Question

Find the solution set of \( x \) for which the expression \( \frac{x\left(3^{x}-1\right)(x+1)^{2}}{(x-3)(x-2)^{4}} \leq 0 \).

Answer 1

\(\frac{x(3^x -1) (x + 1)^2}{(x - 3)(x -2)^4} \le 0\)       ....(1)

For existance of fractional function, we have x -3 \(\ne\) 0 and x - 2 \(\ne\) 0

⇒ \(x \ne 3\) & \(x \ne 2\)        .....(2)

∵ \((x -2)^4\ge 0 \) also \(x -2 \ne0\)

⇒ \((x - 2)^4 >0\)

So, when multiplying expression (1) by (x - 2)⁴ sign remains unaltered

\(\frac{x(3^x -1) (x + 1)^2}{x - 3} \le 0\)

\(\therefore x \in(-\infty,3)-[2]\) or \(x \in(-\infty, 2)\cup (2,3)\) is solution set for given expression.