Question

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

• A. $1 + 2x^2$
• B. $2 + x^2$
• C. $1 + x$
• D. $2 + x$

Answer 1

Answer: (B)

$2 + x^2$

Answer 2

We have, $g (x) =1+ \sqrt {x}$ and $f [g (x)] = 3+ 2 \sqrt{x} + x$ ..(i) Also, $f [g(x)] = f (1+ \sqrt{x})$ ...(ii) By (i) and (ii), we get $f (1+ \sqrt{x}) = 3 + 2 \sqrt{x} + x$ Let $1+ \sqrt{x} = y$ or $x = (y - 1)^2$. $\therefore \, \, f (y) = 3 + 2 (y - 1) + (y - 1)^2$. $= 3 + 2y - 2 + y^2 - 2y + 1 = 2 + y^2$ $\therefore \, f (x) = 2 + x^2$