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JEE Mathematics Solved Question Paper 2004

Multiple Choice Questions

If f(x) is continuous on $[- π, π]$ , where

$f (x) = \{- 2 \sin (x), for - π \leq x \leq - \frac{π}{2} αsin (x) + β, for - \frac{π}{2} < x < \frac{π}{2} \cos (x), for \frac{π}{2} \leq x \leq π\}$

then $α and β$ are

- 1, - 1
1, - 1
1, 1
- 1, 1

2.
If f(x) = $\{\log_{(1 - 3 x)} (1 + 3 x), for x \neq 0 k, for x = 0\}$ continuous at x = 0, then k is equal to

- 2

2

1

- 1

- 1

$Given, \{\log_{(1 - 3 x)} (1 + 3 x), for x \neq 0 k, forx = 0\} Now, \lim_{x \to 0} f (x) = \lim_{x \to 0} \frac{\log (1 + 3 x)}{\log (1 - 3 x)} = - \lim_{x \to 0} \frac{\log (1 + 3 x)}{3 x} . \frac{(- 3 x)}{\log (1 - 3 x)} = - 1 and f (0) = k ∵ f (x) is continuous at x = 0 . ∴ k = - 1$

If $A = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}]$ , then A² - 5A is equal to

2I
- 2I
3I
null matrix

If A = $[\begin{matrix} 2 & 1 \\ - 1 & 2 \end{matrix}], B = [\begin{matrix} 1 & - 2 \\ 2 & 1 \end{matrix}], C = [\begin{matrix} 1 & - 3 \\ 2 & 1 \end{matrix}]$ , then

A + B = B + A and A + (B + C) = (A + B) + C
A + B = B + A and AC = BC
A + B = B + A and AB = BC
AC = BC and A = BC

$A = [\begin{matrix} - 2 & 4 \\ - 1 & 2 \end{matrix}]$ , then A² is equal to

null matrix
unit matrix
$[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$
$[\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]$

If $x = \log (1 + t^{2}) and y = t - \tan^{- 1} (t)$ . Then, $\frac{dy}{dx}$ is equal to

e^x - 1
t² - 1
$\frac{\sqrt{e^{x} - 1}}{2}$
e^x - y

If A = [x y z], B = $[\begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix}] and C = [\begin{matrix} x \\ y \\ z \end{matrix}]$ . Then, ABC = O, if