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1

### JEE Main 2013 (Offline)

The real number $$k$$ for which the equation, $$2{x^3} + 3x + k = 0$$ has two distinct real roots in $$\left[ {0,\,1} \right]$$
A
lies between 1 and 2
B
lies between 2 and 3
C
lies between $$- 1$$ and 0
D
does not exist.

## Explanation

$$f\left( x \right) = 2{x^3} + 3x + k$$

$$f'\left( x \right) = 6{x^2} + 3 > 0$$

$$\forall x \in R$$ $$\,\,\,\,\,\,$$ (as $$\,\,\,\,\,\,$$ $${x^2} > 0$$)

$$\Rightarrow f\left( x \right)$$ is strictly increasing function

$$\Rightarrow f\left( x \right) = 0$$ has only one real root, so two roots are not possible.
2

### AIEEE 2012

The equation $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$ has:
A
infinite number of real roots
B
no real roots
C
exactly one real root
D
exactly four real roots

## Explanation

Given equation is $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$

Put $${e^{{\mathop{\rm sinx}\nolimits} \,}} = t$$ in the given equation,

we get $${t^2} - 4t - 1 = 0$$

$$\Rightarrow t = {{4 \pm \sqrt {16 + 4} } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm \sqrt {20} } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm 2\sqrt 5 } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = 2 \pm \sqrt 5$$

$$\Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5$$ $$\,\,\,\,\,$$ (as $$t = {e^{\sin x}}$$)

$$\Rightarrow {e^{\sin x}} = 2 - \sqrt 5$$ and

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $${e^{\sin x}} = 2 + \sqrt 5$$

$$\Rightarrow {e^{\sin x}} = 2 - \sqrt 5 < 0$$

and $$\,\,\,\,\,\,\sin x = \ln \left( {2 + \sqrt 5 } \right) > 1$$ So, rejected

Hence given equation has no solution.

$$\therefore$$ The equation has no real roots.
3

### AIEEE 2010

If $$\alpha$$ and $$\beta$$ are the roots of the equation $${x^2} - x + 1 = 0,$$ then $${\alpha ^{2009}} + {\beta ^{2009}} =$$
A
$$\, - 1$$
B
$$\, 1$$
C
$$\, 2$$
D
$$\, - 2$$

## Explanation

$${x^2} - x + 1 = 0$$

$$\Rightarrow x = {{1 \pm \sqrt {1 - 4} } \over 2}$$

$$x = {{1 \pm \sqrt 3 i} \over 2}$$

$$\alpha = {1 \over 2} + i{{\sqrt 3 } \over 2} = - {\omega ^2}$$

$$\beta = {1 \over 2} - {{i\sqrt 3 } \over 2} = - \omega$$

$${\alpha ^{2009}} + {\beta ^{2009}} = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}}$$

$$= - {\omega ^2} - \omega = 1$$
4

### AIEEE 2009

If the roots of the equation $$b{x^2} + cx + a = 0$$ imaginary, then for all real values of $$x$$, the expression $$3{b^2}{x^2} + 6bcx + 2{c^2}$$ is :
A
less than $$4ab$$
B
greater than $$-4ab$$
C
less than $$-4ab$$
D
greater than $$4ab$$

## Explanation

Given that roots of the equation

$$b{x^2} + cx + a = 0$$ are imaginary

$$\therefore$$ $${c^2} - 4ab < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Let $$y = 3{b^2}{x^2} + 6bc\,x + 2{c^2}$$

$$\Rightarrow 3{b^2}{x^2} + 6bc\,x + 2{c^2} - y = 0$$

As $$x$$ is real, $$D \ge 0$$

$$\Rightarrow 36{b^2}{c^2} - 12{b^2}\left( {2{c^2} - y} \right) \ge 0$$

$$\Rightarrow 12{b^2}\left( {3{c^2} - 2{c^2} + y} \right) \ge 0$$

$$\Rightarrow {c^2} + y \ge 0$$

$$\Rightarrow y \ge - {c^2}$$

But from eqn. $$(i),$$ $${c^2} < 4ab$$

or $$- {c^2} > - 4ab$$

$$\therefore$$ we get $$y \ge - {c^2} > - 4ab$$

$$y > - 4ab$$

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