MHT CET · Maths · Functions
If \(\mathrm{f}(x)=x^2+1\) and \(\mathrm{g}(x)=\frac{1}{x}\), then the value of \(\mathrm{f}(\mathrm{g}(\mathrm{g}(\mathrm{f}(x))))\) at \(x=1\) is
- A \(4\)
- B \(1\)
- C \(5\)
- D \(3\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
\mathrm{f}(x)=x^2+1, & \mathrm{~g}(x)=\frac{1}{x} \\
\therefore \quad \mathrm{f}(\mathrm{g}(\mathrm{g}(\mathrm{f}(x)))) & =\mathrm{f}\left(\mathrm{g}\left(\mathrm{g}\left(x^2+1\right)\right)\right) \\
& =\mathrm{f}\left(\mathrm{g}\left(\frac{1}{x^2+1}\right)\right) \\
& =\mathrm{f}\left(x^2+1\right) \\
& =\left(x^2+1\right)^2+1
\end{aligned}
\)
\(\therefore \quad\) At \(x=1\), we get the value of above function
\(
\begin{aligned}
& =\left[(1)^2+1\right]^2+1 \\
& =5
\end{aligned}
\)
\begin{aligned}
\mathrm{f}(x)=x^2+1, & \mathrm{~g}(x)=\frac{1}{x} \\
\therefore \quad \mathrm{f}(\mathrm{g}(\mathrm{g}(\mathrm{f}(x)))) & =\mathrm{f}\left(\mathrm{g}\left(\mathrm{g}\left(x^2+1\right)\right)\right) \\
& =\mathrm{f}\left(\mathrm{g}\left(\frac{1}{x^2+1}\right)\right) \\
& =\mathrm{f}\left(x^2+1\right) \\
& =\left(x^2+1\right)^2+1
\end{aligned}
\)
\(\therefore \quad\) At \(x=1\), we get the value of above function
\(
\begin{aligned}
& =\left[(1)^2+1\right]^2+1 \\
& =5
\end{aligned}
\)
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