Solved Examples for JEE: Differentiability | Mock Tests for JEE Main and Advanced 2025 PDF Download

Illustration 1: Let [.] denotes the greatest integer function and f(x) = [tan²x], then does the limit exist or is the function differentiable or continuous at 0? (1999)
Solution: Given f(x) = [tan²x]
Now, -45°< x < 45°
tan(-45°)< tanx < tan45°
-tan 45°< tan x < tan 45°
-1< tan x <1
So, 0 <tan²x < 1
[tan²x] = 0
So, f(x) is zero for all values of x form x = -45° to 45°.
Hence, f is continuous at x =0 and f is also differentiable at 0 and has a value zero.

Illustration 2: A function is defined as follows:
f(x)= x³  , x²< 1
x   , x²≥ 1
Discuss the differentiability of the function at x=1.
Solution:We have R.H.D. = Rf'(1)
= limh→0 (f(1-h)-f(1))/h
= limh→0 (1+h-1)/h = 1
and L.H.D. = Lf'(1)= lim_h→0 (f(1-h)-f(1))/(-h)
= lim_h→0 ((1-h)3-1)/(-h)
= lim_h→0 (3-3h+h²) = 3
?Rf'(1)≠ Lf'(1)⇒ f(x) is not differentiable at x=1.

Illustration 3:If y = (sin-1x)² + k sin-1x, show that (1-x²) (d² y)/dx² - x dy/dx = 2
Solution: Here y = (sin-1x)² + k sin-1x.
Differentiating both sides with respect to x, we have
dy/dx = 2(sin-1 x)/√(1-x² ) + k/√(1-x² )
⇒(1-x² ) (dy/dx)² = 4y + k²
Differentiating this with respect to x, we get
(1-x²) 2 dy/dx.(d² y)/(dx² ) - 2x (dy/dx)² = 4(dy/dx)
⇒(1-x² ) ( d² y)/dx² -x dy/dx = 2