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If $x^{2}+2\left ( a-1 \right )x+\left ( a+5 \right )> 0$ $\forall$ $x\epsilon R$ then

Option 1)
$\left [ -1,4 \right ]$

Option 2)
$(-1,4)$

Option 3)
$(-1,4]$

Option 4)
$[-1,4)$

Answers (1)

best_answer

quadratic is always positive when

$a> 0$ & $D< 0$

here coeff of $x^{2}> 0$ so only $D< 0$ is required.

$\Rightarrow \: 4\left ( a-1 \right )^{2}-4\left ( a+5 \right )< 0$

$\Rightarrow \: a^{2}-3a-4< 0$

$\Rightarrow \: \left ( a-4 \right )\left ( a+4 \right )< 0$

$\Rightarrow \: a\: \epsilon \: \left ( -1,4 \right )$

$\therefore$ Option (B)

Sign of quadratic expression as positive. -

$ax^{2}+bx+c$ will be always positive 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 above x-axis so $ax^{2}+bx+c= 0$ has no real roots.

Option 1)

$\left [ -1,4 \right ]$

This is incorrect

Option 2)

$(-1,4)$

This is correct

Option 3)

$(-1,4]$

This is incorrect

Option 4)

$[-1,4)$

This is incorrect

Posted by

prateek

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