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:

Step 1: Solve the value of $x$ for the given inequality $x^2-3x-10<0$

$x^2–3x–10<0$

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

$\Rightarrow x\in (-2,5)$(If $a<b,$ then$(x–a)(x–b)<0,a<x<b$)

Step 2: Solve for the $10x-x^2-16>0$

$$\begin{gathered} 10x–x^2–16>0 \\ \Rightarrow x^2–10x+16<0 \end{gathered}$$

$$\begin{gathered} \Rightarrow (x–2)(x–8)<0 \\ \Rightarrow x\in (2,8) \end{gathered}$$ ; (If $a<0,$then $(x–a)(x–b)<0,a<x<b$)

$\therefore x\in (-2,5)\cap (2,8)$

$\Rightarrow x\in (2,5)$

Hence, option D is the correct option.