Question

Solve: ${\log }_{{\log }_2\left(\frac{x}{2}\right)}(x^2-10x+22)>0$

• A. $x\in (0,5-\sqrt{3})\cup (7,\infty )$
• B. $x\in (3,5-\sqrt{3})\cup (7,\infty )$
• C. $x\in (-\infty ,5-\sqrt{3})\cup (7,\infty )$
• D. None of these

Answer 1

The correct option is B $x\in (3,5-\sqrt{3})\cup (7,\infty )$

Given inequality

${\log }_{{\log }_2\left(\frac{x}{2}\right)}(x^2-10x+22)>0$ .....(i)
We must have
i. $x^2-10x+22>0$
$\Rightarrow$ $xϵ(-\infty ,5-\sqrt{3})\cup (5-\sqrt{3},\infty )$ .....(ii)
ii. $\frac{x}{2}>0$

$\Rightarrow x>0$ ..... (iii)

Case I: If $0<{\log }_2\left(\frac{x}{2}\right)<1$

$\Rightarrow$ $1<\frac{x}{2}<2$
or $2<x<4$ ..... (iv)
Therefore, from Eq. (i), $x^2-10x+22<1$
or $x^2-10x+21<0$
$\Rightarrow$ $3<x<7$ ..... (v)
From Eqs. (ii), (iii), (iv) and (v), the common solution is $3<x<5-\sqrt{3}$

Case II: If ${\log }_2\left(\frac{x}{2}\right)>1$

$\frac{x}{2}>2$ or $x>4$ (vi)
Therefore, from Eqs. (i), $x^2-10x+22>1$
or $x^2-10x+21>0$
or $x<3$ or $x>7$ ...... (vii)
From Eqs. (ii), (iii), (vi) and (vii), the common solution is $x\in (7,\infty )$
Hence, $x\in (3,5-\sqrt{3})\cup (7,\infty )$
Ans: B