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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Area Under Curves

Consider the region R=\ (x, y): x ≤ y ≤ 9-(11)/(3) x^2, x ≥ 0\ ....

Consider the region R={(x,y):x≤y≤9−113x2,x≥0}R=\left\{(x, y): x \leq y \leq 9-\frac{11}{3} x^2, x \geq 0\right\}. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R , is:

Options

  1. A730119\frac{730}{119}
  2. B625111\frac{625}{111}
  3. C821123\frac{821}{123}
  4. D567121\frac{567}{121}

Correct answer

D. 567121\frac{567}{121}

Step-by-step solution

\mathrm{t} .\left(9-\frac{11 \mathrm{t}^2}{3}-\mathrm{t}\right)A=9t−t2−113t3dAdt=9−2t−11t2⇒11t2+2t−9=011t2+11t−9t−9=0t=−1t=911\begin{aligned} & \mathrm{A}=9 \mathrm{t}-\mathrm{t}^2-\frac{11}{3} \mathrm{t}^3 \\ & \frac{\mathrm{dA}}{\mathrm{dt}}=9-2 \mathrm{t}-11 \mathrm{t}^2 \\ & \Rightarrow 11 \mathrm{t}^2+2 \mathrm{t}-9=0 \\ & 11 \mathrm{t}^2+11 \mathrm{t}-9 \mathrm{t}-9=0 \\ & \mathrm{t}=-1 & \mathrm{t}=\frac{9}{11}\end{aligned}∴ largest area =911(9−113,81121−911)=911⋅6311=567121\begin{aligned} & \therefore \text{ largest area }=\frac{9}{11}\left(9-\frac{11}{3}, \frac{81}{121}-\frac{9}{11}\right) \\ & =\frac{9}{11} \cdot \frac{63}{11}=\frac{567}{121}\end{aligned}$$
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