Relations Question 7
Question: If $ g(x)=x^{2}+x-2 $ and $ \frac{1}{2}gof(x)=2x^{2}-5x+2, $ then which is not a possible f(x)?
Options:
A) $ 2x-3 $
B) $ -2x+2 $
C) $ x-3 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{1}{2}(gof)(x)=2x^{2}-5x+2 $
Or $ \frac{1}{2}g[f(x)]=2x^{2}-5x+2 $
$ \therefore [{{{f(x)}}^{2}}+{f(x)}-2]=[2x^{2}-5x+2] $
Or $ f{{(x)}^{2}}+f(x)-(4x^{2}-10x+6)=0 $
$ \therefore f(x)=\frac{-1\pm \sqrt{1+4(4x^{2}-10x+6)}}{2} $
$ =\frac{-1\pm \sqrt{(16x^{2}-40x+25)}}{2}=\frac{-1\pm (4x-5)}{2} $
$ =2x-3 $ or $ -2x+2 $
BETA
