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  • Please help! - Complex numbers and quadratic equations - JEE Main-5

Let f\left ( x \right )= -x^{2}+ax-a. Then values of 'a' for which f\left ( x \right )= 0 has no real root are 

  • Option 1)

    (0,4)

  • Option 2)

    (-1,5)

  • Option 3)

    (-1,4)

  • Option 4)

    (-2,5)

 

Answers (1)

best_answer

\\*\because f(x)=0\; \; has\; \; no \; \; real\; \; root\; \; so\; \; D< 0\\*\therefore a^{2}-4a< 0\Rightarrow a(a-4)< 0\\*\Rightarrow a\epsilon (0,4)

 

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

No real root of

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

& D= b^{2}-4ac

- wherein

 

 


Option 1)

(0,4)

This is correct

Option 2)

(-1,5)

This is incorrect

Option 3)

(-1,4)

This is incorrect

Option 4)

(-2,5)

This is incorrect

Posted by

Aadil

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