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Mathematics · 2020
JEE Main · 7 January 2020, Shift 1 · Q51
If g (x)=x^2+x-1 and (g ° f)(x)=4 x^2-10 x+5, then f(5/4) is equal to:
If $\displaystyle \mathrm{g}(x)=x^2+x-1$ and
$\displaystyle (g \circ f)(x)=4 x^2-10 x+5$, then $\displaystyle f\left(\frac{5}{4}\right)$ is equal to :
Official answer
(2)
$\displaystyle -\frac{1}{2}$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.