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

We have $$g(x)=x^2+x-1$$ and the composition $$\bigl(g\circ f\bigr)(x)=g\!\bigl(f(x)\bigr)=4x^2-10x+5.$$

Writing $$y=f(x),$$ the definition of composition gives us

$$g\!\bigl(f(x)\bigr)=g(y)=y^2+y-1.$$

But by the statement of the problem the same quantity also equals $$4x^2-10x+5.$$ Hence we must have

$$y^2+y-1=4x^2-10x+5.$$

Substituting back $$y=f(x)$$ we get a quadratic equation in $$f(x):$$

$$\bigl(f(x)\bigr)^2+f(x)-1=4x^2-10x+5.$$

Now we bring every term to the left‐hand side:

$$\bigl(f(x)\bigr)^2+f(x)-1-4x^2+10x-5=0,$$

so

$$\bigl(f(x)\bigr)^2+f(x)-4x^2+10x-6=0.$$

This is a quadratic in the unknown $$f(x).$$ For a quadratic $$at^2+bt+c=0$$ the roots are given by the quadratic formula $$t=\dfrac{-b\pm\sqrt{\,b^2-4ac\,}}{2a}.$$

Here $$a=1,\;b=1,\;c=-4x^2+10x-6,$$ so

$$f(x)=\dfrac{-1\pm\sqrt{\,1-4\bigl(-4x^2+10x-6\bigr)\,}}{2}.$$

Inside the square root we simplify step by step:

$$1-4\bigl(-4x^2+10x-6\bigr)=1+16x^2-40x+24.$$

Combining like terms gives

$$16x^2-40x+25.$$

Recognising a perfect square, we note $$16x^2-40x+25=(4x-5)^2.$$ Hence

$$f(x)=\dfrac{-1\pm|4x-5|}{2}.$$

The two possible expressions that come out are obtained very easily:

If $$|4x-5|=4x-5$$ we get $$f(x)=\dfrac{-1+(4x-5)}{2}=2x-3,$$ and if $$|4x-5|=5-4x$$ we get $$f(x)=\dfrac{-1+(5-4x)}{2}=2-2x.$$

Thus for every $$x$$ the value of $$f(x)$$ is either $$2x-3$$ or $$2-2x.$$ Both of these, when put back into $$g(x),$$ reproduce $$4x^2-10x+5,$$ so the composition condition is satisfied in either case.

Now we specifically need $$f\!\left(\dfrac54\right).$$ Substituting $$x=\dfrac54$$ in either expression:

Using $$f(x)=2x-3,$$ we get $$f\!\left(\dfrac54\right)=2\left(\dfrac54\right)-3=\dfrac{10}{4}-3=\dfrac{5}{2}-3=-\dfrac12.$$

Using $$f(x)=2-2x,$$ we get $$f\!\left(\dfrac54\right)=2-2\left(\dfrac54\right)=2-\dfrac{10}{4}=2-\dfrac{5}{2}=-\dfrac12.$$

Both routes give the same numerical answer, so unambiguously

$$f\!\left(\dfrac54\right)=-\dfrac12.$$

Hence, the correct answer is Option B.