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  • Help me solve this - Complex numbers and quadratic equations - JEE Main-9

f\left ( x \right )= a^{2}+bx+1\; \left ( a,b\, \epsilon \, R \right ) . If a-b< -1 then  f\left ( x \right )= 0 will have roots in 

  • Option 1)

    \left ( -1,0 \right )

  • Option 2)

    \left ( 0,1 \right )

  • Option 3)

    \left ( 1,2 \right )

  • Option 4)

    \left ( 2,3 \right )

 

Answers (1)

f\left ( -1 \right )=a-b+1< 0  (given)

f\left (0 \right )=1> 0

\therefore \: f\left (-1 \right ) and f\left (0 \right ) are of opposite sign

So f\left ( x \right )=0 will have a root in \left ( -1,0 \right )

\therefore Option (A)

 

f (α ) & f (β) are of opposite signs -

Here, f\left ( x \right )= ax^{2}+bx+c

f\left ( x \right )= 0 for one x\in \left ( \alpha ,\beta \right )

- wherein

 

 


Option 1)

\left ( -1,0 \right )

This is correct

Option 2)

\left ( 0,1 \right )

This is incorrect

Option 3)

\left ( 1,2 \right )

This is incorrect

Option 4)

\left ( 2,3 \right )

This is incorrect

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