Functions Question 521
Question: If $ g(x)=x^{2}+x-2 $ and $ \frac{1}{2} $ (gof) $ (x)=2x^{2}-5x+2, $ then $ f(x) $ is equal to
Options:
A) $ 2x-3 $
B) $ 2x+3 $
C) $ 2x^{2}+3x+1 $
D) $ 2x^{2}-3x-1 $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ g(x)=x^{2}+x-2 $ and $ \frac{1}{2}(gof)(x)=2x^{2}-5x+2 $
$ \Rightarrow g(f(x))=4x^{2}-10x+4\Rightarrow {{(f(x))}^{2}}+f(x)-2 $ $ =4x^{2}-10x+4 $
$ \Rightarrow {{(f(x))}^{2}}+f(x)-(4x^{2}-10x+6)=0 $
$ \Rightarrow f(x)=\frac{-1\pm \sqrt{16x^{2}-40x+25}}{2} $ $ =\frac{-1\pm (4x-5)}{2} $ $ =\frac{4x-6}{2} $ or $ \frac{-4x+4}{2}=2x-3, $ or $ -2x+2 $ Hence $ f(x)=2x-3. $
BETA
