\log (\frac{2}{3})$

$\log (\frac{3}{2})$

$\frac{1}{2} \log (\frac{2}{3})$

$\frac{1}{2} \log (\frac{3}{2})$

$\log (\frac{2}{3})$

$We have, \lim_{x \to 0} \frac{4^{x} - 9^{x}}{x (4^{x} + 9^{x})} Using L - hospital' s rule = \lim_{x \to 0} \frac{4^{x} \log (4) - 9^{x} \log (9)}{(4^{x} + 9^{x}) + x (4^{x} \log (4) + 9^{x} \log (9))} = \frac{\log (4) - \log (9)}{2} = \frac{2 \log (\frac{2}{3})}{2} = \log (\frac{2}{3})$