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1

IIT-JEE 2011 Paper 1 Offline

Numerical
The positive integer value of $$n\, > \,3$$ satisfying the equation $${1 \over {\sin \left( {{\pi \over n}} \right)}} = {1 \over {\sin \left( {{{2\pi } \over n}} \right)}} + {1 \over {\sin \left( {{{3\pi } \over n}} \right)}}$$ is
Your Input ________

Answer

Correct Answer is 7
2

IIT-JEE 2010 Paper 2 Offline

Numerical
Two parallel chords of a circle of radius 2 are at a distance $$\sqrt 3 + 1$$ apart. If the chords subtend at the center , angles of $${\pi \over k}$$ and $${{2\pi } \over k},$$ where$$k > 0,$$ then the value of $$\left[ k \right]$$ is

[Note :[k] denotes the largest integer less than or equal to k ]

Your Input ________

Answer

Correct Answer is 3
3

IIT-JEE 2010 Paper 1 Offline

Numerical
The maximum value of the expression $${1 \over {{{\sin }^2}\theta + 3\sin \theta \cos \theta + 5{{\cos }^2}\theta }}$$ is
Your Input ________

Answer

Correct Answer is 2

Explanation

Let

$$f(\theta ) = {1 \over {{{\sin }^2}\theta + 3\sin \theta \cos \theta + 5{{\cos }^2}\theta }}$$

Again let

$$g(\theta ) = {\sin ^2}\theta + 3\sin \theta \cos \theta + 5{\cos ^2}\theta $$

$$ = {{1 - \cos 2\theta } \over 2} + 5\left( {{{1 + \cos 2\theta } \over 2}} \right) + {3 \over 2}\sin 2\theta $$

$$ = 3 + 2\cos 2\theta + {3 \over 2}\sin 2\theta $$

$$\therefore$$ $$g{(\theta )_{\min }} = 3 - \sqrt {4 + {9 \over 4}} = 3 - {5 \over 2} = {1 \over 2}$$

$$\therefore$$ $$f(\theta ) = {1 \over {g{{(\theta )}_{\min }}}} = 2$$

4

IIT-JEE 2010 Paper 1 Offline

Numerical
The number of values of $$\theta $$ in the interval, $$\left( { - {\pi \over 2},\,{\pi \over 2}} \right)$$ such that$$\,\theta \ne {{n\pi } \over 5}$$ for $$n = 0,\, \pm 1,\, \pm 2$$ and $$\tan \,\theta = \cot \,5\theta \,$$ as well as $$\sin \,2\theta = \cos \,4 \theta $$ is
Your Input ________

Answer

Correct Answer is 3

Explanation

Given, $$\tan \theta = \cot 5\theta $$

$$ \Rightarrow \tan \theta = \tan \left( {{\pi \over 2} - 5\theta } \right)$$

$$ \Rightarrow {\pi \over 2} - 5\theta = n\pi + \theta $$

$$ \Rightarrow 6\theta = {\pi \over 2} - {{n\pi } \over 6}$$

$$ \Rightarrow \theta = {\pi \over {12}} - {{n\pi } \over 6}$$

Also $$\cos 4\theta = \sin 2\theta = \cos \left( {{\pi \over 2} - 2\theta } \right)$$

$$ \Rightarrow 4\theta = 2n\pi \, \pm \,\left( {{\pi \over 2} - 2\theta } \right)$$

Taking positive

$$6\theta = 2n\pi + {\pi \over 2} \Rightarrow \theta = {{n\pi } \over 3} + {\pi \over {12}}$$

Taking negative

$$2\theta = 2n\pi - {\pi \over 2} \Rightarrow \theta = n\pi - {\pi \over 4}$$

Above values of $$\theta$$ suggests that there are only 3 common solutions.

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