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be continuous functions which are twice differentiable on the interval (–1, 2). Let the values of f and g at the points –1, 0 and 2 be as given in the following table:
In each of the intervals (–1, 0) and (0, 2) the function (f – 3g)" never vanishes. Then the correct statement(s) is(are)
  • a)
    f;(x) – 3g'(x) = 0 has exactly th r ee solution s in (–1, 0) ∪ (0, 2)
  • b)
    f ' (x) – 3g'(x) = 0 has exactly one solution in (–1, 0)
  • c)
    f ' (x) – 3g'(x) = 0 has exactly one solution in (0, 2)
  • d)
    f ' (x) – 3g'(x) = 0 has exactly two solutions in (–1, 0) and exactly two solutions in (0, 2)
Correct answer is option 'B,C'. Can you explain this answer?
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Let h(x) = f(x) – 3g(x) h(–1) = h(0) = h(2) = 3
∴ By Rolle’s theorem h'(x) = 0 has atleast one solution in (–1, 0) and atleast one solution in (0, 2) But h''(x) never vanishes in (–1, 0) and (0, 2) therefore h'(x) = 0 should have exactly one solution in each interval.
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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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be continuous functions which are twice differentiable on the interval (–1, 2). Let the values of f and g at the points –1, 0 and 2 be as given in the following table:In each of the intervals (–1, 0) and (0, 2) the function (f – 3g)" never vanishes. Then the correct statement(s) is(are)a)f;(x) – 3g'(x) = 0 has exactly th r ee solution s in (–1, 0) ∪ (0, 2)b)f ' (x) – 3g'(x) = 0 has exactly one solution in (–1, 0)c)f ' (x) – 3g'(x) = 0 has exactly one solution in (0, 2)d)f ' (x) – 3g'(x) = 0 has exactly two solutions in (–1, 0) and exactly two solutions in (0, 2)Correct answer is option 'B,C'. Can you explain this answer?
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be continuous functions which are twice differentiable on the interval (–1, 2). Let the values of f and g at the points –1, 0 and 2 be as given in the following table:In each of the intervals (–1, 0) and (0, 2) the function (f – 3g)" never vanishes. Then the correct statement(s) is(are)a)f;(x) – 3g'(x) = 0 has exactly th r ee solution s in (–1, 0) ∪ (0, 2)b)f ' (x) – 3g'(x) = 0 has exactly one solution in (–1, 0)c)f ' (x) – 3g'(x) = 0 has exactly one solution in (0, 2)d)f ' (x) – 3g'(x) = 0 has exactly two solutions in (–1, 0) and exactly two solutions in (0, 2)Correct answer is option 'B,C'. 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 be continuous functions which are twice differentiable on the interval (–1, 2). Let the values of f and g at the points –1, 0 and 2 be as given in the following table:In each of the intervals (–1, 0) and (0, 2) the function (f – 3g)" never vanishes. Then the correct statement(s) is(are)a)f;(x) – 3g'(x) = 0 has exactly th r ee solution s in (–1, 0) ∪ (0, 2)b)f ' (x) – 3g'(x) = 0 has exactly one solution in (–1, 0)c)f ' (x) – 3g'(x) = 0 has exactly one solution in (0, 2)d)f ' (x) – 3g'(x) = 0 has exactly two solutions in (–1, 0) and exactly two solutions in (0, 2)Correct answer is option 'B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for be continuous functions which are twice differentiable on the interval (–1, 2). Let the values of f and g at the points –1, 0 and 2 be as given in the following table:In each of the intervals (–1, 0) and (0, 2) the function (f – 3g)" never vanishes. Then the correct statement(s) is(are)a)f;(x) – 3g'(x) = 0 has exactly th r ee solution s in (–1, 0) ∪ (0, 2)b)f ' (x) – 3g'(x) = 0 has exactly one solution in (–1, 0)c)f ' (x) – 3g'(x) = 0 has exactly one solution in (0, 2)d)f ' (x) – 3g'(x) = 0 has exactly two solutions in (–1, 0) and exactly two solutions in (0, 2)Correct answer is option 'B,C'. Can you explain this answer?.
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