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Continuity and Differentiability

JEE Main 2025 (Online) 29th January Evening ShiftMCQMedium📈 Topic in 18% papers+4−1
Ques. 1JEE Main 2025, 29 January (Evening Shift)

Let the function f(x)= (x2-1 ) |x2-ax+2 |+ |x| be not differentiable at the two points x= =2 and x= . Then the distance of the point ( , ) from the line 12x+5y+10=0 is equal to :

  1. 5
  2. 2
  3. 4
  4. 3

Answer

BONUS

Solution

|g(x)| has a corner where g=0; |x|= x is smooth

x=2 is a zero of x2-ax+2 ⇒ a=3 ⇒ (x-1)(x-2) ⇒ =1

Point (2,1): d= |12(2)+5(1)+10| 122+52= 3913=3

At x=1 the factor (x2-1) vanishes, so f is in fact differentiable there; the data are inconsistent ⇒ question dropped

Answer: Bonus (full marks to all who attempted)

Note: The official NTA answer key gives option (4). Our faculty team finds that no option is fully correct, so this question should be dropped (bonus: full marks to everyone who attempted it).

Keep practising — Continuity and Differentiability

Ques. 2JEE Main 2025, 28 January (Morning Shift)

Let f(x) = cases 3x, & x < 0 \ \1 + x + [x], x + 2[x]\, & 0 x 2 \ 5, & x > 2 cases where [.] denotes greatest integer function. If and are the number of points, where f is not continuous and is not differentiable, respectively, then + equals .

Ques. 3JEE Main 2025, 24 January (Evening Shift)

Let [x] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function f(x)=[x]+|x-2|, -2<x<3, is not continuous and not differentiable. Then m+n is equal to :

  1. 6
  2. 9
  3. 8
  4. 7
Ques. 4JEE Main 2025, 23 January (Morning Shift)

If the function f(x)= cases 2x \ (k1+1)x+ (k2-1)x \, & x<0\\[2mm] 4, & x=0\\[2mm] 2x e ( 2+k1x2+k2x ), & x>0 cases is continuous at x=0, then k12+k22 is equal to :

  1. 5
  2. 10
  3. 20
  4. 8

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