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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

MCQ (Single Correct Answer)
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 $$sin^2\beta=3sin^2\theta$$

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

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

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

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

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

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

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

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

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

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

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

Questions Asked from Trigonometric Functions & Equations

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