Q

# Tell me? - Complex numbers and quadratic equations - JEE Main-2

The values of $'a'$ for which $-x^{2}+x-a< 0$  $\forall$   $x\; \epsilon \; R$  is

• Option 1)

$a> -2$

• Option 2)

$a> -1$

• Option 3)

$a> \frac{1}{4}$

• Option 4)

$a> 0$

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Coeff of $x^{2}=-1< 0$ only required more is  $D< 0\, \Rightarrow \: 1-4a< 0\: \Rightarrow \, a> \frac{1}{4}$

$\therefore$ Option (C)

Value of quadratic expression as negative. -

$ax^{2}+bx+c$ will be always negative, for all  $x\epsilon R$, If $a< 0$  &  $b^{2}-4ac< 0$  $\left ( Where\; a,b,c\; \epsilon\; R \right )$

- wherein

So, graph of  $y= ax^{2}+bx+c$  will be always below x-axis so $ax^{2}+bx+c= 0$  has no real roots.

Option 1)

$a> -2$

This is incorrect

Option 2)

$a> -1$

This is incorrect

Option 3)

$a> \frac{1}{4}$

This is correct

Option 4)

$a> 0$

This is incorrect

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