If |c| ≤ 1/2 and f(x) is a differentiable function at x = 0 given by f(x) = {b sin⁻¹((c+x)/2), -1/2

Q: Why didn't we use the left branch to find $a$?

Great question! The variable $a$ only appears in the right branch. While the left branch is crucial for the full differentiability of the function (finding $b$ and $c$), it is independent of $a$. Focus on the branch that contains your target variable.

Q: Can I just use L'Hôpital's Rule instead of the standard limit?

Absolutely! L'Hôpital's Rule is a powerful tool. If you differentiate the numerator and denominator with respect to $x$, you get $\frac{d}{dx}(e^{ax/2} - 1) / \frac{d}{dx}(x) = \frac{a}{2} e^{ax/2} / 1$. As $x \to 0$, this becomes $a/2$. Both methods lead to the same beautiful truth.

Q: The problem mentions differentiability, but we only used continuity. Did I miss something?

You didn't miss anything! Differentiability implies continuity. We used the 'continuity' part of the definition to solve for $a$. If the question asked for $b$ or $c$, we would then need to equate the derivatives (LHD = RHD). You solved it efficiently by using the necessary condition first.

Question

If $|c| \leq 1/2$ and $f(x)$ is a differentiable function at $x=0$ given by $f(x) = \begin{cases} b \sin^{-1} \left( \frac{c+x}{2} ight), & -\frac{1}{2} < x < 0 \ 1/2, & x = 0 \ \frac{e^{ax/2} - 1}{x}, & 0 < x < \frac{1}{2} \end{cases}$ Find the value of 'a'.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a coordinate plane, looking at a curve that has been stitched together from three different mathematical worlds. This is the essence of a piecewise function—a puzzle where the pieces must fit perfectly, without a single jagged edge.

The problem asks us to find the value of $a$, a constant hidden within the right-hand branch. We are told the function $f(x)$ is differentiable at $x=0$.

In the language of calculus, this is a profound statement. It means that if you were to walk along this curve, you would never encounter a sharp corner or a sudden jump; it is a smooth, continuous path.

### The Bridge of Continuity

The first step in our journey is to acknowledge the hierarchy of calculus. Differentiability is a high-status property; it implies continuity.

If a function is differentiable at a point, it must be continuous at that point. This is our golden key. For the function to be continuous at $x=0$, the value of the function as we approach from the left must equal the value at the point itself, which must also equal the value as we approach from the right.

Mathematically, we write this as:

$$\lim_{x 	o 0^-} f(x) = f(0) = \lim_{x 	o 0^+} f(x)$$

We know $f(0) = 1/2$. This gives us a target: we need the right-hand limit to equal $1/2$.

### The Right-Hand Limit Challenge

Let us isolate the right branch: $f(x) = \frac{e^{ax/2} - 1}{x}$ for $x > 0$. We need to evaluate:

$$\lim_{x 	o 0^+} \frac{e^{ax/2} - 1}{x}$$

This looks intimidating, but it is a classic form. We recall the standard limit result:

$$\lim_{t 	o 0} \frac{e^t - 1}{t} = 1$$

Our expression is almost there, but the denominator is $x$, while the exponent is $ax/2$. To use the standard limit, the denominator must match the exponent exactly.

### Algebraic Surgery

We perform our algebraic surgery by multiplying and dividing the expression by $a/2$. This gives us:

$$\lim_{x 	o 0^+} \left( \frac{e^{ax/2} - 1}{x} ight) \cdot \frac{a/2}{a/2}$$

Rearranging this, we get:

$$\lim_{x 	o 0^+} \left( \frac{e^{ax/2} - 1}{ax/2} ight) \cdot \frac{a}{2}$$

### The Elegant Cancellation

As $x$ approaches zero, the term $ax/2$ also approaches zero. Therefore, the term inside the parenthesis becomes exactly $1$.

We are left with:

$$1 \cdot \frac{a}{2} = \frac{a}{2}$$

Now, we return to our continuity condition. We set our calculated limit equal to the function value at zero:

$$\frac{a}{2} = \frac{1}{2}$$

The twos cancel out with satisfying precision, leaving us with **$a = 1$**.

It is a moment of pure mathematical harmony. By understanding the nature of continuity and the structure of limits, we navigated straight to the heart of the problem to find the constant.

Enter Numerical Value:

Visualized Solution

Visualizing the Piecewise Function

The Continuity Condition

Setting up the Right Hand Limit

The Standard Exponential Limit

Algebraic Manipulation

Evaluating the Limit

Solving for $a$

Checking Differentiability (Optional Context)

Visualizing the Tangent

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Bridge of Continuity

The Right-Hand Limit Challenge

Algebraic Surgery

The Elegant Cancellation

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