Let f1: R → R, f2 : (-π/2, π/2) → R, f3 : (-1, e^(π/2) – 2) → R and f4: R → R be functions defined b

Q: Wait, why is $\sqrt{x^2} = |x|$? Can't I just write it as $x$?

This is a classic trap! The expression $\sqrt{x^2}$ is defined as the non-negative square root. If $x$ is negative, say $x = -2$, then $\sqrt{(-2)^2} = \sqrt{4} = 2$, which is $-x$. Thus, $\sqrt{x^2} = |x|$. In calculus, especially when checking differentiability, this distinction is vital because $|x|$ behaves differently for $x > 0$ and $x < 0$, leading to different left-hand and right-hand derivatives.

Q: Why is $f_4(x) = x^2 \sin(1/x)$ differentiable at $x=0$ if its derivative is not continuous?

Differentiability at a point only requires the derivative to exist at that specific point. For $f_4(x)$, the derivative at $x=0$ is calculated using the first principle: $\lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin(1/h) = 0$. This limit exists. However, for $x \neq 0$, the derivative is $f_4'(x) = 2x \sin(1/x) - \cos(1/x)$. As $x \to 0$, the $\cos(1/x)$ term oscillates infinitely, so $\lim_{x \to 0} f_4'(x)$ does not exist. Thus, the function is differentiable, but the derivative function itself is discontinuous.

Q: How do I handle the Greatest Integer Function $[\sin(\log_e(x+2))]$? It looks so scary!

Don't let the notation intimidate you! The key is to look at the behavior in a tiny neighborhood around $x=0$. Since $\log_e(2) \approx 0.693$, and we know $\sin(0.693)$ is a value between $0$ and $1$, the greatest integer of this value is always $0$. Because the function is constant ($f_3(x) = 0$) in that small neighborhood, it is perfectly continuous and differentiable, and its derivative is also constant ($0$).

Question

Let $f_1: R 	o R$, $f_2 : \left(-\frac{\pi}{2}, \frac{\pi}{2}ight) 	o R$, $f_3 : (-1, e^{\pi/2} – 2) 	o R$ and $f_4: R	o R$ be functions defined by (i) $f_1(x) = \sin \sqrt{1-e^{-x^2}}$ (ii) $f_2(x) = \begin{cases} \frac{\sin x}{	an^{-1} x} & 	ext{if } x 
eq 0 \ 1 & 	ext{if } x = 0 \end{cases}$, where the inverse trigonometric function $	an^{-1}x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$ (iii) $f_3(x) = [\sin(\log_e(x + 2))]$, where, for $t \in R$, $[t]$ denotes the greatest integer less than or equal to $t$, (iv) $f_4(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}ight) & 	ext{if } x 
eq 0 \ 0 & 	ext{if } x = 0 \end{cases}$. Match the functions in List-I with the related properties in List-II.

List-I:

P. The function $f_1$ is

Q. The function $f_2$ is

R. The function $f_3$ is

S. The function $f_4$ is

List-II:

1. NOT continuous at $x = 0$

2. continuous at $x = 0$ and NOT differentiable at $x = 0$

3. differentiable at $x = 0$ and its derivative is NOT continuous at $x = 0$

4. differentiable at $x = 0$ and its derivative is continuous at $x = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are examining the behavior of four distinct functions at the critical point $x=0$. This analysis focuses on continuity and differentiability, which are fundamental concepts in JEE Advanced calculus.

### Function 1: The Sharp Corner Trap

Consider the function $f_1(x) = \sin \sqrt{1-e^{-x^2}}$. At the origin, $f_1(0) = \sin \sqrt{1-1} = 0$, confirming the function is continuous.

To check for differentiability, we evaluate the derivative using the first principle:

$$f_1'(0) = \lim_{h 	o 0} \frac{\sin \sqrt{1-e^{-h^2}}}{h}$$

For very small $h$, we use the approximation $1-e^{-h^2} \approx h^2$. Consequently, the expression simplifies to:

$$\lim_{h 	o 0} \frac{\sin |h|}{h}$$

The Right-Hand Derivative (RHD) is $1$, while the Left-Hand Derivative (LHD) is $-1$. Because the one-sided derivatives clash, $f_1$ possesses a **sharp corner** at $x=0$, rendering it **not differentiable**.

### Function 2: The Jump Discontinuity

Next, we examine $f_2(x) = \frac{|\sin x|}{	an^{-1} x}$. We evaluate the limit as $x$ approaches $0$ from both sides.

As $x 	o 0^+$, $\sin x$ is positive, so the limit is $1$. As $x 	o 0^-$, $\sin x$ is negative, causing $|\sin x|$ to behave as $-\sin x$, resulting in a limit of $-1$.

Because the left-hand and right-hand limits do not match, the function exhibits a **jump discontinuity** at the origin.

### Function 3: The Greatest Integer Illusion

Now, consider $f_3(x) = [\sin(\log_e(x + 2))]$. Near $x=0$, the inner term $\log_e(x+2)$ approaches $\log_e(2) \approx 0.693$.

Since $0 < 0.693 < \pi/2$, the sine of this value is a positive fraction between $0$ and $1$. The greatest integer of any value in the interval $(0, 1)$ is $0$.

Thus, in a small neighborhood around the origin, $f_3(x)$ behaves as the constant function $y=0$. It is **perfectly continuous and differentiable**, making it the smoothest function in this set.

### Function 4: The Oscillating Paradox

Finally, we analyze $f_4(x) = x^2 \sin(1/x)$. By the **Squeeze Theorem**, we know $f_4(x)$ is continuous at $x=0$.

Using the first principle for the derivative at the origin:

$$f_4'(0) = \lim_{h 	o 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h 	o 0} h \sin(1/h) = 0$$

This confirms the function is differentiable at $x=0$. However, for $x 
eq 0$, the product rule yields:

$$f_4'(x) = 2x \sin(1/x) - \cos(1/x)$$

As $x 	o 0$, the term $\cos(1/x)$ oscillates wildly between $-1$ and $1$. Therefore, the limit of the derivative does not exist, which implies that while the function is differentiable, its **derivative is not continuous** at the origin.

List-I

List-II

Select Matching Pairs:

Visualized Solution

The Grand Analysis of Four Functions

Function 1: Continuity

Function 1: Differentiability Setup

Function 1: The Sharp Corner

Function 2: Continuity Check

Function 2: Discontinuity

Function 3: The Greatest Integer Trap

Function 3: Constant Neighborhood

Function 3: Differentiability

Function 4: Squeeze Theorem

Function 4: Differentiability at Origin

Function 4: Continuity of Derivative

Final Conclusion

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Function 1

Function 2

Function 3

Function 4

Similar Questions

Let $f (x) = {x^{2} sin (\frac{1}{x}) 0, x \neq = 0, x = 0$ ; Then at $x = 0$

If $f (x) = ⎩ ⎨ ⎧ - x - \frac{π}{2}, - cos x, x - 1, ln x, x \leq - \frac{π}{2} - \frac{π}{2} < x \leq 0 0 < x \leq 1 x > 1$

The function $f (x) = {\frac{π}{4} + tan^{- 1} x, \frac{1}{2} (∣ x ∣ - 1), ∣ x ∣ \leq 1 ∣ x ∣ > 1$ is:

In this questions there are entries in columns I and II. Each entry in column I is related to exactly one entry in column II. Write the correct letter from column II against the entry number in column I in your answer book.

List-I

List-II

Let $f : [0, \infty) \to [0, 3]$ be a function defined by $f (x) = {max {sin t : 0 \leq t \leq x}, 2 + cos x, 0 \leq x \leq π x > π$ Then which of the following is true ?

Let $g (x) = x f (x)$ , where $f (x) = {x sin \frac{1}{x}, 0, x \neq = 0 x = 0$ . At $x = 0$

Consider the function $f : (0, 2) \to R$ defined by $f (x) = \frac{x}{2} + \frac{2}{x}$ and the function $g (x)$ defined by $g (x) = {min {f (t)}, \frac{3}{2} + x, 0 < t \leq x and 0 < x \leq 1 1 < x < 2$ . Then

If $f (x) = {x e^{- (\frac{1}{∣ x ∣} + \frac{1}{x})}, 0, x \neq = 0 x = 0$ then $f (x)$ is

Let $f (x) = {(x - 1)^{2} sin \frac{1}{( x - 1 )} - ∣ x ∣ - 1 if x \neq = 1 if x = 1$ be a real-valued function. Then the set of points where $f (x)$ is not differentiable is .........

List-I

List-II

Select Matching Pairs:

Visualized Solution

The Grand Analysis of Four Functions

Function 1: Continuity

Function 1: Differentiability Setup

Function 1: The Sharp Corner

Function 2: Continuity Check

Function 2: Discontinuity

Function 3: The Greatest Integer Trap

Function 3: Constant Neighborhood

Function 3: Differentiability

Function 4: Squeeze Theorem

Function 4: Differentiability at Origin

Function 4: Continuity of Derivative

Final Conclusion

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Function 1

Function 2

Function 3

Function 4

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List-I

List-II

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