If $g(x)=x^{2}+x-1$ and $(g \circ f)(x)=4 x^{2}-10 x+5$, then $f\left(\frac{5}{4}\right)$ is equal to:
Correct Option: , 2
$(g \circ f)(x)=g(f(x))=f^{2}(x)+f(x)-1$
$g\left(f\left(\frac{5}{4}\right)\right)=4\left(\frac{5}{4}\right)^{2}-10 \cdot \frac{5}{4}+5=-\frac{5}{4}$ $\left[\because g(f(x))=4 x^{2}-10 x+5\right]$
$g\left(f\left(\frac{5}{4}\right)\right)=f^{2}\left(\frac{5}{4}\right)+f\left(\frac{5}{4}\right)-1$
$-\frac{5}{4}=f^{2}\left(\frac{5}{4}\right)+f\left(\frac{5}{4}\right)-1$
$f^{2}\left(\frac{5}{4}\right)+f\left(\frac{5}{4}\right)+\frac{1}{4}=0$
$\left(f\left(\frac{5}{4}\right)+\frac{1}{2}\right)^{2}=0$
$t\left(\frac{5}{4}\right)=-\frac{1}{2}$
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