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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

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

2

1

3

3

Given curve is,

$\sqrt{\frac{x}{a}} + \sqrt{\frac{y}{b}} = 1 . . . (i) On differentiating w . r . t . to x, we get \frac{1}{\sqrt{a}} . \frac{1}{2 \sqrt{x}} + \frac{1}{\sqrt{b}} . \frac{1}{2 \sqrt{y}} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = - \frac{\sqrt{b} \sqrt{y}}{\sqrt{a} \sqrt{x}} \Rightarrow {[\frac{dy}{dx}]}_{(x_{1}, y_{1})} = \frac{- \sqrt{b} \sqrt{y_{1}}}{\sqrt{a} \sqrt{x_{1}}} ∴ Equation of tangent passing through the point (x_{1}, y_{1}) is (y - y_{1}) = \frac{- \sqrt{b} \sqrt{y_{1}}}{\sqrt{a} \sqrt{x_{1}}} (x - x_{1}) \frac{y}{\sqrt{{by}_{1}}} - \frac{y_{1}}{\sqrt{{by}_{1}}} = - \frac{x}{\sqrt{{ax}_{1}}} + \frac{x_{1}}{\sqrt{{ax}_{1}}} \Rightarrow \frac{x}{\sqrt{{ax}_{1}}} + \frac{y}{\sqrt{{by}_{1}}} = \frac{x_{1}}{\sqrt{{ax}_{1}}} + \frac{y_{1}}{\sqrt{{by}_{1}}} = \sqrt{\frac{x_{1}}{a}} + \sqrt{\frac{y_{1}}{b}} \Rightarrow \frac{x}{\sqrt{{ax}_{1}}} + \frac{y}{\sqrt{{by}_{1}}} = 1 [∵ from Eq . (i)] [∵ at (x_{1}, y_{1}), \sqrt{\frac{x_{1}}{a}} + \sqrt{\frac{y_{1}}{b}} = 1] But, \frac{x}{\sqrt{{ax}_{1}}} + \frac{y}{\sqrt{{by}_{1}}} = k ∴ k = 1$