MHT CET · Maths · Functions
If \(\mathrm{g}(x)=1+\sqrt{x}\) and \(\mathrm{f}(\mathrm{g}(x))=3+2 \sqrt{x}+x\), then \(\mathrm{f}(\mathrm{f}(x))\) is
- A \(x^2+4 x+6\)
- B \(x^4+x^2+6\)
- C \(x^2+x+6\)
- D \(x^4+4 x^2+6\)
Answer & Solution
Correct Answer
(D) \(x^4+4 x^2+6\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & g(x)=1+\sqrt{x} \text { and } \mathrm{f}(\mathrm{g}(x))=3+2 \sqrt{x}+x \\ & \therefore \quad \mathrm{f}(\mathrm{g}(x))=\left[(\sqrt{x})^2+2 \sqrt{x}+1\right]+2 \\ & =(\sqrt{x}+1)^2+2 \\ & =[\mathrm{g}(x)]^2+2 \\ & \Rightarrow \mathrm{f}(x)=x^2+2 \\ & \Rightarrow \mathrm{f}(\mathrm{f}(x))=\left(x^2+2\right)^2+2=x^4+4 x^2+6 \\ & \end{aligned}\)
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