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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

$\frac{π}{4}$
$\frac{π}{2}$
$π$
$2 π$

6.
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

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{6}$

$\frac{π}{2}$

$\frac{π}{6}$

$(2 \sec (θ), 3 \tan (θ)) Given equation of the hyperbola is \frac{x^{2}}{4} - \frac{y^{2}}{9} = 1 On differentiating w . r . t x, we get \frac{2 x}{4} - \frac{2 yy'}{9} = 0 \Rightarrow y' = \frac{2 x}{4} \times \frac{9}{2 y} = \frac{9 x}{4 y} ∴ m_{1} = \frac{9 \times 2 \sec (θ)}{4 \times 3 \tan (θ)} = \frac{3}{2} \csc (θ) and given line is 3 x - y + 4 = 0 ∴ m_{2} = \frac{- 3}{- 1} = 3 Since, both the Iines are parallel ∴ m_{1} = m_{2} \Rightarrow \frac{3}{2} \csc (θ) = 3 \Rightarrow \csc (θ) = 2 \Rightarrow θ = \csc^{- 1} (2) \Rightarrow θ = \frac{π}{6}$

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