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Continuity and Differentiability · Question 5

JEE Advanced 2024 Paper 1MatchingMedium📈 Topic in 33% papers+4−1
1

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)

Answer

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

Solution

(P) g ( 12-x )=2x for 0 x 12, else 0. So h=1 on [0, 12 ], h=0 elsewhere: range \0,1\. P → 5

(Q) h=f: f'(0)= x 0|x| 1x=0 and f is differentiable for x 0, so differentiable on R. f is even (not one-one) and f(x) |x| for large |x|, so bounded below (not onto). Q → 3

(R) h=x-g(x): h=x for x<0 or x> 12, h=3x-1 on [0, 12 ]. Values cover (- ,0) [-1, 12] ( 12, )= R: onto; not one-one (h(- 12)=h( 16)). R → 2

(S) h=g: range [0,1]. S → 4

Answer: (C)

JEE Advanced 2024 Paper 1

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2

Let R denote the set of all real numbers. Let f : R R be an arbitrary function and let g : R R be the function defined by g(x) = xf(x) , for all x R . Then which of the following statements is (are) TRUE ?

  1. The function g is always continuous at x = 0
  2. If f is continuous at x = 0 , then g is differentiable at x = 0
  3. If g is differentiable at x = 0 , then f is continuous at x = 0
  4. If g is differentiable at x = 0 , then x 0 f(x) exists
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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 .

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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 .

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