Mathematics · 2025
JEE Main · 3 April 2025, Shift 1 · Q17
Let f(x)= {(1+a x)^1 / x, x<0; 1+b, x=0; ((x+4)^1 / 2-2)/((x+c)^1 / 3-2), x>0} be continuous at x=0. Then e^a b c is equal to:
Let $\displaystyle f(x)= \begin{cases}(1+a x)^{1 / x} \quad, & x<0 \\ 1+b \quad, & x=0 \\ \frac{(x+4)^{1 / 2}-2}{(x+c)^{1 / 3}-2}, & x>0\end{cases}$
be continuous at $\displaystyle x=0$. Then $\displaystyle e^a b c$ is equal to:
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 48$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.