Let f:[0,1] arrow[0,1] be the function defined by f(x)=\;x³/3-x²+\;5/9x+\;17/36. Consider the square region S=[0,1] ×[0,1]. Let G=\{(x, y) S: y>f(x)\} be called the green region and R=\{(x, y) S: y [JEE Adv 2023 P1]

There exists an h ≤ft[\;1/4, \;2/3] 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 ≤ft[\;1/4, \;2/3] 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 ≤ft[\;1/4, \;2/3] 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 ≤ft[\;1/4, \;2/3] such that the area of the red region above the line L_h equals the area of the green region below the line L_h

Test Index

Functions

Section: Mathematics

Question : 8 of 30

Marks: +1, -0

Let

f:[0,1] \rightarrow[0,1]

f(x)=\;\frac{x^3}{3}-x^2+\;\frac{5}{9}x+\;\frac{17}{36}

S=[0,1] \times[0,1]

G=\{(x, y) \in S: y>f(x)\}

R=\{(x, y) \in S: y< f(x)\}

L_h=\{(x, h) \in S: x \in[0,1]\}

h \in[0,1]

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 $L_h$

Solution:

f(x)=\;\frac{x^3}{3}-x^2+\;\frac{5}{9}x+\;\frac{17}{36}, f'(x)=x^2-2x+\;\frac{5}{9}

For maxima/minima,

f'(x)=0 \Rightarrow x=\;\frac{1}{3}

A_R=\int\limits_{0}^{1} f(x) dx=\;\frac{1}{2} \Rightarrow A_G=\;\frac{1}{2}

(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) dx=\;\frac{1}{2}, \int\limits_{0}^{1} \;\frac{1}{2} dx=\;\frac{1}{2} \Rightarrow \int\limits_{0}^{1} \left(f(x)-\;\frac{1}{2}\right) dx=0

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

(D)

\because

Option (C) is correct

\Rightarrow

option (D) is also correct.

Go to Question:

Prev Question

Next Question