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Continuity and Differentiability · Question 3

JEE Advanced 2026 Paper 2NumericalHard📈 Topic in 17% papers+4
1

For a real number , let [ ] denote the greatest integer less than or equal to . For a finite set S, let |S| denote the number of elements in the set S. Consider the functions f:(-3,3) (- , ) and g:(-3,3) (- , ) defined by f(x)=[x3] e (1+ 2 ( (x-[x]) ) ) and g(x)=x3 2 ( e(1+x-[x]) ). Let A=\x (-3,3): f is discontinuous at x\ and B=\x (-3,3): g is discontinuous at x\. Then the value of |A|+2|B|-|A B| is .

Answer

56

Solution

2( (x-[x]))= 2( x), so h(x)= e(1+ 2 x) is continuous and h=0 only at integers.

[x3] jumps where x3=k, k=-26, ,26: 53 points. At x=0, 1, 2 (x3 integer and x integer) h=0, so f is continuous there.

|A|=53-5=48 (none of them is an integer).

For g: at an integer n, right limit =n3 20=0, left limit =n3 2( e2) 0 unless n=0. Elsewhere g is continuous.

B=\-2,-1,1,2\, |B|=4, A B= .

|A|+2|B|-|A B|=48+8-0=56.

Answer: 56

JEE Advanced 2026 Paper 2

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2

Let R denote the set of all real numbers. Let f: R R and g: R (0,4) be functions defined by f(x)= e(x2+2x+4), and g(x)= 41+e-2x. Define the composite function f g-1 by (f g-1)(x)=f (g-1(x) ), where g-1 is the inverse of the function g. Then the value of the derivative of the composite function f g-1 at x=2 is .

JEE Advanced 2025 Paper 2
3

Let f: R R and g: R R be functions defined by f(x)= cases x|x| ( 1x ), & x 0, \ 0, & x=0, cases and g(x)= cases 1-2x, & 0 x 12, \ 0, & otherwise. cases Let a,b,c,d R. Define the function h: R R by h(x)=a f(x)+b (g(x)+g ( 12-x ) )+c (x-g(x) )+d g(x), x R. Match each entry in List-I to the correct entry in List-II.

  1. (P) → (4) (Q) → (3) (R) → (1) (S) → (2)
  2. (P) → (5) (Q) → (2) (R) → (4) (S) → (3)
  3. (P) → (5) (Q) → (3) (R) → (2) (S) → (4)
  4. (P) → (4) (Q) → (2) (R) → (1) (S) → (3)
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