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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

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

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

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

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

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

6.
If x + y = 8, then maximum value of x²y is

$\frac{2048}{9}$

$\frac{2048}{81}$

$\frac{2048}{3}$

$\frac{2048}{27}$

$\frac{2048}{27}$

$Given, x + y = 8 . . . (i) Let A = x^{2} y = x^{2} (8 - x) [∵ from Eq . (i)] \Rightarrow A = 8 x^{2} - x^{3} \Rightarrow \frac{dA}{dx} = 16 x - 3 x^{2} For maxima or minima, put \frac{dA}{dx} = 0 \Rightarrow 16 x - 3 x^{2} = 0 \Rightarrow x = 0, \frac{16}{3} Now, \frac{d^{2} A}{{dx}^{2}} = 16 - 6 x {(\frac{d^{2} A}{{dx}^{2}})}_{x = \frac{16}{3}} = 16 - 32 ∴ A is Maximum at x = \frac{16}{3} .$

$∴$ Maximum value of A

$= 8 {(\frac{16}{3})}^{2} - {(\frac{16}{3})}^{3} = \frac{2048}{9} (8 - \frac{16}{3}) = \frac{256}{9} \times \frac{8}{3} = \frac{2048}{27}$

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