If g(x)=1+x and f(g(x))=3+2x+x then f(x)=

f(g(x))=f(1+x)f(1+x)=3+2x+xLet 1+x=yx=(y−1)2∴f(y)=3+2(y−1)+(y−1)2=2+y2=f(x)=2+x2

If g(x)=1+x and f(g(x))=3+2x+x then f(x)=

f(g(x))=f(1+x)f(1+x)=3+2x+xLet 1+x=yx=(y−1)2∴f(y)=3+2(y−1)+(y−1)2=2+y2=f(x)=2+x2

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If $g (x) = 1 + \sqrt{x}$ and $f (g (x)) = 3 + 2 \sqrt{x} + x$ then $f (x) =$

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$1 + 2 x^{2}$

$2 + x^{2}$

$1 + x$

$2 + x$

answer is B.

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Detailed Solution

$\begin{array}{l} f (g (x)) = f (1 + \sqrt{x}) \\ f (1 + \sqrt{x}) = 3 + 2 \sqrt{x} + x \\ L e t 1 + \sqrt{x} = y \\ x = {(y - 1)}^{2} ∴ f (y) = 3 + 2 (y - 1) + {(y - 1)}^{2} \\ = 2 + y^{2} = f (x) = 2 + x^{2} \end{array}$

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If g(x)=1+x and f(g(x))=3+2x+x then f(x)=

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If $g (x) = 1 + \sqrt{x}$ and $f (g (x)) = 3 + 2 \sqrt{x} + x$ then $f (x) =$