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Question: Let K be the set of all real values of x where the function \(f\left( x \right)=\sin |x|-|x|+2\left(...

Let K be the set of all real values of x where the function f(x)=sin⁡∣x∣−∣x∣+2(x−π)cos⁡∣x∣f\left( x \right)=\sin |x|-|x|+2\left( x-\pi \right)\cos |x| is not differentiable. Then the set K is equal to:
A. \left\\{ \pi \right\\}
B. \left\\{ 0 \right\\}
C. ϕ\phi (an empty set)
D. \left\\{ 0,\pi \right\\}

Explanation

Solution

We here have been given a function and we have to find the points of its non differentiability. For thus, we will first break the function at 0 as it contains |x|. Then we will find out its RHL given as RHL=lim⁡h→0f(x+h)−f(x)hRHL=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h} and LHL given as LHL=lim⁡h→0f(x−h)−f(x)−hLHL=\displaystyle \lim_{h \to 0}\dfrac{f\left( x-h \right)-f\left( x \right)}{-h} at x=0. Then, we will see if LHL=RHL, then the function will be continuous. Then, we will see if both the functions we formed after breaking f(x) are a combination of two differentiable functions and hence it will be differentiable everywhere else. Following this process, we will get the required answer.

Complete step by step answer:
We here have been given the function f(x)=sin⁡∣x∣−∣x∣+2(x−π)cos⁡∣x∣f\left( x \right)=\sin |x|-|x|+2\left( x-\pi \right)\cos |x| and we have to check for the points of non differentiability. For this, we will first break the function so that the modulus is eliminated.
Now, we know that:
∣x∣=x if x≥0 ∣x∣=−x if x0 \begin{aligned} & |x|=x\text{ if x}\ge \text{0} \\\ & |x|=-x\text{ if x0} \\\ \end{aligned}
Thus, we will get our function as:
f(x)=sin⁡∣x∣−∣x∣+2(x−π)cos⁡∣x∣f\left( x \right)=\sin |x|-|x|+2\left( x-\pi \right)\cos |x|
\begin{aligned} f\left( x \right)=\left\\{ \begin{matrix} \sin \left( x \right)-\left( x \right)+2\left( x-\pi \right)\cos \left( x \right) & \text{ if }x\ge 0 \\\ \sin \left( -x \right)-\left( -x \right)+2\left( -x-\pi \right)\cos \left( x \right) & \text{ if }x<0 \\\ \end{matrix} \right. \end{aligned}
\begin{aligned} f\left( x \right)=\left\\{ \begin{matrix} \sin x-x+2\left( x-\pi \right)\cos x & \text{ if }x\ge 0 \\\ -\sin x+x-2\left( x-\pi \right)\cos x & \text{ if }x<0 \\\ \end{matrix} \right. \\\ \end{aligned}
Now, we can see that the function breaks at x=0. So first, we will check for the differentiability at x=0.
Now, we know that a function is continuous if the function’s right hand limit (RHL) is equal to its left hand limit (LHL) which is equal to its limit at that point of the function.
So, we will first calculate its RHL.
Now, we know that RHL at any point x in the function is given as:
RHL=lim⁡h→0f(x+h)−f(x)hRHL=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}
Here, x=0 and since h→0h \to 0, x+h>0
Thus, we get the required RHL as:
RHL=lim⁡h→0(sin⁡(x+h)−(x+h)+2((x+h)−π)cos⁡(x+h))−(sin⁡x−x+2(x−π)cos⁡x)h ⇒RHL=lim⁡h→0sin⁡(x+h)−x−h+2xcos⁡(x+π)+2hcos⁡(x+π)−2πcos⁡(x+π)−sin⁡x+x−2xcos⁡x+2πcos⁡xh ⇒RHL=lim⁡h→0sin⁡(x+h)−sin⁡x+2(x+π)(cos⁡(x+h)−cos⁡x)+h+2hcos⁡(x+h)h ⇒RHL=lim⁡h→02cos⁡(x+h2)sin⁡h2−4(x+π)sin⁡(x+h2)sin⁡h2+h+2hcos⁡(x+h)h \begin{aligned} & RHL=\displaystyle \lim_{h \to 0}\dfrac{\left( \sin \left( x+h \right)-(x+h)+2\left( \left( x+h \right)-\pi \right)\cos \left( x+h \right) \right)-\left( \sin x-x+2\left( x-\pi \right)\cos x \right)}{h} \\\ & \Rightarrow RHL=\displaystyle \lim_{h \to 0}\dfrac{\sin \left( x+h \right)-x-h+2x\cos \left( x+\pi \right)+2h\cos \left( x+\pi \right)-2\pi \cos \left( x+\pi \right)-\sin x+x-2x\cos x+2\pi \cos x}{h} \\\ & \Rightarrow RHL=\displaystyle \lim_{h \to 0}\dfrac{\sin \left( x+h \right)-\sin x+2\left( x+\pi \right)\left( \cos \left( x+h \right)-\cos x \right)+h+2h\cos \left( x+h \right)}{h} \\\ & \Rightarrow RHL=\displaystyle \lim_{h \to 0}\dfrac{2\cos \left( x+\dfrac{h}{2} \right)\sin \dfrac{h}{2}-4\left( x+\pi \right)\sin \left( x+\dfrac{h}{2} \right)\sin \dfrac{h}{2}+h+2h\cos \left( x+h \right)}{h} \\\ \end{aligned}Now, separating the limits we get:
⇒RHL=lim⁡h→02cos⁡(x+h2)sin⁡h2h−lim⁡h→04(x+π)sin⁡(x+h2)sin⁡h2h+lim⁡h→0h+2hcos⁡(x+h)h\Rightarrow RHL=\displaystyle \lim_{h \to 0}\dfrac{2\cos \left( x+\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{h}-\displaystyle \lim_{h \to 0}\dfrac{4\left( x+\pi \right)\sin \left( x+\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{h}+\displaystyle \lim_{h \to 0}\dfrac{h+2h\cos \left( x+h \right)}{h}
Now, we know that lim⁡x→0sin⁡xx=1\displaystyle \lim_{x\to 0}\dfrac{\sin x}{x}=1
Using this in the RHL, we get:
⇒RHL=lim⁡h→02cos⁡(x+h2)sin⁡h2h−lim⁡h→04(x+π)sin⁡(x+h2)sin⁡h2h+lim⁡h→0h+2hcos⁡(x+h)h ⇒RHL=lim⁡h→02cos⁡(x+h2)sin⁡h22h2−lim⁡h→04(x+π)sin⁡(x+h2)sin⁡h22h2+lim⁡h→0h+2hcos⁡(x+h)h ⇒RHL=lim⁡h→0(cos⁡(x+h2)−2(x+π)sin⁡(x+h2)+1+2cos⁡(x+h)) \begin{aligned} & \Rightarrow RHL=\displaystyle \lim_{h \to 0}\dfrac{2\cos \left( x+\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{h}-\displaystyle \lim_{h \to 0}\dfrac{4\left( x+\pi \right)\sin \left( x+\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{h}+\displaystyle \lim_{h \to 0}\dfrac{h+2h\cos \left( x+h \right)}{h} \\\ & \Rightarrow RHL=\displaystyle \lim_{h \to 0}\dfrac{2\cos \left( x+\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{2\dfrac{h}{2}}-\displaystyle \lim_{h \to 0}\dfrac{4\left( x+\pi \right)\sin \left( x+\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{2\dfrac{h}{2}}+\displaystyle \lim_{h \to 0}\dfrac{h+2h\cos \left( x+h \right)}{h} \\\ & \Rightarrow RHL=\displaystyle \lim_{h \to 0}\left( \cos \left( x+\dfrac{h}{2} \right)-2\left( x+\pi \right)\sin \left( x+\dfrac{h}{2} \right)+1+2\cos \left( x+h \right) \right) \\\ \end{aligned}
Now, putting h=0 in RHL, we get:
⇒RHL=lim⁡h→0(cos⁡(x+h2)−2(x+π)sin⁡(x+h2)+1+2cos⁡(x+h)) ⇒RHL=cos⁡x−2(x+π)sin⁡x+1+2cos⁡x \begin{aligned} & \Rightarrow RHL=\displaystyle \lim_{h \to 0}\left( \cos \left( x+\dfrac{h}{2} \right)-2\left( x+\pi \right)\sin \left( x+\dfrac{h}{2} \right)+1+2\cos \left( x+h \right) \right) \\\ & \Rightarrow RHL=\cos x-2\left( x+\pi \right)\sin x+1+2\cos x \\\ \end{aligned}
Now, putting x=0 in RHL we get:
⇒RHL=cos⁡x−2(x+π)sin⁡x+1+2cos⁡x ⇒RHL=cos⁡0−2(0+π)sin⁡0+1+2cos⁡0 ⇒RHL=1−2π(0)+1+2(1) ⇒RHL=4 \begin{aligned} & \Rightarrow RHL=\cos x-2\left( x+\pi \right)\sin x+1+2\cos x \\\ & \Rightarrow RHL=\cos 0-2\left( 0+\pi \right)\sin 0+1+2\cos 0 \\\ & \Rightarrow RHL=1-2\pi \left( 0 \right)+1+2\left( 1 \right) \\\ & \Rightarrow RHL=4 \\\ \end{aligned}
Thus, the RHL is 4.
Now, since the function is same for RHD and the limit at that point, we will just calculate the LHL.
Now, we know that LHL at any point is given as:
LHL=lim⁡h→0f(x−h)−f(x)−hLHL=\displaystyle \lim_{h \to 0}\dfrac{f\left( x-h \right)-f\left( x \right)}{-h}
Here, x=0 hence, x-h<0.
Thus, we get our LHL as:
LHL=lim⁡h→0(−sin⁡(x−h)+(x−h)+2((x−h)−π)cos⁡(x−h))−(−sin⁡x+x+2(x−π)cos⁡x)−h ⇒LHL=lim⁡h→0−sin⁡(x−h)+x−h+2xcos⁡(x−h)−2hcos⁡(x−h)−2πcos⁡(x−h)+sin⁡x−x−2xcos⁡x+2πcos⁡x−h ⇒LHL=lim⁡h→0−sin⁡(x−h)+sin⁡x+2(x+π)(cos⁡(x−h)−cos⁡x)+h+2hcos⁡(x−h)−h ⇒LHL=lim⁡h→0−2cos⁡(x−h2)sin⁡h2+4(x+π)sin⁡(x−h2)sin⁡h2+h+2hcos⁡(x−h)−h \begin{aligned} & LHL=\displaystyle \lim_{h \to 0}\dfrac{\left( -\sin \left( x-h \right)+(x-h)+2\left( \left( x-h \right)-\pi \right)\cos \left( x-h \right) \right)-\left( -\sin x+x+2\left( x-\pi \right)\cos x \right)}{-h} \\\ & \Rightarrow LHL=\displaystyle \lim_{h \to 0}\dfrac{-\sin \left( x-h \right)+x-h+2x\cos \left( x-h \right)-2h\cos \left( x-h \right)-2\pi \cos \left( x-h \right)+\sin x-x-2x\cos x+2\pi \cos x}{-h} \\\ & \Rightarrow LHL=\displaystyle \lim_{h \to 0}\dfrac{-\sin \left( x-h \right)+\sin x+2\left( x+\pi \right)\left( \cos \left( x-h \right)-\cos x \right)+h+2h\cos \left( x-h \right)}{-h} \\\ & \Rightarrow LHL=\displaystyle \lim_{h \to 0}\dfrac{-2\cos \left( x-\dfrac{h}{2} \right)\sin \dfrac{h}{2}+4\left( x+\pi \right)\sin \left( x-\dfrac{h}{2} \right)\sin \dfrac{h}{2}+h+2h\cos \left( x-h \right)}{-h} \\\ \end{aligned}Now, separating the limits we get:
⇒LHL=lim⁡h→0−2cos⁡(x−h2)sin⁡h2−h+lim⁡h→04(x+π)sin⁡(x−h2)sin⁡h2−h+lim⁡h→0h+2hcos⁡(x−h)−h\Rightarrow LHL=\displaystyle \lim_{h \to 0}\dfrac{-2\cos \left( x-\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{-h}+\displaystyle \lim_{h \to 0}\dfrac{4\left( x+\pi \right)\sin \left( x-\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{-h}+\displaystyle \lim_{h \to 0}\dfrac{h+2h\cos \left( x-h \right)}{-h}
Again, using the property lim⁡x→0sin⁡xx=1\displaystyle \lim_{x\to 0}\dfrac{\sin x}{x}=1 we get:

& \Rightarrow LHL=\displaystyle \lim_{h \to 0}\dfrac{-2\cos \left( x-\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{-h}+\displaystyle \lim_{h \to 0}\dfrac{4\left( x+\pi \right)\sin \left( x-\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{-h}+\displaystyle \lim_{h \to 0}\dfrac{h+2h\cos \left( x-h \right)}{-h} \\\ & \Rightarrow LHL=\displaystyle \lim_{h \to 0}\dfrac{-2\cos \left( x-\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{-2\dfrac{h}{2}}+\displaystyle \lim_{h \to 0}\dfrac{4\left( x+\pi \right)\sin \left( x-\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{-2\dfrac{h}{2}}+\displaystyle \lim_{h \to 0}\dfrac{h+2h\cos \left( x-h \right)}{-h} \\\ & \Rightarrow LHL=\displaystyle \lim_{h \to 0}\left( \cos \left( x-\dfrac{h}{2} \right)-2\left( x+\pi \right)\sin \left( x-\dfrac{h}{2} \right)+1+2\cos \left( x-h \right) \right) \\\ \end{aligned}$$ Now, putting h=0 in LHL we get: $\begin{aligned} & \Rightarrow LHL=\displaystyle \lim_{h \to 0}\left( \cos \left( x-\dfrac{h}{2} \right)-2\left( x+\pi \right)\sin \left( x-\dfrac{h}{2} \right)+1+2\cos \left( x-h \right) \right) \\\ & \Rightarrow LHL=\cos x-2\left( x+\pi \right)\sin x+1+2\cos x \\\ \end{aligned}$ Now, putting x=0 in LHL we get: $\begin{aligned} & \Rightarrow LHL=\cos x-2\left( x+\pi \right)\sin x+1+2\cos x \\\ & \Rightarrow LHL=\cos \left( 0 \right)-2\left( x+\pi \right)\sin \left( 0 \right)+1+2\cos \left( 0 \right) \\\ & \Rightarrow LHL=1-0+1+2 \\\ & \Rightarrow LHL=4 \\\ \end{aligned}$ Hence, the LHL is 4. Now, we can see that LHL=RHL. Hence, the function is differentiable at x=0. Now, we can see that f(x) is a function combined from trigonometric function and polynomial function. We also know that these two functions are continuous and differentiable everywhere. We also know that the combination of two differentiable functions is also differentiable. Hence, f(x) is differentiable everywhere. Thus, $K=\phi $ (an empty set) **Hence, option (C) is the correct option.** **Note:** There is an easy way to check for LHL and RHL. This is done by calculating f’(x) and then keeping x=x-h for LHL and x=x+h for RHL. This will give us the values of RHL and LHL. It is a short method and should be used in competition type questions.