Question

The solution of the inequality $\log _{\frac{25-x^2}{16}}\left(\frac{24-2 x-x^2}{14}\right)>1$ 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 $
$\therefore \frac{24-2 x - x ^2}{14}>\frac{25- x ^2}{16} \text { or } x ^2+16 x -17<0$
$\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{24-2 x - x ^2}{14}<\frac{25- x ^2}{16} $
$\therefore 0< \frac{25- x ^2}{16}<1 \text { and } 0<\frac{24-2 x - x ^2}{14}<\frac{25- x ^2}{16} $
$\Rightarrow 9< x^2< 25 \Rightarrow x^2+2 x-24< 0 \text { and } x^2+16 x-17 >0$
$\Rightarrow -5< x< -3,3< x< 5 \Rightarrow -6< x< 4 \text { and } x< -17, x >1$
$\Rightarrow 3< x< 4 \ldots .(3) \Rightarrow 1< x< 4 $
$\text { from (2), (3) we get } x \in(-3,1) \cup(3,4) $