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### AIEEE 2004

MCQ (Single Correct Answer)
Let $$\alpha ,\,\beta$$ be such that $$\pi < \alpha - \beta < 3\pi$$.
If $$sin{\mkern 1mu} \alpha + \sin \beta = - {{21} \over {65}}$$ and $$\cos \alpha + \cos \beta = - {{27} \over {65}}$$ then the value of $$\cos {{\alpha - \beta } \over 2}$$
A
$${{ - 6} \over {65}}\,\,$$
B
$${3 \over {\sqrt {130} }}$$
C
$${6 \over {65}}$$
D
$$- {3 \over {\sqrt {130} }}$$

## Explanation

Given $$sin{\mkern 1mu} \alpha + \sin \beta = - {{21} \over {65}}$$ .........(1)

and $$\cos \alpha + \cos \beta = - {{27} \over {65}}$$ ........(2)

Square and add (1) and (2) you will get

$$2\left( {1 + \cos \alpha \cos \beta + \sin \alpha \sin \beta } \right)$$$$= {{{{\left( {21} \right)}^2} + {{\left( {27} \right)}^2}} \over {{{\left( {65} \right)}^2}}}$$

$$\Rightarrow$$ $$2\left( {1 + \cos \left( {\alpha - \beta } \right)} \right) = {{1170} \over {{{\left( {65} \right)}^2}}}$$

$$\Rightarrow$$ $$4{\cos ^2}{{\alpha - \beta } \over 2}$$$$= {{1170} \over {{{\left( {65} \right)}^2}}}$$

$$\Rightarrow$$ $${\cos ^2}{{\alpha - \beta } \over 2}$$$$= {9 \over {130}}$$

$$\therefore$$ $$\cos {{\alpha - \beta } \over 2} = \pm {3 \over {\sqrt {130} }}$$

[ But $$\cos {{\alpha - \beta } \over 2} \ne + {3 \over {\sqrt {130} }}$$

as $$\pi < \alpha - \beta < 3\pi$$

$$\Rightarrow$$ $${\pi \over 2} < {{\alpha - \beta } \over 2} < {{3\pi } \over 2}$$

$$\Rightarrow$$ $$\cos {{\alpha - \beta } \over 2} < 0$$ ]

So $$\cos {{\alpha - \beta } \over 2} = - {3 \over {\sqrt {130} }}$$
2

### AIEEE 2004

MCQ (Single Correct Answer)
If $$u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta }$$
then the difference between the maximum and minimum values of $${u^2}$$ is given by
A
$${\left( {a - b} \right)^2}$$
B
$$2\sqrt {{a^2} + {b^2}}$$
C
$${\left( {a + b} \right)^2}$$
D
$$2\left( {{a^2} + {b^2}} \right)$$

## Explanation

Given $$u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta }$$$$+ \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta }$$

$$\therefore$$ $${u^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta + {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta$$
$$+ 2\sqrt {\left( {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } \right)} \times \sqrt {\left( {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } \right)}$$

$$\Rightarrow {u^2} =$$ $${a^2} + {b^2}$$$$+$$$$2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {{{\cos }^4}\theta + {{\sin }^4}\theta } \right)}$$

$$\Rightarrow {u^2} =$$ $${a^2} + {b^2}$$$$+$$$$2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {1 - 2{{\cos }^2}\theta {{\sin }^2}\theta } \right)}$$

$$\Rightarrow {u^2} =$$ $${a^2} + {b^2}$$$$+$$$$2\sqrt {\left( {{a^4} + {b^4} - 2{a^2}{b^2}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}}$$

$$\Rightarrow {u^2} =$$ $${a^2} + {b^2}$$$$+$$$$2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\left( {2\cos \theta \sin \theta } \right)}^2}} \over 4} + {a^2}{b^2}}$$

$$\Rightarrow {u^2} =$$ $${a^2} + {b^2}$$$$+$$$$2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}}$$

We know $$0 \le {\sin ^2}2\theta \le 1$$

$$\therefore$$ $$0 \le {\left( {{a^2} - {b^2}} \right)^2}{{{{\sin }^2}2\theta } \over 4} \le {{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4}$$

$$\Rightarrow$$ $${a^2}{b^2} \le$$ $${\left( {{a^2} - {b^2}} \right)^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}$$$$\le {{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4} + {a^2}{b^2}$$

$$\therefore$$ Min value of $${u^2} = {a^2} + {b^2}$$ $$+ 2\sqrt {{a^2}{b^2}}$$ = $${\left( {a + b} \right)^2}$$

and Max value of $${u^2} = {a^2} + {b^2}$$ $$+ 2\sqrt {{{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4} + {a^2}{b^2}}$$

$$= {a^2} + {b^2}$$ $$+ 2\sqrt {{{{{\left( {{a^2} - {b^2}} \right)}^2} + 4{a^2}{b^2}} \over 4}}$$

$$= {a^2} + {b^2}$$ $$+ 2\sqrt {{{{{\left( {{a^2} + {b^2}} \right)}^2}} \over 4}}$$

$$= {a^2} + {b^2}$$ $$+\, {a^2} + {b^2}$$

$${ = 2\left( {{a^2} + {b^2}} \right)}$$

Max of $${u^2}$$ - Min of $${u^2}$$ = $${2\left( {{a^2} + {b^2}} \right)}$$ - $${\left( {a + b} \right)^2}$$

= $${2\left( {{a^2} + {b^2}} \right)}$$ - $${\left( {{a^2} + {b^2} + 2ab} \right)}$$

= $$\sqrt {{{ = 2\left( {{a^2} + {b^2} - 2ab} \right)} \over 4}}$$

= $${\left( {a - b} \right)^2}$$
3

### AIEEE 2002

MCQ (Single Correct Answer)
Which one is not periodic
A
$$\left| {\sin 3x} \right| + {\sin ^2}x$$
B
$$\cos \sqrt x + {\cos ^2}x$$
C
$$\cos \,4x + {\tan ^2}x$$
D
$$cos\,2x + \sin x$$

## Explanation

$$\sqrt x$$ is non periodic function and $$\cos \left( {something} \right)$$ is a periodic function so here in $$\cos \sqrt x$$ $$\to$$ inside periodic function there is non periodic function which always produce non periodic function.

$${{{\cos }^2}x}$$ is a periodic function with period $$\pi$$

Note :
(1) When $$n$$ is odd then the period of $${\sin ^n}\theta$$, $${\cos ^n}\theta$$, $${\csc ^n}\theta$$, $${\sec ^n}\theta$$ = $$2\pi$$

(2) When $$n$$ is even then the period of $${\sin ^n}\theta$$, $${\cos ^n}\theta$$, $${\csc ^n}\theta$$, $${\sec ^n}\theta$$ = $$\pi$$

(3) When $$n$$ is even/odd then the period of $${\tan ^n}\theta$$, $${\cot ^n}\theta$$ = $$\pi$$

(3) When $$n$$ is even/odd then the period of $$\left| {{{\sin }^n}\theta } \right|$$, $$\left| {{{\cos }^n}\theta } \right|$$, $$\left| {{{\csc }^n}\theta } \right|$$, $$\left| {{{\sec }^n}\theta } \right|$$, $$\left| {{{\tan }^n}\theta } \right|$$, $$\left| {{{\cot }^n}\theta } \right|$$ = $$\pi$$

$$\cos \sqrt x + {\cos ^2}x$$ = non periodic function + periodic function = non periodic function
4

### AIEEE 2002

MCQ (Single Correct Answer)
The number of solution of $$\tan \,x + \sec \,x = 2\cos \,x$$ in $$\left[ {0,\,2\,\pi } \right]$$ is
A
2
B
3
C
0
D
1

## Explanation

Given equation is $$\tan \,x + \sec \,x = 2\cos \,x$$

$$\Rightarrow$$ $${{\sin x} \over {\cos x}}$$$$+ {1 \over {\cos x}}$$ $$= 2\cos x$$

$$\Rightarrow$$ $${{\sin x + 1} \over {\cos x}} = 2\cos x$$

$$\Rightarrow$$ $${\sin x + 1}$$ $$=$$ $$2{\cos ^2}x$$

$$\Rightarrow$$ $${\sin x + 1}$$ $$= 2\left( {1 - {{\sin }^2}x} \right)$$

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

$$\Rightarrow$$ $$\left( {2\sin x - 1} \right)\left( {1 + \sin x} \right)$$$$= 0$$

$$\Rightarrow$$ $${\sin x = {1 \over 2}}$$ and $${\sin x = - 1}$$

When $${\sin x = {1 \over 2}}$$ then possible $$x$$ = $$30^\circ$$, $$150^\circ$$

When $${\sin x = - 1}$$ then possible $$x$$ = $$270^\circ$$

So three solutions possible.

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