Question

Assertion :The solution set of the inequality
${\log }_{0.7}\left({\log }_6\frac{x^2+x}{x+4}\right)<0$ is $(-4,-3)\cup (8,\infty )$ Reason: For $x>0$, ${\log }_{a}x$ is an increasing function, if $a>1$ and a decreasing function, if $0<a<1$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

${\log }_{0.7}\left({\log }_6\frac{x^2+x}{x+4}\right)$ is valid when $\frac{x^2+x}{x+4}>1$

$\Rightarrow \frac{x^2+x-x-4}{x+4}>0$

$\Rightarrow x\in (-4,-2)\bigcup (2,\infty )$ ....(1)
${\log }_{0.7}\left({\log }_6\frac{x^2+x}{x+4}\right)<0$
${\log }_6\frac{x^2+x}{x+4}>1$
$\Rightarrow \frac{x^2+x}{x+4}>6$
$\Rightarrow \frac{x^2+x}{x+4}-6>0$
$\Rightarrow \frac{(x-8)(x+3)}{x+4}>0$
$\Rightarrow x\in (-4,-3)\cup (8,\infty )$
Ans: A