Mathematics · 2026
JEE Main · 23 January 2026, Shift 1 · Q17
Let f(x)= {(a x^2+2 a x+3)/(4 x^2+4 x-3), x ≠-3/2, 1/2; b, x=-3/2, 1/2} be continuous at x=-3/2. If f o f(x)=7/5, then x is equal to:
Let $\displaystyle f(x)= \begin{cases}\frac{\mathrm{a} x^2+2 \mathrm{a} x+3}{4 x^2+4 x-3}, & x \neq-\frac{3}{2}, \frac{1}{2} \\ \mathrm{~b}, & x=-\frac{3}{2}, \frac{1}{2}\end{cases}$
be continuous at $\displaystyle x=-\frac{3}{2}$. If $\displaystyle f o f(x)=\frac{7}{5}$, then $\displaystyle x$ is equal to:
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 1$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.