Limits, Continuity & Differentiability Question - None

Question

Let $ f(x) = x|x| $ and $ g(x) = \sin x $.

Statement-1: $ g \circ f $ is differentiable at $ x = 0 $ and its derivative is continuous at that point.

Statement-2: $ g \circ f $ is twice differentiable at $ x = 0 $.

Accepted Answer

Step 1: Find f(x) and its derivatives.

f(x) = x|x| = {x² (x≥0), -x² (x<0)}

f'(x) = {2x (x>0), -2x (x<0)} = 2|x|

f'(0) = 0 (left and right derivatives both equal 0)

Step 2: Check if g∘f is differentiable at x=0.

By chain rule: (g∘f)'(0) = g'(f(0))·f'(0) = g'(0)·0 = cos(0)·0 = 0 ✓

(g∘f) is differentiable at x=0. Statement-1 first part: TRUE

Step 3: Check continuity of (g∘f)' at x=0.

(g∘f)'(x) = g'(f(x))·f'(x) = cos(x|x|)·2|x|

As x→0: (g∘f)'(x) = cos(x|x|)·2|x| → cos(0)·0 = 0 = (g∘f)'(0) ✓

The derivative is continuous at x=0. Statement-1 second part: TRUE

Step 4: Check if g∘f is twice differentiable at x=0.

For x>0: (g∘f)'(x) = cos(x²)·2x

For x<0: (g∘f)'(x) = cos(x²)·2x

(g∘f)''(x) requires differentiating cos(x²)·2x, which involves f'(x)=2|x|.

Since f'(x)=2|x| is not differentiable at x=0 (left derivative = -2, right derivative = +2), (g∘f)' is not differentiable at x=0. Statement-2: FALSE

∴ Answer: A (Statement-1 is true, Statement-2 is false)

Question:

Step-by-Step Solution

Master Limits, Continuity & Differentiability with Mathbee