Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

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

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

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

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

7.
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

1, - 1

2, - 3

- 1, 1

3, - 2

1, - 1

$Given, A = [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}] and B = [\begin{matrix} 1 & a \\ 4 & b \end{matrix}] A + B = [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}] + [\begin{matrix} 1 & a \\ 4 & b \end{matrix}] = [\begin{matrix} 2 & - 1 + a \\ 6 & - 1 + b \end{matrix}] \Rightarrow {(A + B)}^{2} = [\begin{matrix} 2 & - 1 + a \\ 6 & - 1 + b \end{matrix}] [\begin{matrix} 2 & - 1 + a \\ 6 & - 1 + b \end{matrix}] = [\begin{matrix} 4 - 6 + 6 a & - 2 + 2 a + 1 - b - a + ab \\ 12 - 6 + 6 b & - 6 + 6 a + 1 + b^{2} - 2 b \end{matrix}] = [\begin{matrix} - 2 + 6 a & - 1 + a - b + ab \\ 6 + 6 b & - 5 + 6 a - 2 b + b^{2} \end{matrix}]$

$and A^{2} = [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}] [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}] = [\begin{matrix} 1 - 2 & - 1 + 1 \\ 2 - 2 & - 2 + 1 \end{matrix}] = [\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}] Also, B^{2} = [\begin{matrix} 1 & a \\ 4 & b \end{matrix}] [\begin{matrix} 1 & a \\ 4 & b \end{matrix}] = [\begin{matrix} 1 + 4 a & a + ab \\ 4 + 4 b & 4 a + b^{2} \end{matrix}] Given, {(A + B)}^{2} = A^{2} + B^{2} ∴ [\begin{matrix} - 2 + 6 a & - 1 + a - b + ab \\ 6 + 6 b & - 5 + 6 a - 2 b + b^{2} \end{matrix}] = [\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}] + [\begin{matrix} 1 + 4 a & a + ab \\ 4 + 4 b & 4 a + b^{2} \end{matrix}] \Rightarrow [\begin{matrix} - 2 + 6 a & - 1 + a - b + ab \\ 6 + 6 b & - 5 + 6 a - 2 b + b^{2} \end{matrix}] = [\begin{matrix} 4 a & a + ab \\ 4 + 4 b & - 1 + 4 a + b^{2} \end{matrix}]$