Let f ( x ) = { 1 − ∣ x ∣ , 0 , ∣ x ∣ ≤ 1 ∣ x ∣ 1 and g ( x ) = f ( x − 1 ) + f ( x + 1 ) . Then the number of points where g is not differentiable is

Question

Let f ( x ) = { 1 − ∣ x ∣ , 0 , ​ ∣ x ∣ ≤ 1 ∣ x ∣ > 1 ​ and g ( x ) = f ( x − 1 ) + f ( x + 1 ) . Then the number of points where g is not differentiable is

TestHub · Accepted Answer

f ( x ) = ⎩ ⎨ ⎧ 1 + x 1 − x 0 i f − 1 ≤ x ≤ 0 i f 0 ≤ x ≤ 1 i f x > 1 or x < − 1 g ( x ) = f ( x − 1 ) + f ( x + 1 ) = ⎩ ⎨ ⎧ 0 1 − ∣ x + 1∣ 1 − ∣ x − 1∣ 0 if x < − 2 if − 2 ≤ x ≤ 0 if 0 ≤ x ≤ 2 if x > 2 Points where f is not differentiable are − 2 , − 1 , 0 , 1 and 2 .

Mathematics - Differential Equation Question with Solution | TestHub

Doubts & Discussion

JEE Main (2028)