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JEE Mathematics Solved Question Paper 2005

Multiple Choice Questions

$A = [\begin{matrix} 1 & 2 \\ 3 & 2 \\ - 1 & 0 \end{matrix}], B = [\begin{matrix} 1 & 3 & 2 \\ 4 & - 1 & 3 \end{matrix}]$ , then order of AB is

2 x 2
3 x 3
1 x 3
3 x 2

$\frac{d}{dx} [\sec \{\cos^{- 1} (\frac{x}{8})\}]$ is equal to

$\frac{1}{8}$
$- \frac{1}{8}$
$\frac{8}{x^{2}}$
- $\frac{8}{x^{2}}$

3.
If f(x) = $\sqrt{1 + \cos^{2} (x^{2})}$ , then $f' (\frac{\sqrt{π}}{2})$ is

$\frac{\sqrt{π}}{6}$

$- \sqrt{\frac{π}{6}}$

$\frac{1}{\sqrt{6}}$

$\frac{π}{\sqrt{6}}$

$- \sqrt{\frac{π}{6}}$

$Given, f (x) = \sqrt{1 + \cos^{2} (x^{2})} . . . (i) On differentiating Eq . (i) w . r . t . x, we get f' (x) = \frac{- 2 \sin (x^{2}) \cos (x^{2})}{2 \sqrt{1 + \cos^{2} (x^{2})}} (2 x) \Rightarrow f' (x) = \frac{- \sin (2 x^{2})}{\sqrt{1 + \cos^{2} (x^{2})} (x)} f' (\frac{\sqrt{π}}{2}) = - \frac{\sqrt{π}}{2} . \frac{\sin (2 (\frac{π}{4}))}{\sqrt{1 + \frac{1}{2}}} = - \sqrt{\frac{π}{6}}$

If A + I = $[\begin{matrix} 3 & - 2 \\ 4 & 1 \end{matrix}]$ , then (A + I)(A - I) is equal to

$[\begin{matrix} - 5 & - 4 \\ 8 & - 9 \end{matrix}]$
$[\begin{matrix} - 5 & 4 \\ - 8 & 9 \end{matrix}]$
$[\begin{matrix} 5 & 4 \\ 8 & 9 \end{matrix}]$
$[\begin{matrix} - 5 & - 4 \\ - 8 & - 9 \end{matrix}]$

If y = ${asin}^{3} (θ)$ and x = ${acos}^{3} (θ)$ , then at $θ = \frac{π}{3}, \frac{dy}{dx}$ is equal to

$\frac{\sqrt{π}}{6}$
$- \sqrt{3}$
$\frac{- 1}{\sqrt{3}}$
$\sqrt{3}$

If x + y = 8, then maximum value of x²y is

$\frac{2048}{9}$
$\frac{2048}{81}$
$\frac{2048}{3}$
$\frac{2048}{27}$

If A = $[\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}]$ and B = $[\begin{matrix} 1 & a \\ 4 & b \end{matrix}]$ and (A + B)² = A² + B². Then, a and b are respectively