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

### Analyzing the Setup The function provided is $f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x|$. To determine the differentiability at $x=0$, we must address the behavior of the modulus function $|x|$, which changes its definition at the origin. We split the domain into two distinct realms: the positive realm where $x \ge 0$ and the negative realm where $x < 0$. ### The Positive Realm ($x \ge 0$) In this region, $|x| = x$. The function simplifies to: $$f(x) = \sin x - x + 2(x - \pi) \cos x$$ To find the derivative $f'(x)$, we apply the product rule to the term $2(x - \pi) \cos x$: $$f'(x) = \cos x - 1 + [2 \cos x - 2(x - \pi) \sin x]$$ Combining like terms, we obtain: $$f'(x) = 3 \cos x - 1 - 2(x - \pi) \sin x$$ Evaluating the right-hand derivative at $x=0$: $$f'(0^+) = 3 \cos(0) - 1 - 2(0 - \pi) \sin(0) = 3(1) - 1 - 0 = 2$$ ### The Negative Realm ($x < 0$) In this region, $|x| = -x$. Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$, the function becomes: $$f(x) = -\sin x + x + 2(x - \pi) \cos x$$ Differentiating this expression with respect to $x$: $$f'(x) = -\cos x + 1 + [2 \cos x - 2(x - \pi) \sin x]$$ Simplifying the expression, we get: $$f'(x) = \cos x + 1 - 2(x - \pi) \sin x$$ Evaluating the left-hand derivative at $x=0$: $$f'(0^-) = \cos(0) + 1 - 2(0 - \pi) \sin(0) = 1 + 1 - 0 = 2$$ ### Conclusion Since the left-hand derivative and the right-hand derivative are equal ($f'(0^-) = f'(0^+) = 2$), the function is differentiable at $x=0$. The set $K$ of points where the function is non-differentiable is **empty**.

Wait, why don't we check $x = \pi$? The term $(x - \pi)$ looks suspicious.

That is a brilliant observation! It is easy to get distracted by the $(x - \pi)$ term, but remember: differentiability is about the *smoothness* of the function. The term $(x - \pi)$ is a simple polynomial, which is infinitely differentiable everywhere. The only 'kink' in our function comes from the $|x|$ terms. Always identify the source of the non-differentiability first!

JEE Main 11 Jan 2019 (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $K$ be the set of all real values of $x$ where the function $f (x) = sin ∣ x ∣ - ∣ x ∣ + 2 (x - π) cos ∣ x ∣$ is not differentiable. Then the set $K$ is equal to:

Q: Wait, why don't we check $x = \pi$? The term $(x - \pi)$ looks suspicious.

That is a brilliant observation! It is easy to get distracted by the $(x - \pi)$ term, but remember: differentiability is about the *smoothness* of the function. The term $(x - \pi)$ is a simple polynomial, which is infinitely differentiable everywhere. The only 'kink' in our function comes from the $|x|$ terms. Always identify the source of the non-differentiability first!

Q: I thought $|x|$ is always non-differentiable at $x = 0$. Why is this function differentiable there?

You are absolutely right that $|x|$ itself has a sharp corner at $x = 0$. However, when you add or multiply it with other functions, those 'sharp corners' can sometimes cancel each other out. In this problem, the contributions from the $\sin |x|$, $-|x|$, and $2(x - \pi) \cos |x|$ terms combine perfectly to smooth out the transition. Never assume; always calculate!

Q: How do I handle the product rule with $(x - \pi)$?

Treat it like any other product rule problem! If you have $u(x) = 2(x - \pi)$ and $v(x) = \cos x$, the derivative is $u'(x)v(x) + u(x)v'(x)$. Here, $u'(x) = 2$ and $v'(x) = -\sin x$. So the derivative is $2 \cos x - 2(x - \pi) \sin x$. Just be careful with the signs!

Select Answer:

Visualized Solution

Identifying Potential Points of Non-Differentiability

Given function: $f (x) = sin ∣ x ∣ - ∣ x ∣ + 2 (x - π) cos ∣ x ∣$
The modulus function $∣ x ∣$ changes behavior at $x = 0$ .
Potential point of non-differentiability: $x = 0$ .

Defining $f (x)$ for $x \geq 0$

For $x \geq 0$ , $∣ x ∣ = x$ .
Substitute $∣ x ∣ = x$ into $f (x)$ :
$f (x) = sin x - x + 2 (x - π) cos x$

Calculating Right-Hand Derivative

Differentiating $f (x)$ for $x > 0$ :
$f^{'} (x) = cos x - 1 + 2 [1 \cdot cos x + (x - π) (- sin x)]$
$f^{'} (x) = 3 cos x - 1 - 2 (x - π) sin x$

Evaluating RHD at $x = 0$

At $x \to 0^{+}$ :
$f^{'} (0^{+}) = 3 cos (0) - 1 - 2 (0 - π) sin (0)$
$f^{'} (0^{+}) = 3 (1) - 1 - 0 = 2$

Defining $f (x)$ for $x < 0$

For $x < 0$ , $∣ x ∣ = - x$ .
Substitute $∣ x ∣ = - x$ into $f (x)$ :
$f (x) = sin (- x) - (- x) + 2 (x - π) cos (- x)$
$f (x) = - sin x + x + 2 (x - π) cos x$

Calculating Left-Hand Derivative

Differentiating $f (x)$ for $x < 0$ :
$f^{'} (x) = - cos x + 1 + 2 [1 \cdot cos x + (x - π) (- sin x)]$
$f^{'} (x) = cos x + 1 - 2 (x - π) sin x$

Evaluating LHD at $x = 0$

At $x \to 0^{-}$ :
$f^{'} (0^{-}) = cos (0) + 1 - 2 (0 - π) sin (0)$
$f^{'} (0^{-}) = 1 + 1 - 0 = 2$

Checking Differentiability at $x = 0$

Since $f^{'} (0^{+}) = f^{'} (0^{-}) = 2$ , the function is differentiable at $x = 0$ .
The slope of the tangent at $x = 0$ is $2$ .

Final Conclusion

For all other $x \neq = 0$ , $f (x)$ is a combination of smooth trigonometric and polynomial functions.
Thus, $f (x)$ is differentiable for all $x \in R$ .
The set $K$ of non-differentiable points is the empty set $ϕ$ .

00:00 / 00:00

The Sigma Insight: Differentiability of a Function

Solution Diagram

Analyzing the Setup

The function provided is

f (x) = sin ∣ x ∣ - ∣ x ∣ + 2 (x - π) cos ∣ x ∣

. To determine the differentiability at

x = 0

, we must address the behavior of the modulus function

∣ x ∣

, which changes its definition at the origin.

We split the domain into two distinct realms: the positive realm where

x \geq 0

and the negative realm where

x < 0

The Positive Realm ( $x \geq 0$ )

In this region,

∣ x ∣ = x

. The function simplifies to:

f (x) = sin x - x + 2 (x - π) cos x

To find the derivative

f^{'} (x)

, we apply the product rule to the term

2 (x - π) cos x

f^{'} (x) = cos x - 1 + [2 cos x - 2 (x - π) sin x]

Combining like terms, we obtain:

f^{'} (x) = 3 cos x - 1 - 2 (x - π) sin x

Evaluating the right-hand derivative at

x = 0

f^{'} (0^{+}) = 3 cos (0) - 1 - 2 (0 - π) sin (0) = 3 (1) - 1 - 0 = 2

The Negative Realm ( $x < 0$ )

In this region,

∣ x ∣ = - x

. Since

sin (- x) = - sin x

and

cos (- x) = cos x

, the function becomes:

f (x) = - sin x + x + 2 (x - π) cos x

Differentiating this expression with respect to

x

f^{'} (x) = - cos x + 1 + [2 cos x - 2 (x - π) sin x]

Simplifying the expression, we get:

f^{'} (x) = cos x + 1 - 2 (x - π) sin x

Evaluating the left-hand derivative at

x = 0

f^{'} (0^{-}) = cos (0) + 1 - 2 (0 - π) sin (0) = 1 + 1 - 0 = 2

Conclusion

Since the left-hand derivative and the right-hand derivative are equal (

f^{'} (0^{-}) = f^{'} (0^{+}) = 2

), the function is differentiable at

x = 0

The set

K

of points where the function is non-differentiable is empty.

Common Traps & FAQs

Wait, why don't we check

x = π

? The term

(x - π)

looks suspicious.

That is a brilliant observation! It is easy to get distracted by the

(x - π)

term, but remember: differentiability is about the *smoothness* of the function. The term

(x - π)

is a simple polynomial, which is infinitely differentiable everywhere. The only 'kink' in our function comes from the

∣ x ∣

terms. Always identify the source of the non-differentiability first!

I thought

∣ x ∣

is always non-differentiable at

x = 0

. Why is this function differentiable there?

You are absolutely right that

∣ x ∣

itself has a sharp corner at

x = 0

. However, when you add or multiply it with other functions, those 'sharp corners' can sometimes cancel each other out. In this problem, the contributions from the

sin ∣ x ∣

- ∣ x ∣

, and

2 (x - π) cos ∣ x ∣

terms combine perfectly to smooth out the transition. Never assume; always calculate!

How do I handle the product rule with

(x - π)

Treat it like any other product rule problem! If you have

u (x) = 2 (x - π)

and

v (x) = cos x

, the derivative is

u^{'} (x) v (x) + u (x) v^{'} (x)

. Here,

u^{'} (x) = 2

and

v^{'} (x) = - sin x

. So the derivative is

2 cos x - 2 (x - π) sin x

. Just be careful with the signs!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $K$ be the set of all real values of $x$ where the function $f (x) = sin ∣ x ∣ - ∣ x ∣ + 2 (x - π) cos ∣ x ∣$ is not differentiable. Then the set $K$ is equal to:

Select Answer:

Visualized Solution

Identifying Potential Points of Non-Differentiability

Defining $f (x)$ for $x \geq 0$

Calculating Right-Hand Derivative

Evaluating RHD at $x = 0$

Defining $f (x)$ for $x < 0$

Calculating Left-Hand Derivative

Evaluating LHD at $x = 0$

Checking Differentiability at $x = 0$

Final Conclusion

The Sigma Insight: Differentiability of a Function

Analyzing the Setup

The Positive Realm ( $x \geq 0$ )

The Negative Realm ( $x < 0$ )

Conclusion

Similar Questions

Let $S = {t \in R : f (x) = ∣ x - π ∣ \cdot (e^{∣ x ∣} - 1) sin ∣ x ∣ is not differentiable at t}$ . Then the set $S$ is equal to :

Let $f (x) = {max {∣ x ∣, x^{2}}, 8 - 2∣ x ∣, ∣ x ∣ \leq 2 2 < ∣ x ∣ \leq 4$ . Let S be the set of points in the interval (-4, 4) at which f is not differentiable. Then S:

Let $f : (- 1, 1) \to R$ be a function defined by $f (x) = max {- ∣ x ∣, - 1 - x^{2}}$ . If K be the set of all points at which f is not differentiable, then K has exactly :

Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{2}}$ . Let $S$ denote the set of all points in $R$ , where $f$ is not differentiable. Then :

Let $S$ be the set of all points in $(- π, π)$ at which the function, $f (x) = min {sin x, cos x}$ is not differentiable. Then $S$ is a subset of which of the following?

Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{3}}$ . The set of all points where $f (x)$ is NOT differentiable is

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

Let $f (x) = {x^{2} cos \frac{π}{x} 0 x \neq = 0 x = 0, x \in R$ then $f$ is

The set of points where $f (x) = \frac{x}{1 + ∣ x ∣}$ is differentiable is

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

Select Answer:

Visualized Solution

Identifying Potential Points of Non-Differentiability

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥0

Calculating Right-Hand Derivative

Evaluating RHD at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0

Calculating Left-Hand Derivative

Evaluating LHD at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Checking Differentiability at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Final Conclusion

The Sigma Insight: Differentiability of a Function

Analyzing the Setup

The Positive Realm ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥0)

The Negative Realm ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0)

Conclusion

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={max{∣x∣,x2},8−2∣x∣,​∣x∣≤22<∣x∣≤4​. Let S be the set of points in the interval (-4, 4) at which f is not differentiable. Then S:

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={x2cosxπ​0​x=0x=0​,x∈R then f is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The set of points where f(x)=1+∣x∣x​ is differentiable is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function given by y=∣∣x∣−1∣ is differentiable for all real numbers except the points

Defining $f (x)$ for $x \geq 0$

Evaluating RHD at $x = 0$

Defining $f (x)$ for $x < 0$

Evaluating LHD at $x = 0$

Checking Differentiability at $x = 0$

The Positive Realm ( $x \geq 0$ )

The Negative Realm ( $x < 0$ )

Let $S = {t \in R : f (x) = ∣ x - π ∣ \cdot (e^{∣ x ∣} - 1) sin ∣ x ∣ is not differentiable at t}$ . Then the set $S$ is equal to :

Let $f (x) = {max {∣ x ∣, x^{2}}, 8 - 2∣ x ∣, ∣ x ∣ \leq 2 2 < ∣ x ∣ \leq 4$ . Let S be the set of points in the interval (-4, 4) at which f is not differentiable. Then S:

Let $f : (- 1, 1) \to R$ be a function defined by $f (x) = max {- ∣ x ∣, - 1 - x^{2}}$ . If K be the set of all points at which f is not differentiable, then K has exactly :

Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{2}}$ . Let $S$ denote the set of all points in $R$ , where $f$ is not differentiable. Then :

Let $S$ be the set of all points in $(- π, π)$ at which the function, $f (x) = min {sin x, cos x}$ is not differentiable. Then $S$ is a subset of which of the following?

Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{3}}$ . The set of all points where $f (x)$ is NOT differentiable is

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

Let $f (x) = {x^{2} cos \frac{π}{x} 0 x \neq = 0 x = 0, x \in R$ then $f$ is

The set of points where $f (x) = \frac{x}{1 + ∣ x ∣}$ is differentiable is

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