Functions Question 423
Question: Suppose that $ g(x)=1+\sqrt{x} $ and $ f(g(x))=3+2\sqrt{x}+x $ , then $ f(x) $ is
[MP PET 2000; Karnataka CET 2002]
Options:
A) $ 1+2x^{2} $
B) $ 2+x^{2} $
C) $ 1+x $
D) $ 2+x $
Show Answer
Answer:
Correct Answer: B
Solution:
$ g(x)=1+\sqrt{x} $ and $ f(g(x))=3+2\sqrt{x}+x $ …..(i)
Þ $ f(1+\sqrt{x})=3+2\sqrt{x}+x $ Put $ 1+\sqrt{x}=y $
Þ $ x={{(y-1)}^{2}} $ then, $ f(y)=3+2(y-1)+{{(y-1)}^{2}} $ $ =2+y^{2} $ therefore, $ f(x)=2+x^{2} $ .
BETA
