Question

The solution set of the inequality $\left|9^{x}-3^{x+1}-15\right|< 2.9^{x}-3^{x}$ is

• A. $\left(-\infty, 1\right)$
• B. $\left(1, \infty\right)$
• C. $(-\infty, 1]$
• D. None of these

Answer 1

Answer: (B)

$\left(1, \infty\right)$

Answer 2

Let $3^x=y$, then the inequality is $| y^2 -3y-15| < 2y^2-y\,...(i)$ The inequality holds if $2y^{2}-y>0 \Rightarrow y < 0$ or $y > \frac{1}{2}$ $\because y=3^{x} \le0 \Rightarrow y >\frac{1}{2}$ Now the inequality on solving, $-\left(2y^{2}-y\right) < y^{2}-3y-15 < 2y^{2}-y$ $\Rightarrow 3y^{2}-4y-15>0$ and $y^{2}+2y+15>0$ Solution of first inequality $3y^{2}-4y-15>0$ is $y3$ Solution of second inequality $y^{2}+2y+15>0$ is $y\in R$ The common solution is $y>3 \Rightarrow 3^{x}> x \Rightarrow x>1 \Rightarrow x\in\left(1, \infty\right)$