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Continuity and Differentiability

Multiple Choice Questions

971.

The value of $\lim_{x \to 1} (1 - x) . \tan (\frac{πx}{2})$ will be

$\frac{π}{2}$
$\frac{2}{π}$
$2 π$
$π$

972.

Let f(x) = $\{\frac{x^{2} - 4 x + 3}{x^{2} + 2 x - 3}, x \neq 1 k, x = 1\}$ If f(x) is continuous at x = 1, then the value of k will be

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

973.

Let f(x) = $\{x^{n} . \sin (\frac{1}{x}), x \neq 0 0, x = 0\}$ Then, f(x) is differentiable at x = 0, if

$n \in (0, 1)$
$n \in (1, 2)$
$n \in (1, \infty)$
$n \in (- \infty, \infty)$

974.

In which interval the function $f (x) = \sqrt{\log_{10} (\frac{5 x - x^{2}}{4})}$ is defined ?

[1, 4]
[0, 5)
(0, 1)
(- 1, $\infty$ )

975.

$\lim_{x \to 0} \frac{x . 2^{x} - x}{1 - \cos (x)}$ equals

$\log (2)$
$\frac{1}{2} \log (2)$
$2 \log (2)$
None of these

976.

Let f(2) = 4 and f'(2)= 1. Then, $\lim_{x \to 2} \frac{xf (2) - 2 f (x)}{x - 2}$ is given by

977.

$If y = \log_{\sin (x)} (\tan (x)), then {(\frac{dy}{dx})}_{\frac{π}{4}}$ is equal to

$\frac{4}{\log (2)}$
$- 4 \log (2)$
$- \frac{4}{\log (2)}$
None of these

978.

If y = $\log_{e^{x}} {(x - 2)}^{2}$ for $x \neq 0$ , 2, then y'(3) is equal to

$\frac{1}{3}$
$\frac{2}{3}$
$\frac{4}{3}$
None of these

979.

$\sin^{- 1} (\frac{1 + x^{2}}{2 x})$ is

continuous but not differentiable at x = 1
differentiable at x = 1
neither continuous nor differentiable at x = 1
continuous everywhere

980.
If $\sqrt{x^{2} + y^{2}} = {atan}^{- 1} (\frac{y}{x}), a > 0, then \frac{d^{2} y}{{dx}^{2}}$ at x = 0 is

0

$\frac{2}{a} e^{- \frac{π}{2}}$

$- \frac{2}{a} e^{- \frac{π}{2}}$

None of these

None of these

$\sqrt{x^{2} + y^{2}} = {atan}^{- 1} (\frac{y}{x}), a > 0 . . . (i) Putting x = 0, we get y = {atan}^{- 1} (\infty) y = \frac{aπ}{2} Now, differentiating Eq . (i) w . r . t . x, we get \frac{(2 x + 2 y \frac{dy}{dx})}{2 \sqrt{x^{2} + y^{2}}} = a . \frac{1}{1 + \frac{y^{2}}{x^{2}}} . \{\frac{x \frac{dy}{dx} - y . 1}{x^{2}}\} \Rightarrow \frac{x + y \frac{dy}{dx}}{\sqrt{x^{2} + y^{2}}} = \frac{{ax}^{2}}{x^{2} + y^{2}} \{\frac{x \frac{dy}{dx} - y}{x^{2}}\} \Rightarrow \sqrt{x^{2} + y^{2}} (x + y \frac{dy}{dx}) = a (x \frac{dy}{dx} - y) . . . (ii) At x = 0, y = \frac{aπ}{2}, \frac{aπ}{2} [\frac{aπ}{2} . \frac{dy}{dx}] = a [- \frac{aπ}{2}] \frac{π}{2} . \frac{dy}{dx} = - 1 \Rightarrow \frac{dy}{dx} = - \frac{2}{π}$

$Again, differentiating Eq . (ii), w . r . t . x, we get \frac{(2 x + 2 y \frac{dy}{dx})}{2 \sqrt{x^{2} + y^{2}}} . (x + y \frac{d y}{d x}) + \sqrt{x^{2} + y^{2}} [1 + y \frac{d^{2} y}{{dx}^{2}} + {(\frac{d y}{d x})}^{2}] = a [x \frac{d^{2} y}{{dx}^{2}} + \frac{d y}{d x} - \frac{d y}{d x}] \Rightarrow \frac{{(x + y \frac{d y}{d x})}^{2}}{\sqrt{x^{2} + y^{2}}} + \sqrt{x^{2} + y^{2}} [1 + y \frac{d^{2} y}{{dx}^{2}} + {(\frac{d y}{d x})}^{2}] = ax \frac{d^{2} y}{{dx}^{2}} Putting x = 0, y = \frac{aπ}{2}, \frac{d y}{d x} = - \frac{2}{π}$

$\frac{aπ}{2} {(- \frac{2}{π})}^{2} + \frac{aπ}{2} [1 + \frac{aπ}{2} (\frac{d^{2} y}{{dx}^{2}}) + {(- \frac{2}{π})}^{2}] = 0 \Rightarrow \frac{2 a}{π} + \frac{aπ}{2} + \frac{a^{2} π^{2}}{4} . \frac{d^{2} y}{{dx}^{2}} + \frac{4}{π^{2}} . \frac{aπ}{2} = 0 \Rightarrow \frac{4 a}{π} + \frac{aπ}{2} + \frac{a^{2} π^{2}}{4} . \frac{d^{2} y}{{dx}^{2}} = 0 \Rightarrow \frac{4}{π} + \frac{π}{2} + \frac{{aπ}^{2}}{4} . \frac{d^{2} y}{{dx}^{2}} = 0 \Rightarrow \frac{d^{2} y}{{dx}^{2}} = \frac{- \frac{4}{π} - \frac{π}{2}}{\frac{{aπ}^{2}}{4}} = \frac{- 2 (8 + π^{2})}{{aπ}^{3}}$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

980. If x2 + y2 = atan-1yx, a > 0, then d2ydx2 at x = 0 is 0 2ae- π2 - 2ae- π2 None of these

980.
If $\sqrt{x^{2} + y^{2}} = {atan}^{- 1} (\frac{y}{x}), a > 0, then \frac{d^{2} y}{{dx}^{2}}$ at x = 0 is

0

$\frac{2}{a} e^{- \frac{π}{2}}$

$- \frac{2}{a} e^{- \frac{π}{2}}$

None of these