Functions Question 368
Question: If $ g(x)=x^{2}+x-2 $ and $ \frac{1}{2}gof(x)=2x^{2}-5x+2 $ , then $ f(x) $ is
[Roorkee 1998; MP PET 2002]
Options:
A) $ 2x-3 $
B) $ 2x+3 $
C) $ 2x^{2}+3x+1 $
D) $ 2x^{2}-3x-1 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{1}{2},(gof),(x)=2x^{2}-5x+2 $ or $ \frac{1}{2},g,[f,(x)]=2x^{2}-5x+2 $ $ \underset{x\to 0}{\mathop{\lim }}\frac{(1+nx+{{,}^{n}}C_2x^{2}+…..\text{higher powers of }x\text{ to }x^{n})-1}{x} $
$ \Rightarrow 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+(4x-5)}{2}=2x-3 $ .
BETA
