Verified Solution • 2025–26 Session moderate

QJEEM24Ma8a02: JEE Main 2024 Mathematics - Class 12 JEEM24Ma8a02 Solution: If the area of the region is then the value of is equal to :

If the area of the region \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 :

Multiple Choice Options

  • A 1
  • B 0
  • C 2
  • D −- 1

Step-by-Step Verified Solution

\left\{(x, y): \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0< a<1\right\}

⇒∫12(1x−ax2)dx=∣ln⁡∣x∣+ax∣12\Rightarrow \int\limits_1^2 {\left( {{1 \over x} - {a \over {{x^2}}}} \right)dx = \left| {\ln |x| + {a \over x}} \right|_1^2}

amp;(ln⁡2+a2)−(ln⁡1+a)=ln⁡2−a2amp;⇒a2=17amp;⇒a=27amp;7a−3=27×7−3=−1\begin{aligned}& \left(\ln 2+\frac{a}{2}\right)-(\ln 1+a)=\ln 2-\frac{a}{2} \\& \Rightarrow \quad \frac{a}{2}=\frac{1}{7} \\& \Rightarrow \quad a=\frac{2}{7} \\& \quad 7 a-3=\frac{2}{7} \times 7-3=-1\end{aligned}

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