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JEE Mathematics Solved Question Paper 2011

Multiple Choice Questions

The multiplicative inverse of A = $[\begin{matrix} \cos (θ) & - \sin (θ) \\ \sin (θ) & \cos (θ) \end{matrix}]$ is

$[\begin{matrix} - \cos (θ) & \sin (θ) \\ - \sin (θ) & - \cos (θ) \end{matrix}]$
$[\begin{matrix} \cos (θ) & \sin (θ) \\ - \sin (θ) & \cos (θ) \end{matrix}]$
$[\begin{matrix} - \cos (θ) & - \sin (θ) \\ \sin (θ) & - \cos (θ) \end{matrix}]$
$[\begin{matrix} \cos (θ) & \sin (θ) \\ \sin (θ) & - \cos (θ) \end{matrix}]$

The value of 'a' for which the system of equations

a³x + (a + 1)³y + (a + 2)³z = 0

ax + (a - 1)y + (a + 2)z = 0

x + y + z = 0

1
0
- 1
None of these

3.
If f(x) = $\{{(1 + |\sin (x)|)}^{\frac{a}{|\sin (x)|}}, - \frac{π}{6} < x < 0 e^{\frac{\tan (2 x)}{\tan (3 x)}}, 0 < x < \frac{π}{6}\}$ is continuous at x = 0, find the values of a and b.

3/2, e^3/2

- 2/3, e^{- 3/2}

2/3, e^2/3

None of these

2/3, e^2/3

$We have, \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0} {\{1 + |\sin (x)|\}}^{\frac{a}{|\sin (x)|}} = e^{\lim_{x \to 0} |\sin (x)| . \frac{a}{|\sin (x)|}} = e^{a} and \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} e^{\frac{\tan (2 x)}{\tan (3 x)}} = e^{\lim_{x \to 0} \frac{\tan (2 x)}{2 x} . \frac{3 x}{\tan (3 x)} \times \frac{2}{3}} = e^{\frac{2}{3}}$

For f(x) to be continuous at x = 0, we must have

$\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0) \Rightarrow e^{a} = e^{\frac{2}{3}} = b \Rightarrow a = 2 / 3 and b = e^{\frac{2}{3}}$

If f(x) = e^xg(x), g(0) = 2, g'(0) = 1 then f'(0) is

At the point x = 1, the function

$f (x) = \{x^{3} - 1, 1 < x < \infty x - 1, - \infty < x \leq 1\}$

continuous and differentiable
continuous and not differentiable
discontinuous and differentiable
discontinuous and not differentiable

If x = 2cos(t) - cos(2t), y = 2sin(t) - sin(2t), then the value of ${\frac{d^{2} y}{{dx}^{2}}}_{t = \frac{π}{2}}$

3/2
5/2
5/3
- 3/2

y = $\log (\tan (\frac{x}{2})) + \sin^{- 1} (\cos (x))$ , then dy/dx is

csc(x) - 1
cs(x)
csc(x) + 1
x

For all real x, the minimum value of $\frac{1 - x + x^{2}}{1 + x + x^{2}}$ is

If x + y = k is normal to y² = 12x, then k is

10.

A particle moves along a straight line according to the law s = 16 - 2t + 3t³, where s metres is the distance of the particle from a fixed point at the end of t second. The acceleration of the particle at the end of 2s is