The set of real values of ‘a’ for which the graph of the function f(x)=ax2+2(a+1)x+9a+4x2−8x+32 is always below the x-axis is

f (x) 0,∀x∈R⇒ax2+2(a+1)x+9a+4<0,∀x∈R⇒a<0 and4(a+1)2−4a(9a+4)<0⇒a∈−∞,−12∪14,∞

The set of real values of ‘a’ for which the graph of the function f(x)=ax2+2(a+1)x+9a+4x2−8x+32 is always below the x-axis is

f (x) 0,∀x∈R⇒ax2+2(a+1)x+9a+4<0,∀x∈R⇒a<0 and4(a+1)2−4a(9a+4)<0⇒a∈−∞,−12∪14,∞

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The set of real values of ‘a’ for which the graph of the function $f (x) = \frac{{ax}^{2} + 2 (a + 1) x + 9 a + 4}{x^{2} - 8 x + 32}$ is always below the x-axis is

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$(- \infty, - \frac{1}{2})$

$(- \frac{1}{2}, \frac{1}{2})$

$(- \frac{1}{2}, 2)$

$(- \infty, \frac{1}{2})$

answer is B.

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Detailed Solution

f (x) < 0 and since $x^{2} - 8 x + 32 > 0, \forall x \in R$
$\Rightarrow {ax}^{2} + 2 (a + 1) x + 9 a + 4 < 0, \forall x \in R \Rightarrow a < 0$ and
$\begin{array}{l} 4 (a + 1)^{2} - 4 a (9 a + 4) < 0 \\ \Rightarrow a \in (- \infty, - \frac{1}{2}) \cup (\frac{1}{4}, \infty) \end{array}$