Question

The set of values of $x$ for which the inequalities $x^2-3x-10<0$ and $10x-x^2-16>0$ hold simultaneously, is

• A. $(-2,5)$
• B. $(2,8)$
• C. $(-2,8)$
• D. $(2,5)$

Answer 1

Answer: (D)

$(2,5)$

We have,
$x^2-3x-10<0$
$\Rightarrow (x-5)(x+2)<0$
$\Rightarrow -2<x<5$
Also,
$10x-x^2-16>0$
$x^2-10x+16<0$
$(x-2)(x-8)<0$
Hence, $2<x<8$
The common set for $x$ is $(2,5)$.