JEE Advanced 2001

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Which of the following functions is differentiable at $x = 0$ ?

Q: Wait, why is $\cos(|x|)$ differentiable at $x=0$? Doesn't the absolute value make it sharp?

That is a brilliant observation, but look closer! The cosine function is an 'even' function, meaning $\cos(-x) = \cos(x)$. Because of this symmetry, $\cos(|x|)$ is identical to $\cos(x)$ for all real numbers. Since $\cos(x)$ is a smooth, oscillating wave with no sharp corners, $\cos(|x|)$ inherits that smoothness perfectly.

Q: Why does the linear term cancel out in $\sin(|x|) - |x|$ but not in $\sin(|x|) + |x|$?

Think about the Taylor series expansion: $\sin(x) \approx x - \frac{x^3}{6}$. When you subtract $x$, you get $(x - \frac{x^3}{6}) - x = -\frac{x^3}{6}$. The linear 'sharp' part vanishes! However, if you add them, you get $(x - \frac{x^3}{6}) + x = 2x - \frac{x^3}{6}$. The $2x$ term remains, and its derivative is $2$, which is non-zero, creating a sharp corner.

Q: Can I just look at the graph to solve this?

While visualization is a powerful tool for intuition, in the JEE Advanced, you must back it up with analytical rigor. Always calculate the LHD and RHD to be absolutely certain. The graph can sometimes deceive you if the 'sharpness' is very subtle!

Select Answer:

Visualized Solution

Visualizing Differentiability at $x = 0$

Objective: Identify which of the four given functions is differentiable at $x = 0$ .
Geometric Meaning: A function is differentiable at a point if its graph is smooth (no sharp corners, cusps, or breaks) at that point.
Mathematical Condition: The Left-Hand Derivative (LHD) must equal the Right-Hand Derivative (RHD) and both must be finite: $L H D = R H D = L$ , where $L \in R$ .

Simplifying the $cos (∣ x ∣)$ Term

Let's analyze the term $cos (∣ x ∣)$ present in Options 1 and 2.
Recall the property of the cosine function: it is an even function, meaning $cos (- x) = cos (x)$ .
Therefore, whether $x$ is positive or negative, $cos (∣ x ∣) = cos (x)$ for all $x \in R$ .
Since $cos (x)$ is differentiable everywhere, $cos (∣ x ∣)$ is perfectly differentiable at $x = 0$ .

Analyzing Option 1: $cos (∣ x ∣) + ∣ x ∣$

We can rewrite Option 1 as: $f_{1} (x) = cos (x) + ∣ x ∣$ .
We know $cos (x)$ is differentiable at $x = 0$ .
However, the absolute value function $∣ x ∣$ has a sharp corner at $x = 0$ , making it non-differentiable at $x = 0$ .
Theorem: The sum of a differentiable function and a non-differentiable function is always non-differentiable.

Analyzing Option 2: $cos (∣ x ∣) - ∣ x ∣$

Similarly, rewrite Option 2 as: $f_{2} (x) = cos (x) - ∣ x ∣$ .
$cos (x)$ is differentiable at $x = 0$ , but $- ∣ x ∣$ is non-differentiable at $x = 0$ .
By the same theorem: $Differentiable - Non-Differentiable = Non-Differentiable$ .
Therefore, Option 2 is also non-differentiable at $x = 0$ .

Testing Option 3: $sin (∣ x ∣) + ∣ x ∣$ for $x \geq 0$

Let $f_{3} (x) = sin (∣ x ∣) + ∣ x ∣$ .
For $x \geq 0$ , $∣ x ∣ = x$ , so the function becomes: $f_{3} (x) = sin (x) + x$ .
Differentiating with respect to $x$ : $f_{3}^{'} (x) = cos (x) + 1$ .
The Right-Hand Derivative (RHD) at $x = 0$ is: $f_{3}^{'} (0^{+}) = cos (0) + 1 = 1 + 1 = 2$ .

Testing Option 3: $sin (∣ x ∣) + ∣ x ∣$ for $x < 0$

For $x < 0$ , $∣ x ∣ = - x$ , so the function becomes: $f_{3} (x) = sin (- x) + (- x) = - sin (x) - x$ .
Differentiating with respect to $x$ : $f_{3}^{'} (x) = - cos (x) - 1$ .
The Left-Hand Derivative (LHD) at $x = 0$ is: $f_{3}^{'} (0^{-}) = - cos (0) - 1 = - 1 - 1 = - 2$ .
Since $L H D \neq = R H D$ ( $- 2 \neq = 2$ ), Option 3 is not differentiable at $x = 0$ .

Testing Option 4: $sin (∣ x ∣) - ∣ x ∣$ for $x \geq 0$

Let $f_{4} (x) = sin (∣ x ∣) - ∣ x ∣$ .
For $x \geq 0$ , $∣ x ∣ = x$ , so the function becomes: $f_{4} (x) = sin (x) - x$ .
Differentiating with respect to $x$ : $f_{4}^{'} (x) = cos (x) - 1$ .
The Right-Hand Derivative (RHD) at $x = 0$ is: $f_{4}^{'} (0^{+}) = cos (0) - 1 = 1 - 1 = 0$ .

Testing Option 4: $sin (∣ x ∣) - ∣ x ∣$ for $x < 0$

For $x < 0$ , $∣ x ∣ = - x$ , so the function becomes: $f_{4} (x) = sin (- x) - (- x) = - sin (x) + x$ .
Differentiating with respect to $x$ : $f_{4}^{'} (x) = - cos (x) + 1$ .
The Left-Hand Derivative (LHD) at $x = 0$ is: $f_{4}^{'} (0^{-}) = - cos (0) + 1 = - 1 + 1 = 0$ .
Since $L H D = R H D = 0$ , the function is differentiable at $x = 0$ .

Deep Intuition: Why did Option 4 become smooth?

Let's look at the Taylor series expansion of $sin (x)$ near $x = 0$ :
$sin (x) = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots$
For $x \geq 0$ : $f_{4} (x) = (x - \frac{x ^{3}}{6} + \dots) - x = - \frac{x ^{3}}{6} + \dots$
For $x < 0$ : $f_{4} (x) = - (- x - \frac{- x ^{3}}{6} + \dots) + x = \frac{x ^{3}}{6} - \dots$
The linear terms cancel out, leaving only cubic and higher-order terms, which are naturally smooth at $x = 0$ .

Final Conclusion

Option 1: Non-differentiable (sharp corner at $y = 1$ ).
Option 2: Non-differentiable (sharp corner at $y = 1$ ).
Option 3: Non-differentiable (sharp corner at $y = 0$ ).
Option 4: Differentiable (smooth flat tangent at $y = 0$ ).
Correct Option: 4 (or $sin (∣ x ∣) - ∣ x ∣$ ).

00:00 / 00:00

The Sigma Insight: Differentiability of a Function

Solution Diagram

Analyzing the Concept of Differentiability

When we ask if a function is differentiable at a point, we are essentially asking: "Is this path smooth?" Imagine driving a car along a road defined by a function. If the road has a sharp, sudden turn—a cusp or a corner—you would have to stop instantly to change direction. That is non-differentiability.

Mathematically, this means the slope from the left (the Left-Hand Derivative, or

L H D

) must perfectly match the slope from the right (the Right-Hand Derivative, or

R H D

). If they do not match, the function is not differentiable.

The Cosine Trap

Let us look at the options involving the term

cos (∣ x ∣)

. Many students panic when they see the absolute value sign, fearing a sharp corner. However, the cosine function is an even function, meaning

cos (- x) = cos (x)

Because of this symmetry, the absolute value sign is essentially invisible. Since

cos (x)

is a smooth, oscillating wave,

cos (∣ x ∣)

is perfectly differentiable everywhere.

However, when we add or subtract

∣ x ∣

from it, we combine a smooth function with a sharp one. A theorem in calculus states that the sum or difference of a differentiable function and a non-differentiable function is always non-differentiable. Thus, options involving

cos (∣ x ∣) \pm ∣ x ∣

are sharp at the origin and are eliminated.

The Sine Struggle

Now, we turn our attention to options involving

sin (∣ x ∣)

. Unlike cosine, sine is an odd function, so

sin (∣ x ∣)

is not simply

sin (x)

. We must use the piecewise definition.

Let us analyze the function

f (x) = sin (∣ x ∣) - ∣ x ∣

. For

x \geq 0

, the function is

f (x) = sin (x) - x

. The derivative is:

f^{'} (x) = cos (x) - 1

x = 0

, the

R H D

is:

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

Now, for

x < 0

, the function is

f (x) = sin (- x) - (- x) = - sin (x) + x

. The derivative is:

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

x = 0

, the

L H D

is:

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

Because the

L H D

equals the

R H D

(both are zero), the function is perfectly smooth at

x = 0

The Taylor Series Revelation

Why did the subtraction work so perfectly? Let us look at the Taylor series expansion of

sin (x)

near

x = 0

sin (x) \approx x - \frac{x ^{3}}{6}

When we calculate

sin (x) - x

, we get:

(x - \frac{x ^{3}}{6}) - x = - \frac{x ^{3}}{6}

The linear term

x

—the very term that causes the sharp corner in

∣ x ∣

—is completely cancelled out. We are left with a cubic term, which is incredibly smooth.

This is the beauty of mathematics: when two "sharp" functions interact in just the right way, they can cancel each other's flaws to create something perfectly smooth. Option 4 is our winner.

Common Traps & FAQs

Wait, why is

cos (∣ x ∣)

differentiable at

x = 0

? Doesn't the absolute value make it sharp?

That is a brilliant observation, but look closer! The cosine function is an 'even' function, meaning

cos (- x) = cos (x)

. Because of this symmetry,

cos (∣ x ∣)

is identical to

cos (x)

for all real numbers. Since

cos (x)

is a smooth, oscillating wave with no sharp corners,

cos (∣ x ∣)

inherits that smoothness perfectly.

Why does the linear term cancel out in

sin (∣ x ∣) - ∣ x ∣

but not in

sin (∣ x ∣) + ∣ x ∣

Think about the Taylor series expansion:

sin (x) \approx x - \frac{x ^{3}}{6}

. When you subtract

x

, you get

(x - \frac{x ^{3}}{6}) - x = - \frac{x ^{3}}{6}

. The linear 'sharp' part vanishes! However, if you add them, you get

(x - \frac{x ^{3}}{6}) + x = 2 x - \frac{x ^{3}}{6}

. The

2 x

term remains, and its derivative is

2

, which is non-zero, creating a sharp corner.

Can I just look at the graph to solve this?

While visualization is a powerful tool for intuition, in the JEE Advanced, you must back it up with analytical rigor. Always calculate the LHD and RHD to be absolutely certain. The graph can sometimes deceive you if the 'sharpness' is very subtle!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Which of the following functions is differentiable at $x = 0$ ?

Select Answer:

Visualized Solution

Visualizing Differentiability at $x = 0$

Simplifying the $cos (∣ x ∣)$ Term

Analyzing Option 1: $cos (∣ x ∣) + ∣ x ∣$

Analyzing Option 2: $cos (∣ x ∣) - ∣ x ∣$

Testing Option 3: $sin (∣ x ∣) + ∣ x ∣$ for $x \geq 0$

Testing Option 3: $sin (∣ x ∣) + ∣ x ∣$ for $x < 0$

Testing Option 4: $sin (∣ x ∣) - ∣ x ∣$ for $x \geq 0$

Testing Option 4: $sin (∣ x ∣) - ∣ x ∣$ for $x < 0$

Deep Intuition: Why did Option 4 become smooth?

Final Conclusion

The Sigma Insight: Differentiability of a Function

Analyzing the Concept of Differentiability

The Cosine Trap

The Sine Struggle

The Taylor Series Revelation

Similar Questions

The function $f (x) = (x^{2} - 1) ∣ x^{2} - 3 x + 2∣ + cos (∣ x ∣)$ is NOT differentiable at

Let $f : R \to R$ be differentiable at $c \in R$ and $f (c) = 0$ . If $g (x) = ∣ f (x) ∣$ , then at $x = c$ , $g$ is :

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

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

Let $f : R \to R$ be a function defined by $f (x) = min {x + 1, ∣ x ∣ + 1}$ . Then which of the following is true?

Let $f (x)$ be defined in the interval $[- 2, 2]$ such that $f (x) = {- 1, x - 1, - 2 \leq x \leq 0 0 < x \leq 2$ and $g (x) = f (∣ x ∣) + ∣ f (x) ∣$ . Test the differentiability of $g (x)$ in $(- 2, 2)$ .

If $f (x) = x (x - x + 1)$ , then

Let $f : R \to R$ be a function such that $f (x + y) = f (x) + f (y), \forall x, y \in R$ . If $f (x)$ is differentiable at $x = 0$ , then

Let $f (x) = {(x - 1) sin \frac{1}{x - 1}, 0, if x \neq = 1 if x = 1$ . Then which one of the following is true?

Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following functions is differentiable at x=0?

Select Answer:

Visualized Solution

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

Simplifying the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos(∣x∣) Term

Analyzing Option 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos(∣x∣)+∣x∣

Analyzing Option 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos(∣x∣)−∣x∣

Testing Option 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(∣x∣)+∣x∣ for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥0

Testing Option 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(∣x∣)+∣x∣ for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0

Testing Option 4: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(∣x∣)−∣x∣ for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0

Deep Intuition: Why did Option 4 become smooth?

Final Conclusion

The Sigma Insight: Differentiability of a Function

Analyzing the Concept of Differentiability

The Cosine Trap

The Sine Struggle

The Taylor Series Revelation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=(x2−1)∣x2−3x+2∣+cos(∣x∣) is NOT differentiable at

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be differentiable at c∈R and f(c)=0. If g(x)=∣f(x)∣, then at x=c, g is :

.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 function given by y=∣∣x∣−1∣ is differentiable for all real numbers except the points

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function defined by f(x)=min{x+1,∣x∣+1}. Then which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=x(x​−x+1​), then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function such that f(x+y)=f(x)+f(y),∀x,y∈R. If f(x) is differentiable at x=0, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={(x−1)sinx−11​,0,​if x=1if x=1​. Then which one of the following is true?

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Which of the following functions is differentiable at $x = 0$ ?

Visualizing Differentiability at $x = 0$

Simplifying the $cos (∣ x ∣)$ Term

Analyzing Option 1: $cos (∣ x ∣) + ∣ x ∣$

Analyzing Option 2: $cos (∣ x ∣) - ∣ x ∣$

Testing Option 3: $sin (∣ x ∣) + ∣ x ∣$ for $x \geq 0$

Testing Option 3: $sin (∣ x ∣) + ∣ x ∣$ for $x < 0$

Testing Option 4: $sin (∣ x ∣) - ∣ x ∣$ for $x < 0$

The function $f (x) = (x^{2} - 1) ∣ x^{2} - 3 x + 2∣ + cos (∣ x ∣)$ is NOT differentiable at

Let $f : R \to R$ be differentiable at $c \in R$ and $f (c) = 0$ . If $g (x) = ∣ f (x) ∣$ , then at $x = c$ , $g$ is :

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

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

Let $f : R \to R$ be a function defined by $f (x) = min {x + 1, ∣ x ∣ + 1}$ . Then which of the following is true?

If $f (x) = x (x - x + 1)$ , then

Let $f : R \to R$ be a function such that $f (x + y) = f (x) + f (y), \forall x, y \in R$ . If $f (x)$ is differentiable at $x = 0$ , then

Let $f (x) = {(x - 1) sin \frac{1}{x - 1}, 0, if x \neq = 1 if x = 1$ . Then which one of the following is true?