Let f(x) = begincases - 1, - 2 le x

Question

Let f(x) = begincases - 1, - 2 le x < 0 x2 - 1 , 0 le x le 2 endcases and g(x) = |f (x) + f (| x |)| . Then, in the interval (-2, 2), g is : (A) differentiable at all points (B) not differentiable at two points (C) not continuous (D) not differentiable at one point

Tardigrade Admin · Accepted Answer

|f(x) | = begincases 1 &, - 2 le x < 0 1 - x2 , 0 le x < 1 x2 - 1 &, 1 le x le 2 endcases and f(|x|) = x2 - 1 , x ∈ [-2 , 2 ] Hence g(x) = begincases x2 &, x ∈ [-2, 0) 0 , x ∈ [0,1) 2(x2 - 1) &, x ∈ [1, 2 ] endcases It is not differentiable at x = 1

Q. Let $f (x) = {- 1, - 2 \leq x < 0 x^{2} - 1, 0 \leq x \leq 2$ and $g (x) = ∣ f (x) + f (∣ x ∣) ∣$ . Then, in the interval $(- 2, 2), g$ is :

differentiable at all points

not differentiable at two points

not continuous

not differentiable at one point

$∣ f (x) ∣ = ⎩ ⎨ ⎧ 1 1 - x^{2} x^{2} - 1, - 2 \leq x < 0, 0 \leq x < 1, 1 \leq x \leq 2$
and $f (∣ x ∣) = x^{2} - 1, x \in [- 2, 2]$
Hence $g (x) = ⎩ ⎨ ⎧ x^{2} 0 2 (x^{2} - 1), x \in [- 2, 0), x \in [0, 1), x \in [1, 2]$
It is not differentiable at $x = 1$

Q. Let f(x)={−1,−2≤x<0x2−1,0≤x≤2​ and g(x)=∣f(x)+f(∣x∣)∣ . Then, in the interval (−2,2),g is :