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Continuity and Differentiability

JEE Main 2024 (Online) 29th January Evening ShiftMCQMedium📈 Topic in 12% papers+4−1
Ques. 1JEE Main 2024, 29 January (Evening Shift)

Let y= e ( 1-x21+x2 ), -1<x<1. Then at x= 12, the value of 225 (y'-y'' ) is equal to

  1. 732
  2. 736
  3. 742
  4. 746

Answer

(2) 736

Solution

y= e(1-x2)- e(1+x2)

y'= -2x1-x2- 2x1+x2= -4x1-x4

y''= -4(1-x4)-4x 4x3(1-x4)2= -4(1+3x4)(1-x4)2

At x= 12: 1-x4= 1516, 1+3x4= 1916

y'=- 3215, y''=- 194 256225=- 1216225

225(y'-y'')=-480+1216=736

Answer: (2)

Keep practising — Continuity and Differentiability

Ques. 2JEE Main 2024, 27 January (Morning Shift)

Consider the function. f(x)= cases a (7x-12-x2 )b |x2-7x+12 |, & x<3\\[2ex] 2 (x-3)x-[x], & x>3\\[2ex] b, & x=3, cases where [x] denotes the greatest integer less than or equal to x. If S denotes the set of all ordered pairs (a, b) such that f(x) is continuous at x=3, then the number of elements in S is :

  1. Infinitely many
  2. 4
  3. 2
  4. 1
Ques. 3JEE Main 2024, 27 January (Morning Shift)

Let f(x)=x3+x2 f'(1)+x f''(2)+f'''(3), x . Then f'(10) is equal to .

Ques. 4JEE Main 2024, 27 January (Morning Shift)

Let for a differentiable function f:(0, ) , f(x)-f(y) e ( xy )+x-y, x, y (0, ). Then n=120f' ( 1n2 ) is equal to .

Ques. 5JEE Main 2024, 27 January (Evening Shift)

Consider the function f:(0,2) defined by f(x)= x2+ 2x and the function g(x) defined by g(x)= cases \f(t)\, & 0<t x and 0<x 1\ 32+x, & 1<x<2 cases. Then,

  1. g is continuous but not differentiable at x=1
  2. g is continuous and differentiable for all x (0,2)
  3. g is not continuous for all x (0,2)
  4. g is neither continuous nor differentiable at x=1

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