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1

WB JEE 2010

MCQ (Single Correct Answer)

The value of $${{\sin 55^\circ - \cos 55^\circ } \over {\sin 10^\circ }}$$ is

A
$${1 \over {\sqrt 2 }}$$
B
2
C
1
D
$$\sqrt 2 $$

Explanation

$${{\sin 55^\circ - \cos (90^\circ - 35^\circ )} \over {\sin 10^\circ }} = {{\sin 55^\circ - \cos 35^\circ } \over {\sin 10^\circ }}$$

$$ = {{2\cos 45^\circ \sin 10^\circ } \over {\sin 10^\circ }} = 2{1 \over {\sqrt 2 }} = \sqrt 2 $$

2

WB JEE 2009

MCQ (Single Correct Answer)

If $$5\cos 2\theta + 2{\cos ^2}\theta /2 + 1 = 0$$, when $$(0 < \theta < \pi )$$, then the values of $$\theta$$ are

A
$${\pi \over 3} \pm \pi $$
B
$${\pi \over 3},{\cos ^{ - 1}}(3/5)$$
C
$${\cos ^{ - 1}}(3/5) \pm \pi $$
D
$${\pi \over 3},\pi - {\cos ^{ - 1}}(3/5)$$

Explanation

$$5\cos 2\theta + 2{\cos ^2}{\theta \over 2} + 1 = 0$$

$$ \Rightarrow 5(2{\cos ^2}\theta - 1) + (1 + \cos \theta ) + 1 = 0$$

$$ \Rightarrow 10{\cos ^2}\theta + \cos \theta - 3 = 0$$

$$ \Rightarrow \cos \theta = {1 \over 2}$$ or $$ - {3 \over 5}$$

If $$\cos \theta = {1 \over 2} \Rightarrow \theta = {\pi \over 3}$$,

If $$\cos \theta = - {3 \over 5} \Rightarrow \theta = {\cos ^{ - 1}}\left( { - {3 \over 5}} \right) = \pi - {\cos ^{ - 1}}{3 \over 5}$$

3

WB JEE 2009

MCQ (Single Correct Answer)

Simplest form of $${2 \over {\sqrt {2 + \sqrt {2 + \sqrt {2 + 2\cos 4x} } } }}$$ is

A
sec x/2
B
sec x
C
cosec x
D
1

Explanation

$${2 \over {\sqrt {2 + \sqrt {2 + \sqrt {2 + 2\cos 4x} } } }}$$

$$ = {2 \over {\sqrt {2 + \sqrt {2 + \sqrt {2(1 + 2{{\cos }^2}2x - 1)} } } }}$$

$$ = {2 \over {\sqrt {2 + \sqrt {2(1 + 2{{\cos }^2}x - 1)} } }} = {2 \over {\sqrt {2 + 2\cos x} }}$$

$$ = {2 \over {\sqrt {2(1 + 2{{\cos }^2}{x \over 2} - 1)} }} = {1 \over {\cos {x \over 2}}} = \sec {x \over 2}$$

4

WB JEE 2009

MCQ (Single Correct Answer)

In triangle ABC, a = 2, b = 3 and sin A = 2/3, thne B is equal to

A
30$$^\circ$$
B
60$$^\circ$$
C
90$$^\circ$$
D
120$$^\circ$$

Explanation

Apply $${{\sin A} \over a} = {{\sin B} \over b}$$ (sine Rule)

$${{2/3} \over 2} = {{\sin B} \over 3} \Rightarrow \sin B = 1 = \sin 90^\circ $$

$$ \Rightarrow B = 90^\circ $$

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