1

### JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
Let $$S$$ be the set of all non-zero real numbers $$\alpha$$ such that the quadratic equation $$\alpha {x^2} - x + \alpha = 0$$ has two distinct real roots $${x_1}$$ and $${x_2}$$ satisfying the inequality $$\left| {{x_1} - {x_2}} \right| < 1.$$ Which of the following intervals is (are) $$a$$ subject(s) os $$S$$?
A
$$\left( { - {1 \over 2} - {1 \over {\sqrt 5 }}} \right)$$
B
$$\left( { - {1 \over {\sqrt 5 }},0} \right)$$
C
$$\left( {0,{1 \over {\sqrt 5 }}} \right)$$
D
$$\left( {{1 \over {\sqrt 5 }},{1 \over 2}} \right)$$

## Explanation

Given, x1 and x2 are roots of

$$\alpha {x^2} - x + \alpha = 0$$

$$\therefore$$ $${x_1} + {x_2} = {1 \over \alpha }$$ and $${x_1}{x_2} = 1$$

Also, $$\left| {{x_1} - {x_2}} \right| < 1$$

$$\Rightarrow {\left| {{x_1} - {x_2}} \right|^2} < 1 \Rightarrow {({x_1} - {x_2})^2} < 1$$

or, $${({x_1} + {x_2})^2} - 4{x_1}{x_2} < 1$$

$$\Rightarrow {1 \over {{\alpha ^2}}} - 4 < 1$$ or $${1 \over {{\alpha ^2}}} < 5$$

$$\Rightarrow 5{\alpha ^2} - 1 > 0$$

or, $$(\sqrt 5 \alpha - 1)(\sqrt 5 \alpha + 1) > 0$$

$$\therefore$$ $$\alpha \in \left( { - \infty , - {1 \over {\sqrt 5 }}} \right) \cup \left( {{1 \over {\sqrt 5 }},\infty } \right)$$ .....(i)

Also, $$D > 0$$

$$\Rightarrow 1 - 4{\alpha ^2} > 0$$ or $$\alpha \in \left( { - {1 \over 2},{1 \over 2}} \right)$$ ...... (ii)

$$\alpha \in \left( { - {1 \over 2},{{ - 1} \over {\sqrt 5 }}} \right) \cup \left( {{1 \over {\sqrt 5 }},{1 \over 2}} \right)$$

2

### JEE Advanced 2013 Paper 2 Offline

MCQ (More than One Correct Answer)
If $${3^x}\, = \,{4^{x - 1}},$$ then $$x\, =$$
A
$${{2{{\log }_3}\,2} \over {2{{\log }_3}\,2 - 1}}$$
B
$${2 \over {2 - {{\log }_2}\,3}}$$
C
$${1 \over {1 - {{\log }_4}\,3}}$$
D
$${{2{{\log }_2}\,3} \over {2{{\log }_2}\,3 - 1}}$$
3

### IIT-JEE 1989

MCQ (More than One Correct Answer)
Let a, b, c be real numbers, $$a \ne 0$$. If $$\alpha \,$$ is a root of $${a^2}{x^2} + bx + c = 0$$. $$\beta \,$$ is the root of $${a^2}{x^2} - bx - c = 0$$ and $$0 < \alpha \, < \,\beta$$, then the equation $${a^2}{x^2} + 2bx + 2c = 0$$ has a root $$\gamma$$ that always satisfies
A
$$\gamma = {{\alpha + \beta } \over 2}$$
B
$$\gamma = \alpha + {\beta \over 2}$$
C
$$\gamma = \alpha$$
D
$$\alpha < \gamma < \beta$$
4

### IIT-JEE 1989

MCQ (More than One Correct Answer)
If $$\alpha$$ and $$\beta$$ are the roots of $${x^2}$$+ px + q = 0 and $${\alpha ^4},{\beta ^4}$$ are the roots of $$\,{x^2} - rx + s = 0$$, then the equation $${x^2} - 4qx + 2{q^2} - r = 0$$ has always
A
two real roots
B
two positive roots
C
two negative roots
D
one positive and one negative root.

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