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JEE Mathematics Solved Question Paper 2002

Multiple Choice Questions

11.

The function f(x) = 1 - x³ - x⁵ is decreasing for :

$1 \leq x \leq 5$
$x \leq 1$
$x \geq 1$
all values of x

12.

$\frac{d}{dx} (x^{x})$ is equal to :

$\log (x)$
$\log (e^{x})$
$x^{x} \log (x)$
$x^{x} \log_{e} (x)$

13.
If the displacements of a particle at time t is given by s² = at² + 2bt + c, then acceleration varies as :

$\frac{1}{s^{2}}$

$\frac{1}{s}$

$\frac{1}{s^{3}}$

s³

$\frac{1}{s^{3}}$

$Given that, s = \sqrt{{at}^{2} + 2 bt + c} On differentiating w . r . t . t, we get \frac{ds}{dt} = \frac{1}{2 \sqrt{{at}^{2} + 2 bt + c}} (2 at + 2 b) \Rightarrow \frac{ds}{dt} = \frac{at + b}{\sqrt{{at}^{2} + 2 bt + c}} = \frac{at + b}{s} Again differentiating, we get \frac{d^{2} s}{{dt}^{2}} = \frac{sa - (at + b) \frac{ds}{dt}}{s^{2}} = \frac{sa - (at + b) (\frac{at + b}{s})}{s^{2}} = \frac{s^{2} a - {(at + b)}^{2}}{s^{3}} = \frac{({at}^{2} + 2 bt + c) a - {(at + b)}^{2}}{s^{3}} = \frac{ac - b^{2}}{s^{3}} \Rightarrow Acceleration varies as \frac{1}{s^{3}} .$

14.

If PQ and PR are the two sides of a triangle, then the angle between them which gives maximum area of the triangle, is :

$π$
$\frac{π}{3}$
$\frac{π}{4}$
$\frac{π}{2}$

15.

The function y = a(1 - cos(x)) is maximum when x is equal to

$π$
$\frac{π}{2}$
$- \frac{π}{2}$
- $\frac{π}{6}$

16.

$If \tan^{- 1} (x) + \tan^{- 1} (y) + \tan^{- 1} (z) = π$ then x + y + z is :

xyz
0
1
2xyz

17.

If $θ = \sin^{- 1} [\sin (- 600 °)]$ , then one of the possible values of $θ$ is :

$\frac{π}{3}$
$\frac{π}{2}$
$\frac{2 π}{3}$
$- \frac{2 π}{3}$

18.

A force of magnitude 5 unit acting along the vector $2 \hat{i} - 2 \hat{j} + \hat{k}$ displaces the point of application from (1, 2, 3) to (5, 3, 7). Then the work done is :