The number of points,where the function f: R | vedclass

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

Question

(B) Given function: $f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x^2-5x+4|$.Factorize the quadratic term: $x^2-5x+4 = (x-1)(x-4)$.So,$f(x) = |x-1| \cos

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given function: $f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x^2-5x+4|$.Factorize the quadratic term: $x^2-5x+4 = (x-1)(x-4)$.So,$f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x-1||x-4|$.Factor out $|x-1|$: $f(x) = |x-1| [\cos |x-2| \sin |x-1| + (x-3)|x-4|]$.Let $g(x) = |x-1|$ and $h(x) = \cos |x-2| \sin |x-1| + (x-3)|x-4|$.The function $f(x) = g(x) \cdot h(x)$ is non-differentiable where $g(x)$ is non-differentiable,provided $h(x) 
eq 0$ at those points.$g(x) = |x-1|$ is non-differentiable at $x = 1$.Check $h(1) = \cos |1-2| \sin |1-1| + (1-3)|1-4| = \cos(1) \cdot 0 + (-2) \cdot |-3| = -6 
eq 0$.Thus,$f(x)$ is non-differentiable at $x = 1$.Now check the term $|x-4|$ in $h(x)$. The function $f(x)$ involves $|x-4|$ multiplied by $(x-3)$.At $x = 4$,$f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x-1||x-4|$.Near $x = 4$,the term $|x-1| \cos |x-2| \sin |x-1|$ is differentiable.The term $(x-3)|x-1||x-4|$ is of the form $k(x)|x-4|$,where $k(4) = (4-3)|4-1| = 3 
eq 0$.Since $k(4) 
eq 0$,the function is non-differentiable at $x = 4$.Therefore,the function is non-differentiable at $x = 1$ and $x = 4$. The total number of points is $2$.

The number of points,where the function $f: R \rightarrow R, f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x^2-5x+4|$ is $NOT$ differentiable,is:

Similar Questions

The function given by $y = ||x| - 1|$ is differentiable for all real numbers except the points

Number of points where the function $f(x) = \text{maximum}(\sqrt{2x - x^2}, 2 - x)$ is non-differentiable is:

The function $f(x) = \sin^{-1}(3x - 4x^3)$ is

The value of $m$ for which the function $f(x) = \begin{cases} mx^2, & x \le 1 \\ 2x, & x > 1 \end{cases}$ is differentiable at $x = 1$,is

Suppose that $f(x)$ is a differentiable function such that $f^{\prime}(x)$ is continuous, $f^{\prime}(0)=1$ and $f^{\prime \prime}(0)$ does not exist. Let $g(x)=x f^{\prime}(x)$. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module