Q42 / 160 · AP EAMCET 2024 (18 May Shift 1) · Mathematics

If $f(x)=\sqrt{x}-1$ and $g\{f(x)\}=x+2 \sqrt{x}+1$ then $g(x)=$

  • Correct answer
    A
    $(x+2)^2$
  • B
    $(x-2)^2$
  • C
    $(\sqrt{x}+2)^2$
  • D
    $(\sqrt{x}-2)^2$

Solution

Given, $f(x)=\sqrt{x-1}$ and, $g\{f(x)\}=x+2 \sqrt{x}+1=(\sqrt{x})^2+2 \sqrt{x}+1^2$ $\Rightarrow g\{f(x)\}=(\sqrt{x}+1)^2\ldots$ (i) Let, $f(x)=\sqrt{x}-1=t \Rightarrow \sqrt{x}=t+1$ By (i), $g(\mathrm{t})=(\mathrm{t}+1+1)^2 \Rightarrow g(x)=(x+2)^2$

Hence, the correct answer is option: 'A'.

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