Let the function,
f(x) = {
-3ax²-2, x

Q: Wait, why do we need to check continuity if the problem already says the function is differentiable?

That is a fantastic conceptual question! In the hierarchy of calculus, differentiability is a 'stronger' condition than continuity. A function can be continuous but not differentiable (like a sharp corner), but it is impossible for a function to be differentiable without being continuous. We use the continuity condition as a necessary 'bridge' to lock down the values of our constants at the transition point $x=1$.

Q: Why do we reject $a=1$ when solving the quadratic equation?

In JEE Advanced problems, constraints are not just background noise; they are critical filters. The problem explicitly states $a > 1$. When we solved the quadratic equation $(a-1)(a-2)=0$, we found two potential candidates. Since $a=1$ violates the given domain constraint, we must discard it. Always keep an eye on those domain restrictions!

Q: How do we know the line $y=-20$ is actually below the curve to form an enclosed area?

Excellent intuition! If you look at the first piece of our function, $-6x^2-2$, it is a downward-opening parabola with a vertex at $(0, -2)$. Since $-20$ is significantly lower than the vertex, the horizontal line $y=-20$ will inevitably slice through the parabola, creating a bounded region. Visualizing the 'geometry' of the function is half the battle in calculus.

Question

Let the function,
$f(x) = \begin{cases}-3ax^{2}-2, & x<1\ a^{2}+bx, & x\ge1\end{cases}$
be differentiable for all $x\in R$ where $a>1,$ $b \in R$. If the area of the region enclosed by $y=f(x)$ and the line $y=-20$ is $\alpha+\beta\sqrt{3},$ $\alpha, \beta\in Z$ then the value of $\alpha+\beta$ is

Sigma Percentile · Accepted Answer

### The Geometry of Smoothness: A Journey Through Differentiability

Welcome, future engineers! Today, we are not just solving a math problem; we are exploring the delicate architecture of a function.

We are given a piecewise function:

$$f(x) = \begin{cases} -3ax^{2}-2, & x < 1 \ a^{2}+bx, & x \ge 1 \end{cases}$$

We are told it is differentiable for all real numbers. This is a powerful statement, implying that our function is a single, seamless entity that transitions perfectly at $x=1$.

### Phase 1: The Bridge of Continuity

Before we can talk about slopes, we must ensure the function is continuous. If there were a jump at $x=1$, the function would not be differentiable.

To ensure a smooth transition, the left-hand limit must equal the right-hand limit at $x=1$:

$$-3a(1)^2 - 2 = a^2 + b(1)$$

This simplifies beautifully into our first constraint:

$$a^2 + 3a + b + 2 = 0$$

### Phase 2: The Elegance of Differentiability

Now, we move to the heart of the problem. Differentiability requires that the slope of the curve must be the same from the left and the right as we approach $x=1$.

We differentiate both pieces:
1. The derivative of $-3ax^2 - 2$ is $-6ax$.
2. The derivative of $a^2 + bx$ is $b$.

At $x=1$, these slopes must match:

$$-6a(1) = b \implies b = -6a$$

By substituting $b = -6a$ into our continuity equation, we get:

$$a^2 + 3a - 6a + 2 = 0$$
$$a^2 - 3a + 2 = 0$$

Factoring this, we find $(a-1)(a-2) = 0$. Since the problem constraints imply $a > 1$, we reject $a=1$ and accept **$a=2$**.

With $a=2$, our constant $b$ becomes **$b = -12$**. Our function is now fully revealed:

$$f(x) = \begin{cases} -6x^2 - 2, & x < 1 \ 4 - 12x, & x \ge 1 \end{cases}$$

### Phase 3: The Area Under the Curve

With the function defined, we turn our attention to the area enclosed by $y=f(x)$ and the line $y=-20$. To find the boundaries, we solve $f(x) = -20$.

For the first piece, $-6x^2 - 2 = -20$ leads to $x^2 = 3$. Since we are in the region $x < 1$, we take $x = -\sqrt{3}$.

For the second piece, $4 - 12x = -20$ leads to $x = 2$. We now have our limits of integration: from $-\sqrt{3}$ to $2$.

We split the integral at the transition point $x=1$. The total area $A$ is the sum of two integrals:

$$A = \int_{-\sqrt{3}}^{1} (f(x) + 20) \, dx + \int_{1}^{2} (f(x) + 20) \, dx$$

For the first part, we integrate $(18 - 6x^2)$ from $-\sqrt{3}$ to $1$:

$$\int_{-\sqrt{3}}^{1} (18 - 6x^2) \, dx = [18x - 2x^3]_{-\sqrt{3}}^{1} = (18 - 2) - (-18\sqrt{3} + 6\sqrt{3}) = 16 + 12\sqrt{3}$$

For the second part, we integrate $(24 - 12x)$ from $1$ to $2$:

$$\int_{1}^{2} (24 - 12x) \, dx = [24x - 6x^2]_{1}^{2} = (48 - 24) - (24 - 6) = 24 - 18 = 6$$

Adding these together, we get $22 + 12\sqrt{3}$. Comparing this to the form $\alpha + \beta\sqrt{3}$, we identify $\alpha = 22$ and $\beta = 12$.

The final sum, $\alpha + \beta$, is **$34$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Problem

Continuity Condition at $x = 1$

Differentiability Condition at $x = 1$

Solving for $a$ and $b$

Finding Intersection Points

Intersection for $x < 1$

Intersection for $x \geq 1$

Setting up the Area Integral

Evaluating the First Integral

Evaluating the Second Integral

Final Summation

The Sigma Insight: Relationship Between Continuity and Differentiability

The Geometry of Smoothness

Phase 1

Phase 2

Phase 3

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