JEE Advance - Mathematics (2023 - Paper 1 Online - No. 3)
Let $f:[0,1] \rightarrow[0,1]$ be the function defined by $f(x)=\frac{x^3}{3}-x^2+\frac{5}{9} x+\frac{17}{36}$. Consider the square region $S=[0,1] \times[0,1]$. Let $G=\{(x, y) \in S: y>f(x)\}$ be called the green region and $R=\{(x, y) \in S: y < f(x)\}$ be called the red region. Let $L_h=\{(x, h) \in S: x \in[0,1]\}$ be the horizontal line drawn at a height $h \in[0,1]$. Then which of the following statements is(are) true?
There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $L_h$ equals the area of the green region below the line $L_h$
There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $L_h$ equals the area of the red region below the line $L_h$
There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $L_h$ equals the area of the red region below the line $L_h$
There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $L_h$ equals the area of the green region below the line $L_k$
Explanation
$\begin{aligned} \text { Given } f:[0,1] & \rightarrow[0,1] \\\\ f(x) & =\frac{x^3}{3}-x^2+\frac{5}{9} x+\frac{17}{36} \\\\ f^{\prime}(x) & =\frac{3 x^2}{3}-2 x+\frac{5}{9}\end{aligned}$
$\begin{aligned} & \quad f^{\prime}(x)=0 \\\\ & 9 x^2-18 x+5=0 \\\\ & \Rightarrow 9 x^2-15 x-3 x+5=0 \\\\ & \Rightarrow 3 x(3 x-5)-1(3 x-5)=0 \\\\ & \Rightarrow(3 x-5)(3 x-1)=0 \\\\ & \Rightarrow x=\frac{1}{3} \text { or } \frac{5}{3} \\\\ & f^{\prime \prime}(x)=2 x-2 \\\\ & f^{\prime \prime}\left(\frac{1}{3}\right)=\frac{2}{3}-2<0 \text { point of maxima }\end{aligned}$
$\begin{aligned} \text { Area }_{\text {red }} & =\int_0^1 f(x) d x \\\\ & =\left[\frac{x^4}{12}-\frac{x^3}{3}+\frac{5 x^2}{18}+\frac{17 x}{36}\right]_0^1 \\\\ & =\frac{1}{12}-\frac{1}{3}+\frac{5}{18}+\frac{17}{36} \\\\ & =\frac{3-12+10+17}{36} \\\\ & =\frac{18}{36}=\frac{1}{2}=0.5 \\\\ \therefore (\text { Area })_{\text {green }} & =1-\frac{1}{2}=0.5\end{aligned}$
(A) $1-h=h-\frac{1}{2} \Rightarrow h=\frac{3}{4}, \frac{3}{4}>\frac{2}{3}$ option (A) is incorrect
(B) $h=\frac{1}{2}-h \Rightarrow h=\frac{1}{4} \Rightarrow$ option (B) is correct.
(C) $\int\limits_0^1 f(x) d x=\frac{1}{2}, \int\limits_0^1 \frac{1}{2} d x=\frac{1}{2} \Rightarrow \int\limits_0^1\left(f(x)-\frac{1}{2}\right) d x=0$
$\Rightarrow h=\frac{1}{2} \Rightarrow$ option (C) is correct.
(D) $\because$ Option (C) is correct $\Rightarrow$ option (D) is also correct.
$\begin{aligned} & \quad f^{\prime}(x)=0 \\\\ & 9 x^2-18 x+5=0 \\\\ & \Rightarrow 9 x^2-15 x-3 x+5=0 \\\\ & \Rightarrow 3 x(3 x-5)-1(3 x-5)=0 \\\\ & \Rightarrow(3 x-5)(3 x-1)=0 \\\\ & \Rightarrow x=\frac{1}{3} \text { or } \frac{5}{3} \\\\ & f^{\prime \prime}(x)=2 x-2 \\\\ & f^{\prime \prime}\left(\frac{1}{3}\right)=\frac{2}{3}-2<0 \text { point of maxima }\end{aligned}$
$\begin{aligned} \text { Area }_{\text {red }} & =\int_0^1 f(x) d x \\\\ & =\left[\frac{x^4}{12}-\frac{x^3}{3}+\frac{5 x^2}{18}+\frac{17 x}{36}\right]_0^1 \\\\ & =\frac{1}{12}-\frac{1}{3}+\frac{5}{18}+\frac{17}{36} \\\\ & =\frac{3-12+10+17}{36} \\\\ & =\frac{18}{36}=\frac{1}{2}=0.5 \\\\ \therefore (\text { Area })_{\text {green }} & =1-\frac{1}{2}=0.5\end{aligned}$
(A) $1-h=h-\frac{1}{2} \Rightarrow h=\frac{3}{4}, \frac{3}{4}>\frac{2}{3}$ option (A) is incorrect
(B) $h=\frac{1}{2}-h \Rightarrow h=\frac{1}{4} \Rightarrow$ option (B) is correct.
(C) $\int\limits_0^1 f(x) d x=\frac{1}{2}, \int\limits_0^1 \frac{1}{2} d x=\frac{1}{2} \Rightarrow \int\limits_0^1\left(f(x)-\frac{1}{2}\right) d x=0$
$\Rightarrow h=\frac{1}{2} \Rightarrow$ option (C) is correct.
(D) $\because$ Option (C) is correct $\Rightarrow$ option (D) is also correct.
Comments (0)
