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If the two roots of the equation, $\left ( a-1 \right )\left ( x^{4}+x^{2} +1\right )+\left ( a+1 \right )\left ( x^{2}+x+1 \right )^{2}=0$

are real and distinct, then the set of all values of ‘a’ is :

Option 1)
$\left (- \frac{1}{2},0 \right )$

Option 2)
$\left ( -\infty ,-2 \right )\cup \left ( 2,\infty \right )$

Option 3)
$\left ( -\frac{1}{2},0 \right )\cup \left ( 0,\frac{1}{2} \right )$

Option 4)
$\left ( 0,\frac{1}{2} \right )$

Answers (2)

As we learnt in

Condition for Real and distinct roots of Quadratic Equation -

$D= b^{2}-4ac> 0$

- wherein

$ax^{2}+bx+c= 0$

is the quadratic equation

$(a-1)(x^{4}+x^{2}+1)+(a+1)(x^{2}+x+1)^{2}=0$

$\Rightarrow\ \; (a-1)(x^{2}+2x^{2}+1-x^{2})+(a+1)(x^{2}+x+1)^{2}=0$

$\Rightarrow\ \; (a-1)((x^{2}+1)^{2}-x^{2})+(a+1)(x^{2}+x+1)^{2}=0$

$\Rightarrow\ \; (a-1)(x^{2}+1+x)(x^{2}+1-x)+(a+1)(x^{2}+x+1)^{2}=0$

$\Rightarrow\ \; (x^{2}+x+1)((a-1)(x^{2}-x+1))+(a+1)(x^{2}+x+1)=0$

$\Rightarrow\ \; x^{2}+x+1 \neq0$

So $(a-1)(x^{2}-x+1)+(a+1)(x^{2}+x+1)=0$

$\Rightarrow\ \; 2ax^{2}+2x+2a=0$

$\Rightarrow\ \; ax^{2}+x+a=0$

So for real value of x $D\geq0$

$\therefore\ \;1-4.a.a\geq0$

$1\geq4a^{2}\ \; \Rightarrow\ \;4a^{2}\leq1$

$\therefore\ \; \frac{-1}{2}\leq a \leq \frac{1}{2}$

But $a\neq0$

So $a \epsilon \left(\frac{-1}{2},0 \right )\bigcup \left(0,\frac{1}{\sqrt{2}} \right )$

Correct option is 3.

Option 1)

$\left (- \frac{1}{2},0 \right )$

This is an incorrect option.

Option 2)

$\left ( -\infty ,-2 \right )\cup \left ( 2,\infty \right )$

This is an incorrect option.

Option 3)

$\left ( -\frac{1}{2},0 \right )\cup \left ( 0,\frac{1}{2} \right )$

This is the correct option.

Option 4)

$\left ( 0,\frac{1}{2} \right )$

This is an incorrect option.

Posted by

Sabhrant Ambastha

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