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  • I have a doubt, kindly clarify. - Complex numbers and quadratic equations - JEE Main-3

f\left ( x \right )= x^{2}+2\left (a-1 \right )x+a^{2}  if  f\left ( x \right )= 0  has no real root then least integer for which it is true is 

  • Option 1)

    a= -1

  • Option 2)

    a=0

  • Option 3)

    a=1

  • Option 4)

    a=2

 

Answers (1)

\because f(x)=0  has no real roots so D< 0

\\*\Rightarrow 4(a^{2}-2a+1)-4a^{2}< 0\Rightarrow 2a> 1\Rightarrow a> \frac{1}{2}

\therefore  minimum interger value of a is 1

 

Quadratic Expression Graph when a > 0 & D < 0 -

No Real and Equal root of

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

& D= b^{2}-4ac

- wherein

 

 


Option 1)

a= -1

This is incorrect

Option 2)

a=0

This is incorrect

Option 3)

a=1

This is correct

Option 4)

a=2

This is incorrect

Posted by

Vakul

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