MHT CET · Maths · Functions
Range of the function \(\mathrm{f}(\mathrm{x})=3+2^{\mathrm{x}}+4^{\mathrm{x}}\) is
- A \((3, \infty)\)
- B \((-\infty, \infty)\)
- C \((3, \infty)\)
- D \((-\infty, 3)\)
Answer & Solution
Correct Answer
(A) \((3, \infty)\)
Step-by-step Solution
Detailed explanation
\(
f(x)=3+2^x+4^x=y
\)
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 \((3, \infty)\)
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 \((3, \infty)\)
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