JEE Advanced 2023 · Paper 1 · Mathematics Q3

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MathematicsMultiple correct+4 / -2Question 3 of 51

Let $f : [0,1] \to [0,1]$ be the function defined by $f(x) = \dfrac{x^3}{3} - x^2 + \dfrac59x + \dfrac{17}{36}$ . 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[\dfrac14, \dfrac23\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[\dfrac14, \dfrac23\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[\dfrac14, \dfrac23\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[\dfrac14, \dfrac23\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

Answer and worked solution

Official answer: B, C, D

Size of the two regions.
$\int_0^1 f\,dx=\frac1{12}-\frac13+\frac5{18}+\frac{17}{36}=\frac12,$
so the red and green regions each have area $\tfrac12$ .

Range of $f$ . $f'(x)=x^2-2x+\tfrac59$ vanishes at $x=\tfrac13$ in $[0,1]$ . So $f$ rises from $f(0)=\tfrac{17}{36}\approx0.47$ to $f(\tfrac13)\approx0.56$ , then falls to $f(1)=\tfrac{13}{36}\approx0.36$ . Hence $\tfrac14<f(x)<\tfrac23$ on $[0,1]$ .

Write $G_a,G_b,R_a,R_b$ for the green/red areas above/below $L_h$ .

(A) False. $G_a=G_b$ means $G_a=\tfrac14$ . But for $h\le\tfrac23$ , $G_a\ge$ area of $\{y>\tfrac23\}=\tfrac13>\tfrac14$ .

(B) True. At $h=\tfrac14$ the whole strip below the line is red, so $R_b=\tfrac14$ and $R_a=\tfrac12-\tfrac14=\tfrac14$ .

(C) True. $G_a=(1-h)-R_a$ , so $G_a-R_b=1-h-(R_a+R_b)=\tfrac12-h$ . This is zero at $h=\tfrac12$ .

(D) True. Similarly $R_a-G_b=R_a-(h-R_b)=\tfrac12-h$ , which is zero at $h=\tfrac12$ .

Topics:Area between curves Continuity Properties of definite integrals

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JEE Advanced 2023 · Paper 1, Mathematics question 3