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JEE Mathematics Solved Question Paper 2016

Multiple Choice Questions

The line y = x + $λ$ is tangent to the ellipse 2x² + 3y² = 1. Then, $λ$ is

- 2
1
$\sqrt{\frac{5}{6}}$
$\sqrt{\frac{2}{3}}$

Let S be the set of points, whose abscissae and ordinates are natural numbers. Let p e S, such that the sum of the distance of P from (8, 0) and (0, 12) is minimum among all elemants in S. Then, the number of such points P in S is

The cosine of the angle between any two diagonals of a cube is

$\frac{1}{3}$
$\frac{1}{2}$
$\frac{2}{3}$
$\frac{1}{\sqrt{3}}$

If x is a positive real number different from 1 such that $\log_{a} (x), \log_{b} (x), \log_{c} (x)$ are in AP, then

$b = \frac{a + c}{2}$
$\sqrt{ac}$
$c^{2} = {(ac)}^{\log_{a} (b)}$
None of these

If a, x are real numbers and $|a|$ < 1, $|x|$ < 1, then 1 + (1+ a) x + (1+ a + a²)x² + ... $\infty$ is equal to

$\frac{1}{(1 - a) (1 - ax)}$
$\frac{1}{(1 - a) (1 - x)}$
$\frac{1}{(1 - x) (1 - ax)}$
$\frac{1}{(1 + ax) (1 - a)}$

6.
If $\log_{0.3} (x - 1) < \log_{0.09} (x - 1)$ , then x lies in the interval

$(2, \infty)$

(1, 2)

(- 2, - 1)

None of these

$(2, \infty)$

$\log_{0.3} (x - 1) < \log_{0.09} (x - 1)$

$\Rightarrow \log_{0.3} (x - 1) < \log_{{(0.3)}^{2}} (x - 1)$

$\Rightarrow \log_{0.3} {(x - 1)}^{2} > \log_{0.3} (x - 1)$

$\Rightarrow {(x - 1)}^{2} > x - 1$ $[∵ 0.3 < 1]$

$\Rightarrow x^{2} - 2 x + 1 - x + 1 > 0$

$\Rightarrow x^{2} - 3 x + 2 > 0$

$\Rightarrow (x - 1) (x - 2) > 0$

$\Rightarrow x < 1, x > 2$

$\Rightarrow x > 2$ $[∵ x ∢ 1]$

$∴$ $(2, \infty)$

Hence, x lies in the interval (2, $\infty$ ).

The value of $\sum_{n = 1}^{13} (i^{n} + i^{n + 1}), i = \sqrt{- 1}$ is

i
i - 1
1
0

If $z_{1}, z_{2}, z_{3}$ are imaginary numbers such that $|z_{1}| = |z_{2}| = |z_{3}| = |\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}}|$ = 1, then $|z_{1} + z_{2} + z_{3}|$ is