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.

Continuity and Differentiability

Long Answer Type

611.

Find all points of discontinuity of f, where f is defined as following:

$f (x) = \{\begin{cases} \begin{matrix} |x| + 3, x \leq - 3 \\ - 2 x, - 3 < x < 3 \end{matrix} \\ 6 x + 2, x \geq 3 \end{cases}$

612.

$Find \frac{dy}{dx}, if y = {(cosx)}^{x} + {(sinx)}^{\frac{1}{x}}$

613.
Find the value of ‘a’ for which the function f defined as
$f (x) = \{\begin{cases} a \sin \frac{π}{2} (x + 1), x \leq 0 \\ \frac{\tan x - \sin x}{x^{3}}, x > 0 \end{cases}$
is continuous at x = 0.

$f (x) = \{\begin{cases} a \sin \frac{π}{2} (x + 1), x \leq 0 \\ \frac{\tan x - \sin x}{x^{3}}, x > 0 \end{cases} The given function f is defined for all x \in R . It is known that a function f is continuous at x = 0, if \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0) \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0} [a \sin \frac{π}{2} (x + 1)] = a \sin \frac{π}{2} = a (1) = a \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} \frac{\tan x - \sin x}{x^{3}} = \lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - \sin x}{x^{3}}$

$= \lim_{x \to 0} \frac{\sin x (1 - \cos x)}{x^{3}} = \lim_{x \to 0} \frac{\sin x . 2 \sin^{2} \frac{x}{2}}{x^{3} \cos x} = 2 \lim_{x \to 0} \frac{1}{\cos x} x \lim_{x \to 0} \frac{\sin x}{x} x \lim_{x \to 0} {[\frac{\sin \frac{x}{2}}{x}]}^{2} = 2 x 1 x 1 x \frac{1}{4} x \lim_{\frac{x}{2} \to 0} {[\frac{\sin \frac{x}{2}}{\frac{x}{2}}]}^{2} = 2 x 1 x 1 x \frac{1}{4} x 1 = \frac{1}{2} Now, f (0) = a \sin \frac{π}{2} (0 + 1) = a \sin \frac{π}{2} = a x 1 = a Since f is continuous at x = 0, a = \frac{1}{2}$