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JEE Mathematics Solved Question Paper 2016

Multiple Choice Questions

In how many ways can 5 keys be put in a ring ?

4!
5!
$\frac{4!}{2}$
$\frac{5!}{2}$

The value of ${(\sqrt{5} + 1)}^{5} - {(\sqrt{5} - 1)}^{5}$ is

The sum of the series $\frac{1}{2 . 3} + \frac{1}{4 . 5} + \frac{1}{6 . 7} + . . .$ is

$\log (\frac{e}{2})$
$\log (\frac{2}{e})$
$\frac{e}{2}$
$\frac{2}{e}$

The maximum value of $3 \cos (θ) + 4 \sin (θ)$ is

3
4
5
None of these

The period of $\sin (θ) - \sqrt{3} \cos (θ)$ is

$\frac{π}{4}$
$\frac{π}{2}$
$π$
$2 π$

If the tangent at the point $(2 \sec (θ), 3 \tan (θ))$ of the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$ is parallel to 3x - y + 4 = 0, then the value of $θ$ is

$\frac{π}{4}$
$\frac{π}{3}$
$\frac{π}{6}$
$\frac{π}{2}$

If $\cos (θ) = \frac{\cos^{2} (α) + \cos^{2} (β) + \cos^{2} (γ)}{\sin^{2} (α) + \sin^{2} (β) + \sin (γ)}$ , where $α, β, γ$ are the angles made by a line with the positive directions of the axes of reference, then the measure of $θ$ is

$60 °$
$90 °$
$30 °$
$45 °$

8.
The value of $\lim_{x \to 0} \frac{\sin (x^{4}) - x^{4} \cos (x^{4}) + x^{20}}{x^{4} (e^{2 x^{4}} - 1 - 2 x^{4})}$ is equal to

0

$- \frac{1}{6}$

$\frac{1}{6}$

Does not exist

$\frac{1}{6}$

$\lim_{x \to 0} \frac{\sin (x^{4}) - x^{4} \cos (x^{4}) + x^{20}}{x^{4} (e^{2 x^{4}} - 1 - 2 x^{4})} Using L' Hospital Rule = \lim_{x \to 0} \frac{4 x^{7} \sin (x^{4}) + 20 x^{19}}{20 e^{2 x^{4}} x^{7} - 16 x^{7} + 4 e^{2 x^{4}} - 4 x^{3}} = \lim_{x \to 0} \frac{4 x^{3} (x^{4} \sin (x^{4}) + 5 x^{16})}{4 x^{3} (2 e^{2 x^{4}} x^{4} - 4 x^{4} + e^{2 x^{4}} - 1)} Usin L' Hospital' s rule = \lim_{x \to 0} \frac{4 x^{3} \sin (x^{4}) + 4 x^{7} \cos (x^{4}) + 80 x^{15}}{16 e^{2 x^{4}} x^{7} + 16 e^{2 x^{4}} x^{3} - 16 x^{3}} Using L' Hospital Rule = \lim_{x \to 0} \frac{4 x^{3} \cos (x^{4}) + 4 x^{3} \cos (x^{4}) - 4 x^{7} \sin (x^{4}) + 240 x^{11}}{32 e^{2 x^{4}} x^{7} + 48 e^{2 x^{4}} . x^{3}} = \lim_{x \to 0} \frac{4 x^{3} (2 \cos (x^{4}) - x^{4} \sin (x^{4}) + 60 x^{11})}{4 x^{3} (8 e^{2 x^{4}} . x^{4} + 12 e^{2 x^{4}})} = \lim_{x \to 0} \frac{2 \cos (x^{4}) - x^{4} \sin (x^{4}) + 60 x^{11}}{e^{2 x^{4}} (8 x^{4} + 12)} = \frac{2 \cos (0^{4}) - {(0)}^{4} \sin (0^{4}) + 60 {(0)}^{11}}{e^{2 {(0)}^{2}} [8 {(0)}^{4} + 12]} = \frac{2}{12} = \frac{1}{6}$