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1

### JEE Main 2014 (Offline)

Let $$\alpha$$ and $$\beta$$ be the roots of equation $$p{x^2} + qx + r = 0,$$ $$p \ne 0.$$ If $$p,\,q,\,r$$ in A.P. and $${1 \over \alpha } + {1 \over \beta } = 4,$$ then the value of $$\left| {\alpha - \beta } \right|$$ is :
A
$${{\sqrt {34} } \over 9}$$
B
$${{2\sqrt 13 } \over 9}$$
C
$${{\sqrt {61} } \over 9}$$
D
$${{2\sqrt 17 } \over 9}$$

## Explanation

Let $$p,q,r$$ are in $$AP$$

$$\Rightarrow 2q = p + r\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Given $${1 \over \alpha } + {1 \over \beta } = 4 \Rightarrow {{\alpha + \beta } \over {\alpha \beta }} = 4$$

We have $$\alpha + \beta = - q/p$$ and $$\alpha \beta = {r \over p}$$

$$\Rightarrow {{ - {q \over p}} \over {{r \over p}}} = 4 \Rightarrow q = - 4r\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

From $$(i),$$ we have

$$2\left( { - 4r} \right) = p + r \Rightarrow p = - 9r$$

$$q = - 4r$$

Now $$\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta }$$

$$= \sqrt {{{\left( {{{ - q} \over p}} \right)}^2} - {{4r} \over p}}$$

$$= {{\sqrt {{q^2} - 4pr} } \over {\left| p \right|}}$$

$$= {{\sqrt {16{r^2} + 36{r^2}} } \over {\left| { - 9r} \right|}}$$

$$= {{2\sqrt {13} } \over 9}$$
2

### JEE Main 2014 (Offline)

If $$a \in R$$ and the equation $$- 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$$ (where [$$x$$] denotes the greater integer $$\le x$$) has no integral solution, then all possible values of a lie in the interval :
A
$$\left( { - 2, - 1} \right)$$
B
$$\left( { - \infty , - 2} \right) \cup \left( {2,\infty } \right)$$
C
$$\left( { - 1,0} \right) \cup \left( {0,1} \right)$$
D
$$\left( {1,2} \right)$$

## Explanation

Given, $$- 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$$
As we know, $$\left[ x \right] + \left\{ x \right\} = x$$
where $$\left[ x \right]$$ is integral part and $$\left\{ x \right\}$$ is fractional part.
$$\therefore$$$$\left\{ x \right\} = x - \left[ x \right]$$
Now put $$\left\{ x \right\}$$ inplace of $$x - \left[ x \right]$$ in the equation.
The new equation is $$- 3{\left\{ x \right\}^2} + 2\left\{ x \right\} + {a^2} = 0$$

[Note : Question says this equation has no integral solution, it means $$\left\{ x \right\} \ne$$ 0. So, $$x$$ is not a integer.]

$$\therefore$$ $$\left\{ x \right\}$$ = $${{ - 2 \pm \sqrt {4 - 4 \times \left( { - 3} \right){a^2}} } \over { - 6}}$$
= $${{ - 2 \pm \sqrt {4 + 12{a^2}} } \over { - 6}}$$
As $$\left\{ x \right\}$$ is fractional part so it is lies between 0 to 1($$0 \le \left\{ x \right\} < 1$$).

By considering positive sign, we get

$$0 \le {{ - 2 + \sqrt {4 + 12{a^2}} } \over { - 6}} < 1$$

$$\Rightarrow$$$$0 \ge - 2 + \sqrt {4 + 12{a^2}} > - 6$$

$$\Rightarrow$$$$2 \ge + \sqrt {4 + 12{a^2}} > - 4$$

$$\because$$$$+ \sqrt {4 + 12{a^2}}$$ is always positive which is greater than any negative no. So can ignore the inequality $$+ \sqrt {4 + 12{a^2}} > - 4$$

Consider this inequality,
$$2 \ge + \sqrt {4 + 12{a^2}}$$
$$\Rightarrow$$ $$4 \ge 4 + 12{a^2}$$
$$\Rightarrow$$ $$12{a^2} \le 0$$
$$\Rightarrow$$ $${a^2} \le 0$$
$$\Rightarrow$$ $${a^2} = 0$$
$$\Rightarrow$$ $${a} = 0$$

If $$a$$ = 0 then $$- 3{\left\{ x \right\}^2} + 2\left\{ x \right\} = 0$$ so $$\left\{ x \right\}$$ becomes 0 but question says $$\left\{ x \right\}$$ $$\ne$$ 0.
So $$a$$ can't be 0.

Now by considering negative sign, we get

$$0 \le {{ - 2 - \sqrt {4 + 12{a^2}} } \over { - 6}} < 1$$

$$\Rightarrow$$$$0 \ge - 2 - \sqrt {4 + 12{a^2}} > - 6$$

$$\Rightarrow$$$$2 \ge - \sqrt {4 + 12{a^2}} > - 4$$

As 2 is always greater than $${ - \sqrt {4 + 12{a^2}} }$$. Ignore this inequality.

Now consider this inequality,

$$- \sqrt {4 + 12{a^2}} > - 4$$

$$\Rightarrow$$ $$\sqrt {4 + 12{a^2}} < 4$$

$$\Rightarrow$$ $$4 + 12{a^2} < 16$$

$$\Rightarrow$$ $$12{a^2} < 12$$

$$\Rightarrow$$ $${a^2} < 1$$

$$\Rightarrow$$ $$\left( {{a^2} - 1} \right) < 0$$

$$\Rightarrow$$ $$\left( {a + 1} \right)\left( {a - 1} \right) < 0$$

$$\Rightarrow$$ $$- 1 < a < 1$$

But earlier we found that $$a$$ $$\ne$$ 0.

So, the range of $$a$$ is = $$\left( { - 1,0} \right) \cup \left( {0,1} \right)$$
3

### JEE Main 2013 (Offline)

If the equations $${x^2} + 2x + 3 = 0$$ and $$a{x^2} + bx + c = 0,$$ $$a,\,b,\,c\, \in \,R,$$ have a common root, then $$a\,:b\,:c\,$$ is
A
$$1:2:3$$
B
$$3:2:1$$
C
$$1:3:2$$
D
$$3:1:2$$

## Explanation

Given equations are

$$\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + 2x + 3 = 0\,\,\,\,\,...\left( i \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,a{x^2} + bx + c = 0\,\,\,...\left( {ii} \right)$$

Roots of equation $$(i)$$ are imaginary roots.

According to the question $$(ii)$$ will also have both roots same as $$(i).$$

Thus $${a \over 1} = {b \over 2} = {c \over 3} = \lambda \left( {say} \right)$$

$$\Rightarrow a = \lambda ,b = 2\lambda ,c = 3\lambda$$

Hence, required ratio is $$1:2:3$$
4

### JEE Main 2013 (Offline)

The number of values of $$k$$, for which the system of equations : $$\matrix{ {\left( {k + 1} \right)x + 8y = 4k} \cr {kx + \left( {k + 3} \right)y = 3k - 1} \cr }$$\$
has no solution, is
A
infinite
B
1
C
2
D
3

## Explanation

From the given system, we have

$${{k + 1} \over k} = {8 \over {k + 3}} \ne {{4k} \over {3k - 1}}$$

( as System has no solution)

$$\Rightarrow {k^2} + 4k + 3 = 8k$$

$$\Rightarrow k = 1,3$$

If $$k = 1$$ then $${8 \over {1 + 3}} \ne {{4.1} \over 2}$$ which is false

And if $$k = 3$$

Then $${8 \over 6} \ne {{4.3} \over {9 - 1}}$$ which is true, therefore $$k=3$$

Hence for only one value of $$k.$$ System has no solution.

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