Let f(x) = {x2| cosπx |,x≠00,x=0,x∈R, then f is

Let f(x) = \left{\right. x^{2} \left|\right. \textrm{ } cos \frac{\pi}{x} \textrm{ } \left|\right. , x \neq 0 \\ 0 , x = 0 , x \in R , then f is

Engineering

Mathematics

Concept of Derivative

Question

Let f(x) = ${\begin{cases} x^{2} | \cos \frac{π}{x} |, x \neq 0 \\ 0, x = 0 \end{cases}, x \in R,$

then f is

differentiable at x = 0 but not differentiable at x = 2

differentiable both at x = 0 and at x = 2

not differentiable at x = 0 but differentiable at x = 2

differentiable neither at x = 0 nor at x = 2

$f (x) = {\begin{cases} x^{2} | \cos \frac{π}{x} |, x \neq 0 \\ 0, x = 0 \end{cases}, x \in R,$

At x = 0

$LHD = \underset{h \to 0}{Lim} \frac{f (0 - h) - f (0)}{- h} = \underset{h \to 0}{Lim} \frac{h^{2} | \cos \frac{π}{h} |}{- h} = 0$

$RHD = \underset{h \to 0}{Lim} \frac{f (0 + h) - f (0)}{h} = \underset{h \to 0}{Lim} \frac{h^{2} | \cos \frac{π}{h} |}{h} = 0$

$f (x) = {\begin{cases} - x^{2} \cos \frac{π}{x}, 2 - h < x < 2 \\ x^{2} \cos \frac{π}{x}, 2 < x < 2 + h \end{cases}$

$f' (x) = {\begin{cases} - 2 xcos \frac{π}{x} - πsin \frac{π}{x}, 2 - h < x < 2 \\ 2 xcos \frac{π}{x} + πsin \frac{π}{x}, 2 < x < 2 + h \end{cases}$

LHD = – $π$ ; RHD = $π$

$∴$ f(x) is differentiable at x = 0 but not derivable at x = 2

Understanding the Problem

We are given a piecewise function:

$f (x) = {\begin{cases} x^{2} | \cos (\frac{π}{x}) | & if x \neq 0 \\ 0 & if x = 0 \end{cases}, for all x \in R$

We need to determine if this function is differentiable at the points x = 0 and x = 2.

A function $f$ is differentiable at a point $a$ if the following limit exists:

$f^{'} (a) = \lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$

This limit is called the derivative of $f$ at $a$ .

Step 1: Analyze Differentiability at x = 0

We apply the definition of the derivative at x = 0.

$f^{'} (0) = \lim_{h \to 0} \frac{f (0 + h) - f (0)}{h} = \lim_{h \to 0} \frac{f (h) - 0}{h} = \lim_{h \to 0} \frac{h^{2} \cdot | \cos (\frac{π}{h}) |}{h} = \lim_{h \to 0} h \cdot | \cos (\frac{π}{h}) |$

Now, we analyze this limit. The cosine term, $| \cos (\frac{π}{h}) |$ , oscillates rapidly between 0 and 1 as $h$ approaches 0. However, it is bounded:

$0 \leq | \cos (\frac{π}{h}) | \leq 1$

We also have the term $h$ which approaches 0. We can use the Squeeze Theorem.

$0 \leq | h \cdot | \cos (\frac{π}{h}) | | = | h | \cdot | \cos (\frac{π}{h}) | \leq | h | \cdot 1 = | h |$

Since $\lim_{h \to 0} 0 = 0$ and $\lim_{h \to 0} | h | = 0$ , by the Squeeze Theorem:

$\lim_{h \to 0} h \cdot | \cos (\frac{π}{h}) | = 0$

Therefore, the derivative at x=0 exists and $f^{'} (0) = 0$ . The function is differentiable at x = 0.

Step 2: Analyze Differentiability at x = 2

For any point other than 0, we can use standard differentiation rules. At x=2, the function is defined by $f (x) = x^{2} \cdot | \cos (\frac{π}{x}) |$ .

We need to check if the derivative exists here. The potential issue is the absolute value function, which is not differentiable where its argument is zero. Let's find where $\cos (\frac{π}{x}) = 0$ .

$\cos (\frac{π}{x}) = 0 \Rightarrow \frac{π}{x} = \frac{π}{2} + n π,$

Detailed Solution

Understanding the Problem

Step 1: Analyze Differentiability at x = 0

Step 2: Analyze Differentiability at x = 2

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