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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
- $\frac{1}{2}$

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

2
$\frac{2}{\sqrt{3}}$
$\frac{2}{\sqrt{5}}$
$\frac{1}{2}$

9.
The value of c in (0, 2) satisfying the mean value theorem for the function f(x) = x(x - 1)², x $\in$ [0, 2] is equal to

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{4}{3}$

Given curve is

f(x) = x(x - 1)², x $\in$ [0,2]

$\Rightarrow f' (x) = x (3 x - 4) + 1 ∴ f' (c) = c (3 c - 4) + 1 Also, f (0) = 0, f (2) = 2 By mean value theorem f (0) - f (2) = f' (c) (0 - 2) \Rightarrow 0 - 2 = (c (3 c - 4) + 1) (- 2) \Rightarrow 3 c^{2} - 4 c + 1 = 1 \Rightarrow c (3 c - 4) = 0 \Rightarrow c = 0, \frac{4}{3}$