The function f(x) = a sin|x| + be|x| is differentiable at x = 0 when

Question

The function f(x) = a sin|x| + be|x| is differentiable at x = 0 when (A) 3a + b = 0 (B) 3a - b = 0 (C) a + b = 0 (D) a - b = 0

Tardigrade Admin · Accepted Answer

Given, f(x)=a sin |x|+b e|x| We know that sin |x| and e|x| is not differentiable at x=0 Therefore, for f(x) to differentiable at x=0, we must have a=b=0 ∴ a+b=0

Q. The function $f (x) = a sin ∣ x ∣ + b e^{∣ x ∣}$ is differentiable at $x = 0$ when

$3 a + b = 0$

$3 a - b = 0$

$a + b = 0$

$a - b = 0$

Given, $f (x) = a sin ∣ x ∣ + b e^{∣ x ∣}$
We know that $sin ∣ x ∣$ and $e^{∣ x ∣}$ is not differentiable at $x = 0$
Therefore, for $f (x)$ to differentiable at $x = 0$ ,
we must have $a = b = 0$
$∴ a + b = 0$

Q. The function f(x)=asin∣x∣+be∣x∣ is differentiable at x=0 when

3a+b=0

3a−b=0

a+b=0

a−b=0