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QJEEAdv23Madb5d: JEE Advanced 2023 Mathematics - Class 11 JEEAdv23Madb5d Solution: Let be the function defined by . Consider the square region .…

Let f:[0,1]→[0,1]f:[0,1] \rightarrow[0,1] be the function defined by f(x)=x33−x2+59x+1736f(x)=\frac{x^3}{3}-x^2+\frac{5}{9} x+\frac{17}{36} . Consider the square region S=[0,1]×[0,1]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 Lh={(x,h)∈S:x∈[0,1]}L_h=\{(x, h) \in S: x \in[0,1]\} be the horizontal line drawn at a height h∈[0,1]h \in[0,1] . Then which of the following statements is(are) true?

Multiple Choice Options

  • A There exists an h∈[14,23]h \in\left[\frac{1}{4}, \frac{2}{3}\right] such that the area of the green region above the line LhL_h equals the area of the green region below the line LhL_h
  • B There exists an h∈[14,23]h \in\left[\frac{1}{4}, \frac{2}{3}\right] such that the area of the red region above the line LhL_h equals the area of the red region below the line LhL_h
  • C There exists an h∈[14,23]h \in\left[\frac{1}{4}, \frac{2}{3}\right] such that the area of the green region above the line LhL_h equals the area of the red region below the line LhL_h
  • D There exists an h∈[14,23]h \in\left[\frac{1}{4}, \frac{2}{3}\right] such that the area of the red region above the line LhL_h equals the area of the green region below the line LkL_k

Step-by-Step Verified Solution

 Given f:[0,1]amp;→[0,1]f(x)amp;=x33−x2+59x+1736f′(x)amp;=3x23−2x+59\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}

amp;f′(x)=0amp;9x2−18x+5=0amp;⇒9x2−15x−3x+5=0amp;⇒3x(3x−5)−1(3x−5)=0amp;⇒(3x−5)(3x−1)=0amp;⇒x=13 or 53amp;f′′(x)=2x−2amp;f′′(13)=23−2lt;0 point of maxima \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}

JEE Advanced 2023 Paper 1 Online Mathematics - Functions Question 4 English Explanation
 Area red amp;=∫01f(x)dxamp;=[x412−x33+5x218+17x36]01amp;=112−13+518+1736amp;=3−12+10+1736amp;=1836=12=0.5∴( Area )green amp;=1−12=0.5\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=12−h⇒h=14⇒h=\frac{1}{2}-h \Rightarrow h=\frac{1}{4} \Rightarrow option (B) is correct.

(C) ∫01f(x)dx=12,∫0112dx=12⇒∫01(f(x)−12)dx=0\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

⇒h=12⇒\Rightarrow h=\frac{1}{2} \Rightarrow option (C) is correct.

(D) ∵\because Option (C) is correct ⇒\Rightarrow option (D) is also correct.

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