Mathematics · 2024
JEE Main · 6 April 2024, Shift 2 · Q11
If the area of the region {(x, y): a/x^2 ≤ y ≤ 1/x, 1 ≤ x ≤ 2,0<a<1} is ( log_e 2)-1/7 then the value of 7 a-3 is equal to:
If the area of the region $\displaystyle \left\{(x, y): \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0<a<1\right\}$ is $\displaystyle \left(\log _e 2\right)-\frac{1}{7}$ then the value of $\displaystyle 7 a-3$ is equal to:
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle -1$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.