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JEE Mathematics Solved Question Paper 2010

Multiple Choice Questions

Let f(x) = $\frac{{(e^{x} - 1)}^{2}}{\sin (\frac{x}{a}) \log (1 + \frac{x}{4})}$ for $x \neq 0$ and f(0) = 12. If f is continuous at x = 0, then the value of a is equal to

If $y = \sin^{- 1} (\sqrt{1 - x})$ , then $\frac{dy}{dx}$ is equal to

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

The derivative of $\sin^{- 1} (2 x \sqrt{1 - x^{2}})$ with respect to $\sin^{- 1} (3 x - 4 x)$ is

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

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

$\frac{x^{2} - 1}{x^{2} + 1}$
$π$
0
1

If $f (x) = |x - 2| + |x + 1| - x$ , then $f' (- 10)$ is equal to

If $x = a (1 + \cos (θ)), y = a (θ + \sin (θ))$ , then $\frac{d^{2} y}{{dx}^{2}}$ at $θ = \frac{π}{2}$ is

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

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

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

8.
The distance between the origin and the normal to the curve y = e^2x + x² at x = 0 is

2

$\frac{2}{\sqrt{3}}$

$\frac{2}{\sqrt{5}}$

$\frac{1}{2}$

$\frac{2}{\sqrt{5}}$

$Given curve is y = e^{2 x} + x^{2} ∴ x = 0 \Rightarrow y = 1 \Rightarrow \frac{dy}{dx} = 2 e^{2 x} + 2 x = 2 (e^{2 x} + x) ∴ {(\frac{dy}{dx})}_{(0, 1)} = 2 ∴ Equation of normal at (0, 1) is y - 1 = - \frac{1}{2} (x - 0) \Rightarrow 2 y + x = 2$