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JEE Mathematics Solved Question Paper 2013

Multiple Choice Questions

11.

If ab > - 1, bc > - 1 and ca > - 1, then is

- 1
$\cot^{- 1} (abc)$
0

12.

If the function $f (x) = \{\frac{x^{2} - (k + 2) x + 2 k}{x - 2} for x \neq 2 2 for x = 2\}$ is continuous at x = 2, then k is equal to

$- \frac{1}{2}$
- 1
0
$\frac{1}{2}$

13.
If xe^xy + ye^{- xy} = $\sin^{2} (x)$ , then $\frac{dy}{dx}$ at x = 0 is

2y² - 1

2y

y² - y

y² - 1

y² - 1

$Given, {xe}^{xy} + {ye}^{- xy} = \sin^{2} (x) On differentiating w . r . t . x, we get e^{xy} + {xe}^{xy} \{x \frac{dy}{dx} + y\} + \frac{dy}{dx} e^{- xy} - {ye}^{- xy} \{x \frac{dy}{dx} + y\} = 2 \sin (x) . \cos (x) \Rightarrow \{x^{2} e^{xy} + e^{- xy} - {xe}^{y} e^{- xy}\} \frac{dy}{dx} + \{e^{xy} + {xye}^{xy} - y^{2} e^{- xy}\} = \sin (2 x) On putting x = 0, we get \{0 + e^{- 0} - 0\} {(\frac{dy}{dx})}_{(x = 0)} + \{e^{0} + 0 - y^{2} e^{- 0}\} = \sin (0) \Rightarrow {[(1) \frac{dy}{dx}]}_{(x = 0)} + (1 - y^{2}) = 0 \Rightarrow {[\frac{dy}{dx}]}_{x = 0} = y^{2} - 1$

14.

If y = $\tan^{- 1} (\frac{2 x - 1}{1 + x - x^{2}})$ , then $\frac{dy}{dx}$ at x = 1 is equal to

$\frac{1}{2}$
$\frac{2}{3}$
1
$\frac{3}{2}$

15.

If f(x) = $\cos^{- 1} (\frac{2 \cos (x + 3 \sin (x))}{\sqrt{13}})$ , then [f'(x)]² is equal to

$\sqrt{1 + x}$
1 + 2x
2
1

16.

If u = $\tan^{- 1} (\frac{\sqrt{1 - x^{2}} - 1}{x})$ and v = $\sin^{- 1} (x)$ , then $\frac{du}{dv}$ is equal to

$\sqrt{1 - x^{2}}$
$- \frac{1}{2}$
1
- x

17.

If y = $\frac{1}{1 + x + x^{2}}$ , then $\frac{dy}{dx}$ is equal to

y²(2 + 2x)
$\frac{- (1 + 2 x)}{y^{2}}$
$\frac{1 + 2 x}{y^{2}}$
- y²(1 + 2x)

18.

If g(x) is the inverse of f(x) and f'(x) = $\frac{1}{1 + x^{3}}$ , then g'(x) is equal to

g(x)
1 + g(x)
1 + {g(x)}³
$\frac{1}{1 + {\{g (x)\}}^{3}}$

19.

The equation of the tangent to the curve $\sqrt{\frac{x}{a}} + \sqrt{\frac{y}{b}}$ = 1 at the point (x₁, y₁) is $\frac{x}{\sqrt{{ax}_{1}}} + \frac{y}{\sqrt{{by}_{1}}} = k$ . Then, the value of k is