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JEE Mathematics Solved Question Paper 2008

Multiple Choice Questions

The inverse matrix of $[\begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix}]$ , is

$[\begin{matrix} \frac{1}{2} & - \frac{1}{2} & \frac{1}{2} \\ - 4 & 3 & - 1 \\ \frac{5}{2} & - \frac{3}{2} & \frac{1}{2} \end{matrix}]$
$[\begin{matrix} \frac{1}{2} & - 4 & \frac{5}{2} \\ 1 & - 6 & 3 \\ 1 & 2 & - 1 \end{matrix}]$
$\frac{1}{2} [\begin{matrix} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 4 & 2 & 3 \end{matrix}]$
$\frac{1}{2} [\begin{matrix} 1 & - 1 & - 1 \\ - 8 & 6 & - 2 \\ 5 & - 3 & 1 \end{matrix}]$

If $f (x) = \{xsin (\frac{1}{x}), x \neq 0 k, x = 0\}$ is continuous at x = 0, then the value of k is

3.
If x = $\frac{1 - t^{2}}{1 + t^{2}}$ and y = $\frac{2 at}{1 + t^{2}}$ , then $\frac{dy}{dx}$ is equal to

$\frac{a (1 - t^{2})}{2 t}$

$\frac{a (t^{2} - 1)}{2 t}$

$\frac{a (t^{2} + 1)}{2 t}$

$\frac{a (t^{2} - 1)}{t}$

$\frac{a (t^{2} - 1)}{2 t}$

$Given x = \frac{1 - t^{2}}{1 + t^{2}} and y = \frac{2 at}{1 + t^{2}} On differentiating w . r . t . t, respectively, we get \frac{dx}{dt} = \frac{(1 + t^{2}) (0 - 2 t) - (1 - t^{2}) (0 + 2 t)}{{(1 + t^{2})}^{2}} = \frac{- 4 t}{{(1 + t^{2})}^{2}} and \frac{dy}{dt} = \frac{(1 + t^{2}) 2 a - 2 at (2 t)}{{(1 + t^{2})}^{2}} = \frac{2 a (1 - t^{2})}{{(1 + t^{2})}^{2}} ∴ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a (1 - t^{2})}{- 2 t} \Rightarrow \frac{dy}{dx} = \frac{a (t^{2} - 1)}{2 t}$

The velocity of a particle at time t is given by the relation v = 6t - $\frac{t^{2}}{6}$ . The distance traveled in 3 s is, if s = 0 at t = 0

$\frac{39}{2}$
$\frac{57}{2}$
$\frac{51}{2}$
$\frac{33}{2}$

The maximum value of function x³ - 12x² + 36x + 17 in the interval [1, 10] is

17
177
77
None of these

The value of $∆ \log (f (x)) + ∆^{2} (3^{x})$ is

$\log [1 + \frac{∆ f (x)}{f (x)}] + 4 . 3^{x}$
$\log [1 + \frac{∆ f (x)}{f (x)}] + 3^{x}$
$\log [\frac{∆ f (x)}{1 + f (x)}] + 4 . 3^{x}$
$\log [\frac{∆ f (x)}{1 + f (x)}] + 3^{x}$

The solution of the equation

$[\begin{matrix} 1 & 0 & 1 \\ - 1 & 1 & 0 \\ 0 & - 1 & 1 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}]$ is (x, y, z) =

(1, 1, 1)
(0, - 1, 2)
(- 1, 2, 2)
(- 1, 0, 2)

The abscissae of the points, where the tangent is to curve y = x³ - 3x² - 9x + 5 is parallel to x-axis, are

x = 0 and 0
x = 1 and - 1
x = 1 and - 3
x = - 1 and 3

If $\vec{a} . \hat{i} = 4, then (\vec{a} \times \hat{j}) . (2 \hat{j} - 3 \hat{k})$ is equal to

12
2
0
- 12

10.

The line joining the points $6 \vec{a} - 4 \vec{b} + 4 \vec{c}, - 4 \vec{c}$ and the line joining the points $- \vec{a} - 2 \vec{b} - 3 \vec{c}, \vec{a} + 2 \vec{b} - 5 \vec{c}$ intersect at