Range of the function $\mathrm{f}(\mathrm{x})=3+2^{\mathrm{x}}+4^{\mathrm{x}}$ is
Range of the function $\mathrm{f}(\mathrm{x})=3+2^{\mathrm{x}}+4^{\mathrm{x}}$ is
- $(3, \infty)$
- $(-\infty, \infty)$
- $(3, \infty)$
- $(-\infty, 3)$
Solution
$f(x)=3+2^x+4^x=y$
Let $2^{\mathrm{x}}=\mathrm{a} \Rightarrow 4^{\mathrm{x}}=\mathrm{a}^2$
$\therefore \mathrm{a}^2+\mathrm{a}+(3-\mathrm{y})=0$
As $\mathrm{a} \in \mathrm{R}$, we write
$\begin{aligned}
& (1)^2-4(1)(3-\mathrm{y}) \geq 0 \\
& \therefore 1-12+4 \mathrm{y} \geq 0 \quad \Rightarrow 4 \mathrm{y} \geq 11 \quad \Rightarrow \mathrm{y} \geq \frac{11}{4}
\end{aligned}$
Also $2^x+4^x>0 \Rightarrow y \neq 3$
$\therefore$ Range from given option is $(3, \infty)$
Asked in: MHT CET 2021 (23 Sep Shift 2)
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