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JEE Mathematics Solved Question Paper 2017

Multiple Choice Questions

1.
Let $∆ = |\begin{matrix} Ax & x^{2} & 1 \\ By & y^{2} & 1 \\ Cz & z^{2} & 1 \end{matrix}| and ∆_{1} = |\begin{matrix} A & B & C \\ x^{2} & y^{2} & z^{2} \\ 1 & 1 & 1 \end{matrix}|, then |\begin{matrix} Ax & By & Cz \\ x^{2} & y^{2} & z^{2} \\ 1 & 1 & 1 \end{matrix}|$

$∆_{1} = 2 ∆$

$∆_{1} = - ∆$

$∆_{1} = ∆$

$∆_{1} \neq ∆$

$∆_{1} = ∆$

$We have, ∆ = |\begin{matrix} Ax & x^{2} & 1 \\ By & y^{2} & 1 \\ Cz & z^{2} & 1 \end{matrix}| and ∆_{1} = |\begin{matrix} A & B & C \\ x^{2} & y^{2} & z^{2} \\ 1 & 1 & 1 \end{matrix}| Now, ∆_{1} = |\begin{matrix} A & B & C \\ x^{2} & y^{2} & z^{2} \\ 1 & 1 & 1 \end{matrix}| On applying C_{1} \to {xC}_{1}, C_{2} \to {yC}_{2}, C_{3} \to {yC}_{3}, we get = \frac{1}{xyz} |\begin{matrix} Ax & By & Cz \\ x^{2} & y^{2} & z^{2} \\ xyz & xyz & xyz \end{matrix}| Takmg common xyz from R_{3}, we get = \frac{xyz}{xyz} |\begin{matrix} Ax & By & Cz \\ x^{2} & y^{2} & z^{2} \\ 1 & 1 & 1 \end{matrix}| = |\begin{matrix} Ax & By & Cz \\ x^{2} & y^{2} & z^{2} \\ 1 & 1 & 1 \end{matrix}| = |\begin{matrix} Ax & x^{2} & 1 \\ By & y^{2} & 1 \\ Cz & z^{2} & 1 \end{matrix}| [∵ |A'| = |A|] ∴ ∆_{1} = ∆$

The function f(x) = x² + 2x - 5 is strictly increasing in the interval

$(- \infty, - 1)$
$[- 1, \infty)$
$(- \infty, 1]$
$(- 1, \infty)$

The point on the curve y² = x where the tangent makes an angle of $π$ /4 with X-axis is

(4, 2)
$(\frac{1}{2}, \frac{1}{4})$
$(\frac{1}{4}, \frac{1}{2})$
(1, 1)

If $|\begin{matrix} 3 & x \\ x & 1 \end{matrix}| = |\begin{matrix} 3 & 2 \\ 4 & 1 \end{matrix}|$ , then x is equal to

4
8
2
$\pm 2 \sqrt{2}$

If $2 |\begin{matrix} 1 & 3 \\ 0 & x \end{matrix}| + |\begin{matrix} y & 0 \\ 1 & 2 \end{matrix}| = |\begin{matrix} 5 & 6 \\ 1 & 8 \end{matrix}|$ , then the value of x and y are

x = 3, y = 3
x = - 3, y = 3
x = 3, y = - 3
x = - 3, y = - 3

The range of $\sec^{- 1} (x)$ is

$[0, π] - \{\frac{π}{2}\}$
$[- \frac{π}{2}, \frac{π}{2}]$
$\{- \frac{π}{2}, \frac{π}{2}\}$
$[0, π]$

If y = $|\begin{matrix} f (x) & g (x) & h (x) \\ l & m & n \\ a & b & c \end{matrix}|$ , then $\frac{dy}{dx}$ is equal to

$|\begin{matrix} f' (x) & g' (x) & h' (x) \\ l & m & n \\ a & b & c \end{matrix}|$
$|\begin{matrix} l & m & n \\ f (x) & g (x) & h (x) \\ a & b & c \end{matrix}|$
$|\begin{matrix} f' (x) & l & a \\ g' (x) & m & b \\ h' (x) & n & c \end{matrix}|$
$|\begin{matrix} l & m & n \\ a & b & c \\ f' (x) & g' (x) & h' (x) \end{matrix}|$

If A is a square matrix of order 3 x 3, then $|KA|$ is equal to

$k^{2} |A|$
$K |A|$
$3 K |A|$
$k^{3} |A|$

If A = $\frac{1}{π} |\begin{matrix} \sin^{- 1} (πx) & \tan^{- 1} (\frac{x}{π}) \\ \sin^{- 1} (\frac{x}{π}) & \cot^{- 1} (πx) \end{matrix}|, B = |\begin{matrix} - \cos^{- 1} (πx) & \tan^{- 1} (\frac{x}{π}) \\ \sin^{- 1} (\frac{x}{π}) & - \tan^{- 1} (πx) \end{matrix}|$ , then A - B is