UPSC NDA Math Model Paper 10

Here, f(x) = |x - 3|
f(x) = x - 3 when x - 3 ≥ 0 ⇒ x ≥ 3
And f(x) = 3 - x when x - 3 < 0 ⇒ x < 3
For, x = 0, f(x) = 3 - x
Function is not changing at x = 0 (∵ function is linear)
So, f(x) is continuous at x = 0
Now, x = 0, f(x) = 3 - x
⇒ f'(x) = -1
f'(0‌−) = -1 and f'(0‌+) = -1
So, f(x) is differentiable at x = 0
Hence, option (3) is correct.