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.

N neha

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