Question

If $g\left(x\right)=x^2+x-1$ and $\left(gof\right)\left(x\right)=4x^2+10x+5$, then $f\left(\frac{5}{4}\right)$ is equal to

• A. $\frac{-3}{2}$
• B. $\frac{-1}{2}$
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is B

$\frac{-1}{2}$

Explanation for the correct option:

Finding the value of $f\left(\frac{5}{4}\right)$:

Given that,

$g\left(x\right)=x^2+x-1$,

Now,

$\begin{array}{rcl}\left(gof\right)\left(x\right)& =& 4x^2+10x+5\\ & \Rightarrow & g\left(f\left(x\right)\right)=4x^2+10x+5\\ & \Rightarrow & f^2\left(x\right)+f\left(x\right)-1=4x^2+10x+5\end{array}$

Substitute $x=\frac{5}{4}$in the above equation,

$$\begin{gathered} f^2\left(\frac{5}{4}\right)+f\left(\frac{5}{4}\right)-1=4{\left(\frac{5}{4}\right)}^2-10\left(\frac{5}{4}\right)+5 \\ \Rightarrow f^2\left(\frac{5}{4}\right)+f\left(\frac{5}{4}\right)-1=\frac{4\times 25}{16}-\frac{50}{4}+5 \\ \Rightarrow f^2\left(\frac{5}{4}\right)+f\left(\frac{5}{4}\right)-6=\frac{-25}{4} \\ \Rightarrow f^2\left(\frac{5}{4}\right)+f\left(\frac{5}{4}\right)=6-\frac{25}{4} \\ \Rightarrow f^2\left(\frac{5}{4}\right)+f\left(\frac{5}{4}\right)+\frac{1}{4}=0 \\ \Rightarrow {\left(f\left(\frac{5}{4}\right)+\frac{1}{2}\right)}^2=0\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right] \\ \Rightarrow f\left(\frac{5}{4}\right)=\frac{-1}{2} \end{gathered}$$

Hence, the correct option is B.