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1

MCQ (Single Correct Answer)

The sum of all real values of $$x$$ satisfying the equation $${\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}}\, = 1$$ is :

A

$$6$$

B

$$5$$

C

$$3$$

D

$$-4$$

Given equation,
$${\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}} = 1$$

**Case 1 : **When x^{2} - 5x + 5 = 1 and x^{2} + 4x - 60 is any real no then this equation satisfy.

**Note :** When we put any real number as a power of 1 the value stays always 1 (1^{ any real no} = 1).

x^{2} - 5x + 5 = 1

(x - 1)(x - 4) = 0

$$\therefore$$ x = 1, 4

**Case 2 : **When x^{2} - 5x + 5 is a real no and x^{2} + 4x - 60 = 0 then the given equation satisfy. As we know if power of any real no is zero then it will become 1((any real number)^{0} = 1).

For, x^{2} + 4x - 60 = 0

(x - 6)(x + 10) = 0

$$\therefore$$ x = 6, -10

**Case 3 :** When x^{2} - 5x + 5 = -1 and x^{2} + 4x - 60 is even this equation satisfy. As we know (-1)^{even} = 1.

For, x^{2} - 5x + 5 = -1

(x - 2)(x - 3) = 0

$$\therefore$$ x = 2, 3

But x can't be 3 because when x = 3 the value of x^{2} + 4x - 60 becomes 3^{2} + 4.3 - 60 = - 39 which is an odd number, then (-1)^{-39} = -1. So for x = 3 equation does not satisfy.

x

(x - 1)(x - 4) = 0

$$\therefore$$ x = 1, 4

For, x

(x - 6)(x + 10) = 0

$$\therefore$$ x = 6, -10

For, x

(x - 2)(x - 3) = 0

$$\therefore$$ x = 2, 3

But x can't be 3 because when x = 3 the value of x

$$\therefore$$ The sum of all the real values = 1 + 4 + 6 + (-10) + 2 = 3

2

MCQ (Single Correct Answer)

Let $$\alpha $$ and $$\beta $$ be the roots of equation $${x^2} - 6x - 2 = 0$$. If $${a_n} = {\alpha ^n} - {\beta ^n},$$ for $$n \ge 1,$$ then the value of $${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$$ is equal to :

A

$$3$$

B

$$ - 3$$

C

$$6$$

D

$$ - 6$$

Given equation, x^{2} - 6x - 2 = 0

Roots are $$\alpha $$ and $$\beta $$.

So, $$\alpha + \beta = 6$$ and $$\alpha \beta = - 2$$

In the question given, $${a_n} = {\alpha ^n} - {\beta ^n}$$

$$\therefore$$ $${a_8} = {\alpha ^8} - {\beta ^8}$$

and $${a_9} = {\alpha ^9} - {\beta ^9}$$

and $${a_{{10}}} = {\alpha ^{{10}}} - {\beta ^{10}}$$

Now, the given equation

$${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$$

= $${{{\alpha ^{10}} - {\beta ^{10}} - 2\left( {{\alpha ^8} - {\beta ^8}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$

=$${{{\alpha ^{10}} - {\beta ^{10}} + \alpha \beta \left( {{\alpha ^8} - {\beta ^8}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$ (as $$\alpha \beta = - 2$$)

=$${{{\alpha ^{10}} - {\beta ^{10}} + {\alpha ^9}\beta - \alpha {\beta ^9}} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$

= $${{{\alpha ^9}\left( {\alpha + \beta } \right) - {\beta ^9}\left( {\alpha + \beta } \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$

= $${{\left( {\alpha + \beta } \right)\left( {{\alpha ^9} - {\beta ^9}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$

= $${{\left( {\alpha + \beta } \right)} \over 2}$$

= $${6 \over 2}$$ (as $${ {\alpha + \beta } }$$ = 6)

= 3

Roots are $$\alpha $$ and $$\beta $$.

So, $$\alpha + \beta = 6$$ and $$\alpha \beta = - 2$$

In the question given, $${a_n} = {\alpha ^n} - {\beta ^n}$$

$$\therefore$$ $${a_8} = {\alpha ^8} - {\beta ^8}$$

and $${a_9} = {\alpha ^9} - {\beta ^9}$$

and $${a_{{10}}} = {\alpha ^{{10}}} - {\beta ^{10}}$$

Now, the given equation

$${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$$

= $${{{\alpha ^{10}} - {\beta ^{10}} - 2\left( {{\alpha ^8} - {\beta ^8}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$

=$${{{\alpha ^{10}} - {\beta ^{10}} + \alpha \beta \left( {{\alpha ^8} - {\beta ^8}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$ (as $$\alpha \beta = - 2$$)

=$${{{\alpha ^{10}} - {\beta ^{10}} + {\alpha ^9}\beta - \alpha {\beta ^9}} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$

= $${{{\alpha ^9}\left( {\alpha + \beta } \right) - {\beta ^9}\left( {\alpha + \beta } \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$

= $${{\left( {\alpha + \beta } \right)\left( {{\alpha ^9} - {\beta ^9}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$

= $${{\left( {\alpha + \beta } \right)} \over 2}$$

= $${6 \over 2}$$ (as $${ {\alpha + \beta } }$$ = 6)

= 3

3

MCQ (Single Correct Answer)

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

$${{2\sqrt 17 } \over 9}$$

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

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

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

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

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

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

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

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

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

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

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

4

MCQ (Single Correct Answer)

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

$$\left( {1,2} \right)$$

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

= $${{ - 2 \pm \sqrt {4 + 12{a^2}} } \over { - 6}}$$

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 = $$\left( { - 1,0} \right) \cup \left( {0,1} \right)$$

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.]

= $${{ - 2 \pm \sqrt {4 + 12{a^2}} } \over { - 6}}$$

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 = $$\left( { - 1,0} \right) \cup \left( {0,1} \right)$$

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