JEE Mains · Maths · STD 12 - 8. Application and integration
If the area of the region \(\left\{(\mathrm{x}, \mathrm{y}): \frac{\mathrm{a}}{\mathrm{x}^2} \leq \mathrm{y} \leq \frac{1}{\mathrm{x}}, 1 \leq \mathrm{x} \leq 2,0<\mathrm{a}<1\right\}\) is \(\left(\log _e 2\right)-\frac{1}{7}\) then the value of \(7 a-3\) is equal to :
- A \(2\)
- B \(0\)
- C \(-1\)
- D \(1\)
Answer & Solution
Correct Answer
(C) \(-1\)
Step-by-step Solution
Detailed explanation
\( \operatorname{area} \int_1^2\left(\frac{1}{x}-\frac{a}{x^2}\right) d x \) \( {\left[\ln x+\frac{a}{x}\right]_1^2} \) \( \ell \operatorname{nn} 2+\frac{a}{2}-a=\log _e 2-\frac{1}{7} \) \( \frac{-a}{2}=-\frac{1}{7} \) \( a=\frac{2}{7} \) \( 7 a=2 \) \( 7 a-3=-1\)
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