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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Area Under Curves

If the area of the region \ (x, y): a x^2 ≤ y ≤ (1)/(x), 1 ≤ x ≤...

If the area of the region {(x,y):ax2≤y≤1x,1≤x≤2,0<a<1}\left\{(x, y): \frac{\mathrm{a}}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0 < \mathrm{a} < 1\right\} is (log⁡e2)−17\left(\log _{\mathrm{e}} 2\right)-\frac{1}{7} then the value of 7a−37 \mathrm{a}-3 is equal to:

Options

  1. A0
  2. B2
  3. C-1
  4. D1

Correct answer

C. -1

Step-by-step solution

 area ∫12(1x−ax2)dx[ℓnx+ax]12ℓnn⁡2+a2−a=log⁡e2−17−a2=−17a=277a=27a−3=−1\begin{aligned} & \text{ area } \int_1^2\left(\frac{1}{x}-\frac{a}{x^2}\right) d x \\ & {\left[\ell n x+\frac{a}{x}\right]_1^2} \\ & \ell \operatorname{nn} 2+\frac{a}{2}-a=\log _e 2-\frac{1}{7} \\ & \frac{-a}{2}=-\frac{1}{7} \\ & a=\frac{2}{7} \\ & 7 a=2 \\ & 7 a-3=-1\end{aligned}
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