If the area of the region \ ( x,y ):frac a x 2 le yle 1/x,1le xle 2,0<a<1 \ is ( log e 2 )-1/7 then the value of 7a-3 is equal to :

Mathematics · Area under the Curves

JEE Main 2024 — 6 April, Shift 2 — Question 18

If the area of the region $\left\{ \left( x,y \right):\frac{a}{{{x}^{2}}}\le y\le \frac{1}{x},1\le x\le 2,0<a<1 \right\}$ is $\left( {{\log }_{e}}2 \right)-\frac{1}{7}$ then the value of $7a-3$ is equal to :

Option A:
2
Option B:
0
Option C:
-1
Correct
Option D:
1

Answer: C

Step-by-step solution

$\operatorname{area} \int_{1}^{2}\left(\frac{1}{x}-\frac{a}{x^{2}}\right) d x$

$\left[\ln x+\frac{\mathrm{a}}{\mathrm{x}}\right]_{1}^{2}$

$\log _{e} 2+\frac{a}{2}-a=\log _{e} 2-\frac{1}{7}$

$\frac{-\mathrm{a}}{2}=-\frac{1}{7}$

$\mathrm{a}=\frac{2}{7}$

$7 \mathrm{a}=2$

It can be written as ;

$\boxed{7a-3=-1}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Mathematics
Chapter: Area under the Curves
Topic: Determination of Parameter(s) in area based problems