Q63 / 100 · BITSAT 2022 · Mathematics

If $g(x)=x^2+x-2$ and $\frac{1}{2} g o f(x)=2 x^2-5 x+2$ then $f(x)$ is equal to

  • Correct answer
    A
    $2 x-3$
  • B
    $2 x+3$
  • C
    $2 x^2+3 x+1$
  • D
    $2 x^2-3 x+1$

Solution

We have, $\frac{1}{2} go f(x)=2 x^2-5 x+2$ $\begin{aligned} & \Rightarrow g(f(x))=4 x^2-10 x+4 \\ & \Rightarrow(f(x))^2+f(x)-2=4 x^2-10 x+4 \\ & \Rightarrow(f(x))^2+f(x)-\left(4 x^2-10 x+6\right)=0 \\ & \Rightarrow f(x)=\frac{-1 \pm \sqrt{1+4\left(4 x^2-10 x+6\right)}}{2} \\ & \Rightarrow f(x)=\frac{-1 \pm \sqrt{16 x^2-40 x+25}}{2} \\ & \Rightarrow f(x)=\frac{-1 \pm(4 x-5)}{2}=2 x-3,-2 x+2 \end{aligned}$ Hence, $f(x)=2 x-3$

Hence, the correct answer is option: 'A'.

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