Continuity and Differentiability

JEE Main 2024 (Online) 27th January Morning ShiftNumericalHard📈 Topic in 24% papers+4

Ques. 1JEE 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 .

Answer

2890

Solution

Divide by x-y:

x>y: f(x)-f(y)x-y x- yx-y+1; y x ⇒ f'(x) 1x+1

x<y: sign reverses; y x ⇒ f'(x) 1x+1

∴ f'(x)= 1x+1 ⇒ f' ( 1n2 )=n2+1

n=120(n2+1)= 20 21 416+20=2870+20=2890

Answer: 2890

Keep practising — Continuity and Differentiability

Ques. 2JEE 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,

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

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