Let f:[0,1]→[0,1] be the function defined by f(x)=x33−x2+59x+1736Consider 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<f(x)} be called the red region. Let Lh={(x,h)∈S:x∈[0,1]} be the horizontal line drawn at a height h∈[0,1]. Then which of the following statements is(are) true?

f(x)=x33−x2+5x9+1736 f'(x)=x2−2x+59 f'(x)=0 at x=13 in [0,1]AR = Area of Red regionAG= Area of Green region AR=∫01f(x)dx=12Total area = 1⇒AG=12 ∫01f(x)dx=12 AG=AR f(0)=1736 f(1)=1336≈0.36 f(13)=181324≈0.558 (A) Correct when h=34 but h∈[14,23] ⇒(A) is incorrect(B) is correct (C) When h=181324,AR=12,AG<12 h=1336,AR<12,AG=12⇒AR=AG for some h∈(1336,181324)⇒(C) is correct(D) Option (D) is remaining coloured part of option (C), hence option (D) is also correct

Let f:[0,1]→[0,1] be the function defined by f(x)=x33−x2+59x+1736Consider 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<f(x)} be called the red region. Let Lh={(x,h)∈S:x∈[0,1]} be the horizontal line drawn at a height h∈[0,1]. Then which of the following statements is(are) true?

f(x)=x33−x2+5x9+1736 f'(x)=x2−2x+59 f'(x)=0 at x=13 in [0,1]AR = Area of Red regionAG= Area of Green region AR=∫01f(x)dx=12Total area = 1⇒AG=12 ∫01f(x)dx=12 AG=AR f(0)=1736 f(1)=1336≈0.36 f(13)=181324≈0.558 (A) Correct when h=34 but h∈[14,23] ⇒(A) is incorrect(B) is correct (C) When h=181324,AR=12,AG<12 h=1336,AR<12,AG=12⇒AR=AG for some h∈(1336,181324)⇒(C) is correct(D) Option (D) is remaining coloured part of option (C), hence option (D) is also correct

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Let $f : [0, 1] \to [0, 1]$ 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?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

There exists an $h \in [\frac{1}{4}, \frac{2}{3}]$ 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 $h \in [\frac{1}{4}, \frac{2}{3}]$ 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 $h \in [\frac{1}{4}, \frac{2}{3}]$ such that the area of the red region above the line $L_{h}$ equals the area of the green region below the line $L_{h}$

There exists an $h \in [\frac{1}{4}, \frac{2}{3}]$ such that the area of the green region above the line $L_{h}$ equals the area of the green region below the line $L_{h}$

answer is B, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f (x) = \frac{x^{3}}{3} - x^{2} + \frac{5 x}{9} + \frac{17}{36} f' (x) = x^{2} - 2 x + \frac{5}{9} f' (x) = 0 a t x = \frac{1}{3} i n [0, 1]$

$A_{R}$ = Area of Red region
$A_{G} =$ Area of Green region
$A_{R} = \int_{0}^{1} f (x) d x = \frac{1}{2}$
Total area = 1
$\Rightarrow A_{G} = \frac{1}{2} \int_{0}^{1} f (x) d x = \frac{1}{2} A_{G} = A_{R} f (0) = \frac{17}{36} f (1) = \frac{13}{36} \approx 0.36 f (\frac{1}{3}) = \frac{181}{324} \approx 0.558$

(A) Correct when $h = \frac{3}{4}$ but $h \in [\frac{1}{4}, \frac{2}{3}]$
$\Rightarrow (A)$ is incorrect
(B) is correct
(C) When $h = \frac{181}{324}, A_{R} = \frac{1}{2}, A_{G} < \frac{1}{2}$
$h = \frac{13}{36}, A_{R} < \frac{1}{2}, A_{G} = \frac{1}{2}$
$\Rightarrow A_{R} = A_{G}$ for some $h \in (\frac{13}{36}, \frac{181}{324})$
$\Rightarrow (C)$ is correct
(D) Option (D) is remaining coloured part of option (C), hence option (D) is also correct