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1

### AIEEE 2006

The number of values of $$x$$ in the interval $$\left[ {0,3\pi } \right]\,$$ satisfying the equation $$2{\sin ^2}x + 5\sin x - 3 = 0$$ is
A
4
B
6
C
1
D
2

## Explanation $$2{\sin ^2}x + 5\sin x - 3 = 0$$

$$\Rightarrow \left( {\sin x + 3} \right)\left( {2\sin x - 1} \right) = 0$$

$$\sin x = {1 \over 2}$$ and $$\,\,\sin x \ne - 3$$

Given that $$x \in \left[ {0,3\pi } \right]$$

So possible values of x are $$30^\circ$$, $$150^\circ$$, $$390^\circ$$, $$510^\circ$$. That means x have 4 values.
2

### AIEEE 2006

If $$0 < x < \pi$$ and $$\cos x + \sin x = {1 \over 2},$$ then $$\tan x$$ is
A
$${{\left( {1 - \sqrt 7 } \right)} \over 4}$$
B
$${{\left( {4 - \sqrt 7 } \right)} \over 3}$$
C
$$- {{\left( {4 + \sqrt 7 } \right)} \over 3}$$
D
$${{\left( {1 + \sqrt 7 } \right)} \over 4}$$

## Explanation

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

$$\Rightarrow {\left( {\cos x + {\mathop{\rm sinx}\nolimits} } \right)^2} = {1 \over 4}$$

$$\Rightarrow {\cos ^2}x + {\sin ^2}x + 2\cos x\sin x = {1 \over 4}$$
$$\left[ \because {{{\cos }^2}x + {{\sin }^2}x = 1\, \,and \,\,2\cos x\sin x = \sin 2x} \right]$$

$$\Rightarrow 1 + \sin 2x = {1 \over 4}$$

$$\Rightarrow \sin 2x = - {3 \over 4},$$ so $$x$$ is obtuse and

$${{2\tan x} \over {1 + {{\tan }^2}x}} = - {3 \over 4}$$

$$\Rightarrow 3{\tan ^2}x + 8\tan x + 3 = 0$$

$$\therefore$$ $$\tan x = {{ - 8 \pm \sqrt {64 - 36} } \over 6}$$

$$= {{ - 4 \pm \sqrt 7 } \over 3}$$

as $$\tan x < 0\,$$

$$\therefore$$ $$\tan x = {{ - 4 - \sqrt 7 } \over 3}$$
3

### AIEEE 2004

A line makes the same angle $$\theta$$, with each of the $$x$$ and $$z$$ axis.
If the angle $$\beta \,$$, which it makes with y-axis, is such that $$\,{\sin ^2}\beta = 3{\sin ^2}\theta ,$$ then $${\cos ^2}\theta$$ equals
A
$${2 \over 5}$$
B
$${1 \over 5}$$
C
$${3 \over 5}$$
D
$${2 \over 3}$$

## Explanation

Concept : If a line makes the angle $$\alpha ,\beta ,\gamma$$ with x, y, z axis respectively then $${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1$$\$

In this question given that the line makes angle θ with x and z-axis and β with y−axis.

$$\therefore\: cos^2\theta+cos^2\beta+cos^2\theta=1$$

$$\Rightarrow\:2cos^2\theta=1-cos^2\beta$$

$$\Rightarrow 2{\cos ^2}\theta = {\sin ^2}\beta$$

But given that $$sin^2\beta=3sin^2\theta$$

$$\therefore$$ $$2{\cos ^2}\theta = 3{\sin ^2}\theta$$

$$\Rightarrow 2{\cos ^2}\theta = 3\left( {1 - {{\cos }^2}\theta } \right)$$

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

$$\Rightarrow 5{\cos ^2}\theta = 3$$

$$\Rightarrow {\cos ^2}\theta = {3 \over 5}$$
4

### AIEEE 2004

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

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

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

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

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