Question

The solution of the inequality ${\log }_{\frac{25-x^2}{16}}\left(\frac{24-2x-x^2}{14}\right)>1$ is

• A. $(-3,3)$
• B. $(-\infty ,17)\cup (-3,4)\cup (1,\infty )$
• C. $(-3,1)\cup (3,4)$
• D. $(-17,1)\cup (3,4)$

Answer 1

The correct option is C $(-3,1)\cup (3,4)$

$\frac{24-2x-x^2}{14}>0$
$\Rightarrow x^2+2x-24<0$
$\Rightarrow x^2+6x-4x-24<0$
$\Rightarrow x(x+6)-4(x+6)<0$
$\Rightarrow (x-4)(x+6)<0$
$\Rightarrow -6<x<4$

CASE I : If $\frac{25-x^2}{16}>1$
$\Rightarrow x^2<9$
$\Rightarrow -3<x<3\cdots \left(1\right)$
$\frac{24-2x-x^2}{14}>\frac{25-x^2}{16}$
or $x^2+16x-17<0$
$\Rightarrow -17<x<1$
$\therefore$ From $\left(1\right),x\in (-3,1)\cdots \left(2\right)$

CASE II:
$0<\frac{25-x^2}{16}<1$
$\Rightarrow 9<x^2<25$
$\Rightarrow -5<x<-3,\text{or}3<x<5\cdots \left(3\right)$
$\frac{24-2x-x^2}{14}<\frac{25-x^2}{16}$
$\Rightarrow x^2+16x-17>0$
$\Rightarrow x<-17\text{or}x>1\cdots \left(4\right)$
From $\left(3\right)$ and $\left(4\right),$
$3<x<5\cdots \left(5\right)$

But $-6<x<4$
$\therefore$ From $\left(2\right)$ and $\left(5\right),$ we get
$x\in (-3,1)\cup (3,4)$