Let S be the set of all points in (– \pi , \pi ) at which the function, f(x) = minsin x, cos x is not differentiable. Then S is a subset of which of the following ?

\ - 3\pi \over 4, - \pi \over 4,3\pi \over 4,\pi \over 4 \

\ - \pi \over 2, - \pi \over 4,\pi \over 4,\pi \over 2 \

\ - 3\pi \over 4, - \pi \over 2,\pi \over 2,3\pi \over 4 \

\ - \pi \over 4,0,\pi \over 4 \

Limits Continuity and Differentiability PYQ — JEE Main 2019 · 12 Jan · Shift 1 · Q37 with Solution

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Correct answer: D

We need points in $(-\pi,\pi)$ where $f(x)=\min\{\sin x,\cos x\}$ is not differentiable.
The minimum of two differentiable functions can fail to be differentiable only at points where the two functions are equal and the smaller branch switches.

So first solve: $\sin x=\cos x$ $\tan x=1$ Hence, $x=\frac{\pi}{4}+n\pi$ In the interval $(-\pi,\pi)$ , this gives $x=-\frac{3\pi}{4},\ \frac{\pi}{4}.$

Now determine which function is the minimum on either side of these points.

At $x=\frac{\pi}{4}$

For $x<\frac{\pi}{4}$ , take $x=0$ : $\sin 0=0,\quad \cos 0=1$ so $\min\{\sin x,\cos x\}=\sin x$ near the left side.
For $x>\frac{\pi}{4}$ , take $x=\frac{\pi}{2}$ : $\sin \frac{\pi}{2}=1,\quad \cos \frac{\pi}{2}=0$ so $\min\{\sin x,\cos x\}=\cos x$ near the right side.

Thus the branch changes from $\sin x$ to $\cos x$ at $x=\frac{\pi}{4}$ .

Left derivative: $f'_-(\tfrac{\pi}{4})=(\sin x)'_{x=\pi/4}=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}$ Right derivative: $f'_+(\tfrac{\pi}{4})=(\cos x)'_{x=\pi/4}=-\sin\frac{\pi}{4}=-\frac{1}{\sqrt{2}}$ These are unequal, so $f$ is not differentiable at $x=\frac{\pi}{4}$ .

At $x=-\frac{3\pi}{4}$

For $x< -\frac{3\pi}{4}$ , take $x=-\pi$ (or nearby): $\sin(-\pi)=0,\quad \cos(-\pi)=-1$ so the minimum is $\cos x$ .
For $x> -\frac{3\pi}{4}$ , take $x=-\frac{\pi}{2}$ : $\sin\left(-\frac{\pi}{2}\right)=-1,\quad \cos\left(-\frac{\pi}{2}\right)=0$ so the minimum is $\sin x$ .

Thus the branch changes from $\cos x$ to $\sin x$ at $x=-\frac{3\pi}{4}$ .

Left derivative:

\left(-\tfrac{3\pi}{4}\right)=(\cos x)'_{x=-3\pi/4}=-\sin\left(-\frac{3\pi}{4}\right)=\frac{1}{\sqrt{2}}$$ Right derivative: $$f'_+\left(-\tfrac{3\pi}{4}\right)=(\sin x)'_{x=-3\pi/4}=\cos\left(-\frac{3\pi}{4}\right)=-\frac{1}{\sqrt{2}}$$ These are unequal, so $f$ is **not differentiable** at $x=-\frac{3\pi}{4}$. 5. Therefore, $$S=\left\{-\frac{3\pi}{4},\frac{\pi}{4}\right\}$$ Now check which option contains this set as a subset. - A: $\left\{-\frac{\pi}{2},-\frac{\pi}{4},\frac{\pi}{4},\frac{\pi}{2}\right\}$ does **not** contain $-\frac{3\pi}{4}$. - B: $\left\{-\frac{3\pi}{4},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{4}\right\}$ does **not** contain $\frac{\pi}{4}$. - C: $\left\{-\frac{\pi}{4},0,\frac{\pi}{4}\right\}$ does **not** contain $-\frac{3\pi}{4}$. - D: $\left\{-\frac{3\pi}{4},-\frac{\pi}{4},\frac{3\pi}{4},\frac{\pi}{4}\right\}$ contains both $-\frac{3\pi}{4}$ and $\frac{\pi}{4}$. Hence the correct option is **D**.

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View written solutionFree

Correct answer: D

We need points in $(-\pi,\pi)$ where $f(x)=\min\{\sin x,\cos x\}$ is not differentiable.
The minimum of two differentiable functions can fail to be differentiable only at points where the two functions are equal and the smaller branch switches.

So first solve: $\sin x=\cos x$ $\tan x=1$ Hence, $x=\frac{\pi}{4}+n\pi$ In the interval $(-\pi,\pi)$ , this gives $x=-\frac{3\pi}{4},\ \frac{\pi}{4}.$

Now determine which function is the minimum on either side of these points.

At $x=\frac{\pi}{4}$

For $x<\frac{\pi}{4}$ , take $x=0$ : $\sin 0=0,\quad \cos 0=1$ so $\min\{\sin x,\cos x\}=\sin x$ near the left side.
For $x>\frac{\pi}{4}$ , take $x=\frac{\pi}{2}$ : $\sin \frac{\pi}{2}=1,\quad \cos \frac{\pi}{2}=0$ so $\min\{\sin x,\cos x\}=\cos x$ near the right side.

Thus the branch changes from $\sin x$ to $\cos x$ at $x=\frac{\pi}{4}$ .

At $x=-\frac{3\pi}{4}$

For $x< -\frac{3\pi}{4}$ , take $x=-\pi$ (or nearby): $\sin(-\pi)=0,\quad \cos(-\pi)=-1$ so the minimum is $\cos x$ .
For $x> -\frac{3\pi}{4}$ , take $x=-\frac{\pi}{2}$ : $\sin\left(-\frac{\pi}{2}\right)=-1,\quad \cos\left(-\frac{\pi}{2}\right)=0$ so the minimum is $\sin x$ .

Thus the branch changes from $\cos x$ to $\sin x$ at $x=-\frac{3\pi}{4}$ .

Left derivative:

\left(-\tfrac{3\pi}{4}\right)=(\cos x)'_{x=-3\pi/4}=-\sin\left(-\frac{3\pi}{4}\right)=\frac{1}{\sqrt{2}}$$ Right derivative: $$f'_+\left(-\tfrac{3\pi}{4}\right)=(\sin x)'_{x=-3\pi/4}=\cos\left(-\frac{3\pi}{4}\right)=-\frac{1}{\sqrt{2}}$$ These are unequal, so $f$ is **not differentiable** at $x=-\frac{3\pi}{4}$. 5. Therefore, $$S=\left\{-\frac{3\pi}{4},\frac{\pi}{4}\right\}$$ Now check which option contains this set as a subset. - A: $\left\{-\frac{\pi}{2},-\frac{\pi}{4},\frac{\pi}{4},\frac{\pi}{2}\right\}$ does **not** contain $-\frac{3\pi}{4}$. - B: $\left\{-\frac{3\pi}{4},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{4}\right\}$ does **not** contain $\frac{\pi}{4}$. - C: $\left\{-\frac{\pi}{4},0,\frac{\pi}{4}\right\}$ does **not** contain $-\frac{3\pi}{4}$. - D: $\left\{-\frac{3\pi}{4},-\frac{\pi}{4},\frac{3\pi}{4},\frac{\pi}{4}\right\}$ contains both $-\frac{3\pi}{4}$ and $\frac{\pi}{4}$. Hence the correct option is **D**.

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