Question

If the graph of $f\left(x\right)=16x^2+8(a+5)x-7a-5$ is strictly , above the $x$-axis, then '$a$' must satisfy the interval

• A. $(-2,-1)$
• B. $(-15,-2)$
• C. $(5,7)$
• D. None of these

Answer 1

Answer: (B)

$(-15,-2)$

We have, $f\left(x\right)=16x^2+8(a+5)x-7a-5$
For $f$ to be strictly above x-axis
Discriminant of quadratic $16x^2+8(a+5)x-7a-5=0$ must be negative
$\Rightarrow 8^2(a+5{)}^2-4(16\left)\right(-7a-5)<0$
$\Rightarrow (a+5{)}^2+(7a+5)<0$
$\Rightarrow a^2+17a+30<0$
$\Rightarrow (a+2)(a+15)<0\Rightarrow a\in (-15,-2)$