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

Answer: (C)

Case-I: If $\frac{25-x^2}{16}>1\Rightarrow x^2<9\Rightarrow -3<x<3$
$\therefore \frac{24-2x-x^2}{14}>\frac{25-x^2}{16}\text{ or }{x}^2+16x-17<0$
$\Rightarrow -17<x<1\therefore \text{ from }(1),x\in (-3,1)$
$\text{ Case-II: }0<\frac{25-x^2}{16}<1;\therefore 0<\frac{24-2x-x^2}{14}<\frac{25-x^2}{16}$
$\therefore 0<\frac{25-x^2}{16}<1\text{ and }0<\frac{24-2x-x^2}{14}<\frac{25-x^2}{16}$
$\Rightarrow 9<x^2<25\Rightarrow x^2+2x-24<0\text{ and }{x}^2+16x-17>0$
$\Rightarrow -5<x<-3,3<x<5\Rightarrow -6<x<4\text{ and }x<-17,x>1$
$\Rightarrow 3<x<4\dots .(3)\Rightarrow 1<x<4$
$\text{ from (2), (3) we get }x\in (-3,1)\cup (3,4)$