Question

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

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

Answer 1

The correct option is D

$(2,5)$

Explanation for the correct option:

Solving ${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathit{x}\mathbf{-}\mathbf{10}\mathbf{<}\mathbf{0}$

$$\begin{gathered} x^2-3x-10<0 \\ \Rightarrow x^2-5x+2x-10<0 \\ \Rightarrow x\left(x-5\right)+2\left(x-5\right)<0 \\ \Rightarrow \left(x+2\right)\left(x-5\right)<0 \end{gathered}$$

$\Rightarrow x\in \left(-2,5\right)...\left(1\right)$

Solving $\mathbf{10}\mathit{x}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{16}\mathbf{>}\mathbf{0}$

$$\begin{gathered} 10x-x^2-16>0 \\ \Rightarrow x^2-10x+16<0 \\ \Rightarrow x^2-8x-2x+16<0 \\ \Rightarrow x\left(x-8\right)-2\left(x-8\right)<0 \\ \Rightarrow \left(x-2\right)\left(x-8\right)<0 \end{gathered}$$

$\Rightarrow x\in \left(2,8\right)...\left(2\right)$

Finding the set of values of $\mathit{x}$:

From $\left(1\right)$ and $\left(2\right)$, we get

$$\begin{gathered} \Rightarrow x\in \left(-2,5\right)\cap \left(2,8\right) \\ \Rightarrow x\in \left(2,5\right) \end{gathered}$$

Hence, the correct answer is option D.