Let the functions f: (-1, 1) → ℝ and g: (-1, 1) → (-1, 1) be defined by f(x) = |2x - 1| + |2x + 1| a

Q: Wait, why is $g(x) = x + 1$ for $x \in (-1, 0)$? Shouldn't it just be $x$?

That is a classic trap! Remember the definition of the fractional part function: $g(x) = x - [x]$. For $x$ in the interval $(-1, 0)$, the greatest integer $[x]$ is $-1$. Therefore, $g(x) = x - (-1) = x + 1$. Always check the value of the greatest integer function carefully in negative intervals!

Q: Why is $x = 0$ a point of discontinuity? I thought we just check the boundaries of the intervals.

Great question. We check boundaries because that is where the definition of the function changes. At $x = 0$, the function $g(x)$ jumps from $1$ (as $x \to 0^-$) to $0$ (at $x = 0$). Since $f(g(x))$ depends on $g(x)$, this jump in $g(x)$ propagates to $f(g(x))$. Always evaluate the left-hand limit, right-hand limit, and the function value at the transition points.

Q: Do I really need to check differentiability at the points where the function is continuous? Isn't continuity enough?

Absolutely not! Continuity is a necessary condition for differentiability, but it is not sufficient. A function can be continuous but still have a 'sharp corner' or 'kink' where the derivative is undefined. In this problem, at $x = -1/2$ and $x = 1/2$, the function is continuous, but the slope changes abruptly from $0$ to $4$. That sharp turn makes it non-differentiable.

Question

Let the functions $f: (-1, 1) 	o \mathbb{R}$ and $g: (-1, 1) 	o (-1, 1)$ be defined by $f(x) = |2x - 1| + |2x + 1|$ and $g(x) = x - [x]$, where $[x]$ denotes the greatest integer less than or equal to $x$. Let $f \circ g: (-1, 1) 	o \mathbb{R}$ be the composite function defined by $(f \circ g)(x) = f(g(x))$. Suppose $c$ is the number of points in the interval $(-1, 1)$ at which $f \circ g$ is NOT continuous, and suppose $d$ is the number of points in the interval $(-1, 1)$ at which $f \circ g$ is NOT differentiable. Then the value of $c + d$ is ____.

Sigma Percentile · Accepted Answer

Welcome, my dear students, to a journey through the elegant world of composite functions. Today, we are not just solving a problem; we are peeling back the layers of a mathematical onion.

We have two functions, $f(x) = |2x - 1| + |2x + 1|$ and $g(x) = x - [x]$, and we are tasked with analyzing their composition, $f(g(x))$, over the interval $(-1, 1)$. This is a classic JEE Advanced problem that tests your ability to handle piecewise definitions, absolute values, and the subtle interplay between continuity and differentiability.

### Phase 1: Decoding the Fractional Part

The first step is to understand our inner function, $g(x) = x - [x]$. This is the fractional part function, often denoted as $\{x\}$.

For $x \in (-1, 0)$, the greatest integer $[x]$ is $-1$, so $g(x) = x - (-1) = x + 1$.

For $x \in [0, 1)$, the greatest integer $[x]$ is $0$, so $g(x) = x - 0 = x$. Notice the range of $g(x)$ here: it is always in the interval $[0, 1)$. This is a crucial observation that will simplify our lives immensely in the next step.

### Phase 2: The Modulus Simplification

Now, we look at the composite function:
$$f(g(x)) = |2g(x) - 1| + |2g(x) + 1|$$

Since $0 \le g(x) < 1$, multiplying by $2$ gives $0 \le 2g(x) < 2$. Adding $1$ to this inequality, we get $1 \le 2g(x) + 1 < 3$.

This means the expression inside the second absolute value, $2g(x) + 1$, is always positive. We can drop those absolute value bars without a second thought. Our function becomes:
$$f(g(x)) = |2g(x) - 1| + 2g(x) + 1$$

The critical point for the first modulus is where $2g(x) - 1 = 0$, which means $g(x) = 1/2$. This splits our analysis into two cases: $0 \le g(x) \le 1/2$ and $1/2 < g(x) < 1$.

### Phase 3: The Piecewise Construction

In Case 1 ($0 \le g(x) \le 1/2$), the expression $2g(x) - 1$ is non-positive. Thus, $|2g(x) - 1| = -(2g(x) - 1) = -2g(x) + 1$.

Adding the rest of our function, we get:
$$f(g(x)) = (-2g(x) + 1) + (2g(x) + 1) = 2$$

In Case 2 ($1/2 < g(x) < 1$), the expression $2g(x) - 1$ is positive. Thus, $|2g(x) - 1| = 2g(x) - 1$.

Adding the rest, we get:
$$f(g(x)) = (2g(x) - 1) + (2g(x) + 1) = 4g(x)$$

### Phase 4: Mapping Back to $x$

Now, we must translate these conditions back to our original variable $x$.

For $x \in (-1, 0)$, $g(x) = x + 1$. The condition $g(x) \le 1/2$ becomes $x + 1 \le 1/2$, or $x \le -1/2$. 
- For $x \in (-1, -1/2]$, $f(g(x)) = 2$. 
- For $x \in (-1/2, 0)$, $f(g(x)) = 4(x + 1) = 4x + 4$.

For $x \in [0, 1)$, $g(x) = x$. The condition $g(x) \le 1/2$ becomes $x \le 1/2$. 
- For $x \in [0, 1/2]$, $f(g(x)) = 2$. 
- For $x \in (1/2, 1)$, $f(g(x)) = 4x$.

### Phase 5: The Final Analysis

We have our piecewise function. Now, let us check for continuity and differentiability.

At $x = -1/2$, the left limit is $2$ and the right limit is $4(-1/2) + 4 = 2$. It is continuous.

At $x = 1/2$, the left limit is $2$ and the right limit is $4(1/2) = 2$. It is continuous.

At $x = 0$, the left limit is $4(0) + 4 = 4$, but the function value is $2$. This is a jump discontinuity. Thus, $c = 1$.

The function is non-differentiable at $x = 0$ because it is discontinuous. At $x = -1/2$, the slope on the left is $0$ and the slope on the right is $4$, creating a sharp corner. Similarly, at $x = 1/2$, the slope on the left is $0$ and the slope on the right is $4$, creating another sharp corner.

Therefore, the function is non-differentiable at $x = 0, -1/2, 1/2$. That gives us $d = 3$.

The final sum is $c + d = 1 + 3 = \mathbf{4}$.

Enter Numerical Value:

Visualized Solution

The Given Functions $f (x)$ and $g (x)$

Simplifying $g (x)$

The Composite Function $f (g (x))$

Simplifying the Second Modulus

Splitting the First Modulus

Evaluating Case $1$

Evaluating Case $2$

Mapping to $x \in (- 1, 0)$

Mapping to $x \in [0, 1)$

Checking Continuity

Number of Discontinuous Points ( $c$ )

Checking Differentiability

Final Answer for $c + d$

The Sigma Insight: Relationship Between Continuity and Differentiability

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Enter Numerical Value:

Visualized Solution

The Given Functions .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

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

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