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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

$\frac{1}{2}$

$Given, f (x) = \{\frac{x^{2} - (k + 2) x + 2 k}{x - 2} for x \neq 2 2 for x = 2\} Since, f (x) is continuous at x = 2 . \Rightarrow \lim_{x \to 2} \frac{x^{2} - (k + 2) x + 2 k}{x - 2} = 2 Apply L' - hospital rule, we get \lim_{x \to 2} \frac{2 x - (k + 2)}{1} = 2 \Rightarrow 2 (2) - (k + 2) = 2 \Rightarrow k + 2 = 2 ∴ k = 0$

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