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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2003

MCQ (Single Correct Answer)
The number of real solutions of the equation $${x^2} - 3\left| x \right| + 2 = 0$$ is
A
3
B
2
C
4
D
1

Explanation

$${x^2} - 3\left| x \right| + 2 = 0$$

$$ \Rightarrow {\left| x \right|^2} - 3\left| x \right| + 2 = 0$$

$$\left( {\left| x \right| - 2} \right)\left( {\left| x \right| - 1} \right) = 0$$

$$\left| x \right| = 1,2$$ or $$x = \pm 1, \pm 2$$

$$\therefore$$ No. of solution $$=4$$
2

AIEEE 2003

MCQ (Single Correct Answer)
The value of '$$a$$' for which one root of the quadratic equation $$$\left( {{a^2} - 5a + 3} \right){x^2} + \left( {3a - 1} \right)x + 2 = 0$$$
is twice as large as the other is
A
$$ - {1 \over 3}$$
B
$$ {2 \over 3}$$
C
$$ - {2 \over 3}$$
D
$$ {1 \over 3}$$

Explanation

Let the roots of given equation be $$\alpha $$ and $$2$$$$\alpha $$ then

$$\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}$$

and $$\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}$$

$$ \Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a + 3} \right)}}$$

$$\therefore$$ $$2\left[ {{1 \over 9}{{{{\left( {1 - 3a} \right)}^2}} \over {{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right]$$

$$ = {2 \over {{a^2} - 5a + 3}}$$

$${{{{\left( {1 - 3a} \right)}^2}} \over {\left( {{a^2} - 5a + 3} \right)}} = 9$$

or $$9{a^2} - 6a + 1$$

$$ = 9{a^2} - 45a + 27$$

or $$39a = 26$$ or $$a = {2 \over 3}$$
3

AIEEE 2003

MCQ (Single Correct Answer)
If the sum of the roots of the quadratic equation $$a{x^2} + bx + c = 0$$ is equal to the sum of the squares of their reciprocals, then $${a \over c},\,{b \over a}$$ and $${c \over b}$$ are in
A
Arithmetic - Geometric Progression
B
Arithmetic Progression
C
Geometric Progression
D
Harmonic Progression

Explanation

$$a{x^2} + bx + c = 0,$$ $$\alpha + \beta = {{ - b} \over a},\alpha \beta = {c \over a}$$

As for given condition, $$\alpha + \beta = {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}}$$

$$\alpha + \beta = {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} - {b \over a}$$

$$ = {{{{{b^2}} \over {{a^2}}} - {{2c} \over a}} \over {{{{c^2}} \over {{a^2}}}}}$$

On simplification $$2{a^2}c = a{b^2} + b{c^2}$$

$$ \Rightarrow {{2a} \over b} = {c \over a} + {b \over c}$$

$$ \Rightarrow {c \over a},{a \over b},{b \over c}$$ are in $$A.P.$$

$$\therefore$$ $${a \over c},{b \over a},\,\,\& \,\,$$ are in $$H.P.$$
4

AIEEE 2002

MCQ (Single Correct Answer)
If $$a,\,b,\,c$$ are distinct $$ + ve$$ real numbers and $${a^2} + {b^2} + {c^2} = 1$$ then $$ab + bc + ca$$ is
A
less than 1
B
equal to 1
C
greater than 1
D
any real no.

Explanation

As $$\,\,\,\,\,{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} > 0$$

$$ \Rightarrow 2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) > 0$$

$$ \Rightarrow 2 > 2\left( {ab + bc + ca} \right)$$

$$ \Rightarrow ab + bc + ca < 1$$

Questions Asked from Quadratic Equation and Inequalities

On those following papers in MCQ (Single Correct Answer)
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