PASSAGE - 2Consider the functions defined implicitly by the equation y3 andndash; 3y + x = 0 on various intervals in the real line. If x andisin;(-andinfin;, - 2) andcup; (2,andinfin;) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x andisin;(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q. a)2g (andndash;1)b)0c)andndash;2g (1)d)2g (1)Correct answer is option 'D'. Can you explain this answer?

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PASSAGE - 2

Consider the functions defined implicitly by the equation y³ – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.

a)
2g (–1)
b)
0
c)
–2g (1)
d)
2g (1)

Correct answer is option 'D'. Can you explain this answer?

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Test: Comprehension Based Questions: Definite Integrals and Applications of Integrals

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PASSAGE - 2Consider the functions defined implicitly by the equation y...

For y = g(x), we have y³ – 3y + x = 0
⇒ [g (x) ]3 - 3[ g (x)] + x=0 ...(1)
Putting x = –x, we get
⇒ [g (-x)]3 - 3[ g (-x)] - x = 0 ...(2)
Adding equations (1) and (2) we get

For g(0) = 0, we should have g(x) + g(–x) = 0
[∵ From other factor we get g(0) = ±√3]

⇒ g(x) is an odd function

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PASSAGE - 2Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q.

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PASSAGE - 2Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q.The area of the region bounded by the curve y = f (x), the x-axis, and the lines x = a and x = b, where -∞ < a <b <-2 , is

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Not differentiable then S 2019 won is an empty set to equals minus 2 minus 1 0 1 2 3 equals minus 2 minus 1 1 2 4 equals minus 2 2 ANS 2 solution FX is not differentiable at negative 2 minus 1 and 2 s equals minus 2 minus 1 0 1 2 qu 21 let F be a differentiable function such than FXX greater than 0 and F14 then 2019 1 exists and equals 4/7 2 exists and equals 4 3 does not exist 4 exists and equals 0 ANS 2 solutions reply equals FX solution of differential equation 22 let's denote the greatest integer less than or equal to 10 spend 2019 equals pi plus 1 11 equals 0 and 1 solution just since it does not exist You 23 2019-102 like four for one ANS 4 solution huge 24 let F equals RB differentiable at CR and F C equals 0 If GX equals FX then at X equals C G is 20191 not differentiable if FC equals 0 to differentiable if FC03 differentiable if FC equals 04 not differentiable ANS 3 solution Q25 is continuous at x equals 0 than the ordered pair PQ is equal to 2019 ANS 3 solution Q26 LED FRRB a continuously differentiable function such that F2 equals 6 and F2 equals 148 2019 1 18/24 3/12/436 ANS 1 solution Q27 2019/1424 square root 2 3/8 square root 248 A for solution Q28 2019 ANS 2 solution huge 29 if the function has to find on is continuous then K is equal to 2019 1 2 2 1 have 3 1 4 1 / square Benz FX is continuous then Q30 Let FX equals 15 minus X 10 X are then the set of all values of X at This is not differentiable is 2019-15 to 10 15 3 5 10 15 24 10 ANS 1 solution Since FX equals 15 minus 10X GX equals FFX equals 15 minus 10 minus 15 minus 10X equals 15 10X 5 then the points were function GX is non-differential or 10 X equals 0 and 10 X equals 5X equals 10 and X 10 equals plus minus 5 X equals 10 and X equals where it's denotes the greatest integer function then 2019 one F is continuous at X equals 4 ANI One solution Q32 if the function is continuous at x equals 5 then the value of A B is 2019 ANS 4 solution function is continuous at x equals 5 LHL equals RHL5 minus pi B plus 3 equals 5 minus pi A plus 1 Q3 and 1914 square root 22 square root 232 square root 244 ANS 1 solution Q34 let FRRB a differential function satisfying F3 plus F2 equals 0 then is equal to 2019 1 1 2 E 1 3 4 E2ANS 1 solution Q35 if FRR is a differentiable function and F2 equals 6 then 2019/124F222F230412F2 ANS 4 solution Using L hospital rule and Live NetSerum we get putting X = 2 2 F2 F2 = 12 F2 F2 = 6 Q36 2019/18/33 seconds 4/3 AMS One solution GX = FX + FX Then in the Interval - 2/2 G is one differentiable at all points two not continuous three not differentiable at two points four not differentiable at one point ANS 4 solution GX is non-differentiable at x equals 1 GX is not differentiable at one point Q38 let KB the set of all real values of x were the function fx = sin x minus x plus 2x minus pi cos x is not differentiable Then the set k is equal to 2019 empty set 2 pi 3 0 4 0 pi ANS 1 solution Then function f x is differentiable for l Related: JEE Main Previous year questions (2016-20): Limits, Continuity and Differentiability?

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PASSAGE - 2Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q. a)2g (–1)b)0c)–2g (1)d)2g (1)Correct answer is option 'D'. Can you explain this answer?

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PASSAGE - 2Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q. a)2g (–1)b)0c)–2g (1)d)2g (1)Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about PASSAGE - 2Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q. a)2g (–1)b)0c)–2g (1)d)2g (1)Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for PASSAGE - 2Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q. a)2g (–1)b)0c)–2g (1)d)2g (1)Correct answer is option 'D'. Can you explain this answer?.

Solutions for PASSAGE - 2Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q. a)2g (–1)b)0c)–2g (1)d)2g (1)Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of PASSAGE - 2Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q. a)2g (–1)b)0c)–2g (1)d)2g (1)Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of PASSAGE - 2Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q. a)2g (–1)b)0c)–2g (1)d)2g (1)Correct answer is option 'D'. Can you explain this answer?, a detailed solution for PASSAGE - 2Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q. a)2g (–1)b)0c)–2g (1)d)2g (1)Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of PASSAGE - 2Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q. a)2g (–1)b)0c)–2g (1)d)2g (1)Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice PASSAGE - 2Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q. a)2g (–1)b)0c)–2g (1)d)2g (1)Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.

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