JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Area Under Curves

1. (A)There exists an $\mathrm{h}\in \left[\frac{1}{4},\frac{2}{3}\right]$ such that the area of the green region above the line ${\mathrm{L}}_{\mathrm{h}}$ equals the area of the green region below the line ${\mathrm{L}}_{\mathrm{h}}$
2. (B)There exists an $\mathrm{h}\in \left[\frac{1}{4},\frac{2}{3}\right]$ such that the area of the red region above the line ${\mathrm{L}}_{\mathrm{h}}$ equals the area of the red region below the line ${\mathrm{L}}_{\mathrm{h}}$✓ Correct
3. (C)There exists an $\mathrm{h}\in \left[\frac{1}{4},\frac{2}{3}\right]$ such that the area of the green region above the line ${\mathrm{L}}_{\mathrm{h}}$ equals the area of the red region below the line ${\mathrm{L}}_{\mathrm{h}}$✓ Correct
4. (D)There exists an $\mathrm{h}\in \left[\frac{1}{4},\frac{2}{3}\right]$ such that the area of the red region above the line ${\mathrm{L}}_{\mathrm{h}}$ equals the area of the green region below the line ${\mathrm{L}}_{\mathrm{h}}$✓ Correct

Answer: (B), (C), (D)

Solution

$f^{\prime }(x)=x^2-2x+\frac{5}{9}$ has a critical point at $x=\frac{1}{3}$ in $[0,1]$.

$f(0)=\frac{17}{36}$, $f(1)=\frac{13}{36}\approx 0.36$, $f\left(\frac{1}{3}\right)=\frac{181}{324}\approx 0.558$

The area of red region $A_R={\int }_0^{1}f(x)dx=\frac{1}{2}$, and area of green region $A_G=1-\frac{1}{2}=\frac{1}{2}$.

(A) For equal green areas above and below $L_h$, we need $h=\frac{3}{4}$, which is outside $\left[\frac{1}{4},\frac{2}{3}\right]$. (A) is incorrect.

(B) For equal red areas above and below $L_h$, we need $h=\frac{1}{4}\in \left[\frac{1}{4},\frac{2}{3}\right]$. (B) is correct.

(C) By intermediate value theorem, there exists $h\in \left(\frac{13}{36},\frac{181}{324}\right)\subset \left[\frac{1}{4},\frac{2}{3}\right]$ where green area above equals red area below. (C) is correct.

(D) This is the complementary coloring of (C), so it is also correct. (D) is correct.

Answer: (B), (C), (D)

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · One or more correct options