Number of points where the function f(x) = (x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The function is $f(x) = (x^2 - 1) |(x - 2)(x + 1)| + \sin(|x|)$.First,analyze the points where the absolute value expressions are zero: $x = 2, -1

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The function is $f(x) = (x^2 - 1) |(x - 2)(x + 1)| + \sin(|x|)$.First,analyze the points where the absolute value expressions are zero: $x = 2, -1, 0$.$1$. At $x = 2$: $f(x) = (x^2 - 1) |x - 2| |x + 1| + \sin(|x|)$. Since $|x - 2|$ is not differentiable at $x = 2$ and the other terms are non-zero at $x = 2$,$f(x)$ is not differentiable at $x = 2$.$2$. At $x = -1$: $f(x) = (x^2 - 1) |x - 2| |x + 1| + \sin(|x|) = (x - 1)(x + 1) |x - 2| |x + 1| + \sin(|x|) = (x - 1) |x - 2| (x + 1) |x + 1| + \sin(|x|)$. Since $(x + 1) |x + 1|$ is differentiable at $x = -1$,the product is differentiable at $x = -1$.$3$. At $x = 0$: The term $\sin(|x|)$ is not differentiable at $x = 0$ because the left-hand derivative is $-\cos(0) = -1$ and the right-hand derivative is $\cos(0) = 1$. The other part $(x^2 - 1) |x^2 - x - 2|$ is differentiable at $x = 0$. Thus,$f(x)$ is not differentiable at $x = 0$.Therefore,the function is not differentiable at $x = 0$ and $x = 2$. The total number of points is $2$.

Number of points where the function $f(x) = (x^2 - 1) | x^2 - x - 2 | + \sin(|x|)$ is not differentiable,is

Similar Questions

Let $f(x) = 15 - |x - 10|; x \in R$. Then the set of all values of $x$,at which the function $g(x) = f(f(x))$ is not differentiable,is

If $f$ is a real-valued differentiable function satisfying $|f(x) - f(y)| \le (x - y)^2$ for all $x, y \in R$ and $f(0) = 0$,then $f(1)$ is equal to:

Assertion $(A)$: If $f(x)$ is not continuous at $x=a$,then it is not differentiable at $x=a$.
Reason $(R)$: If $f(x)$ is differentiable at a point,then it is continuous at that point.

Let $f(x) = x |\sin x|$,$x \in R$. Then,

Let $f(x) = |x - \alpha| + |x - \beta|$,where $\alpha$ and $\beta$ are the roots of the equation $x^2 - 3x + 2 = 0$. Then the number of points in $[\alpha, \beta]$ at which $f$ is not differentiable is

Vedclass Test Series

Exam Paper Generator

Online Exam Module