Let f(x) be a continuous function given by f(x) = {2x, |x| ≤ 1; x² + ax + b, |x| 1}. Find the area

Q: Why do we have to split the integral at $x=-1$? Can't we just integrate the whole thing?

That is a fantastic question! Think of integration as the sum of infinite vertical strips. The area is bounded below by the function $f(x)$. If the 'floor' of your region changes its mathematical definition—which it does here, shifting from a parabola to a line at $x=-1$—you cannot use a single formula. You must respect the boundary where the function definition changes. If you don't split it, you are essentially trying to measure a room with two different floor heights using only one ruler!

Q: Why did we choose the negative root for $y = -\sqrt{-\frac{x}{2}}$?

Remember, we are strictly in the 3rd quadrant! In this quadrant, both $x$ and $y$ are negative. The equation $x = -2y^2$ implies $y^2 = -\frac{x}{2}$, so $y = \pm \sqrt{-\frac{x}{2}}$. If you chose the positive root, you would be calculating area in the 2nd quadrant. Always check your quadrant constraints before proceeding; it is the most common place to lose marks in coordinate geometry.

Q: Is there a trick to avoid the messy arithmetic with the square roots?

In JEE Advanced, sometimes the 'mess' is the point! Notice how the $\frac{\sqrt{2}}{3}$ terms cancelled out in the final summation? That is the examiner's way of rewarding you for staying calm and precise. If you find yourself doing complex arithmetic, don't panic—often, the terms are designed to simplify beautifully at the very end. Trust the process and keep your calculations organized.

Question

Let $f(x)$ be a continuous function given by $f(x)=\left\{\begin{array}{ll}2 x, & |x| \leq 1 \ x^{2}+a x+b, & |x|>1\end{array}ight.$. Find the area of the region in the third quadrant bounded by the curves $x=-2 y^{2}$ and $y=f(x)$ lying on the left of the line $8 x+1=0$.

Sigma Percentile · Accepted Answer

### The Detective Story of Continuity and Curves

Welcome, student. Today, we are not just solving a math problem; we are embarking on a detective story. We have a function, $f(x)$, that is hiding its true identity behind two constants, $a$ and $b$.

Our mission is to unmask this function, visualize the geometry of the 3rd quadrant, and calculate the area of a region trapped between two distinct curves. Take a deep breath. Let us begin.

### Phase 1: The Continuity Detective

We are given that $f(x)$ is continuous. In the language of calculus, this means there are no jumps, no holes, and no breaks in the graph. It is a smooth, unbroken path.

The function is defined as $f(x) = 2x$ when $|x| \leq 1$ and $f(x) = x^2 + ax + b$ when $|x| > 1$. To find $a$ and $b$, we must look at the boundaries where the function changes its definition: $x = 1$ and $x = -1$.

For the function to be continuous at $x = 1$, the limit from the left must equal the limit from the right. Approaching from the left, we use $2x$, which gives us $2(1) = 2$. Approaching from the right, we use $x^2 + ax + b$, which gives $1^2 + a(1) + b = 1 + a + b$.

Equating these, we get our first clue:
$$a + b = 1$$

Now, we repeat this logic at $x = -1$. The left-hand limit uses the quadratic piece, $x^2 + ax + b$, yielding $(-1)^2 + a(-1) + b = 1 - a + b$. The right-hand limit uses $2x$, yielding $2(-1) = -2$.

Equating these, we get $1 - a + b = -2$, or:
$$b - a = -3$$

We now have a system of two linear equations. Adding them, the $a$ terms vanish, leaving $2b = -2$, so $b = -1$. Substituting back, we find $a = 2$.

The mystery is solved: our function is $f(x) = x^2 + 2x - 1$ for $|x| > 1$ and $f(x) = 2x$ for $|x| \leq 1$.

### Phase 2: Visualizing the Battlefield

Now that we have our function, let us step into the 3rd quadrant. Here, both $x$ and $y$ are negative. We are bounded by the line $8x + 1 = 0$ (or $x = -1/8$) and the curve $x = -2y^2$.

Let us analyze the curve $x = -2y^2$. This is a parabola opening to the left. Since we are in the 3rd quadrant, $y$ must be negative.

Solving for $y$, we get $y = -\sqrt{-\frac{x}{2}}$. This is our upper boundary. For any given $x$ in our region, this curve sits above the other functions.

We also have our function $f(x)$ acting as the lower boundary. Look at the definition of $f(x)$: for $x < -1$, it is a parabola ($x^2 + 2x - 1$), and for $-1 \leq x \leq 0$, it is a straight line ($2x$).

This means the 'floor' of our region changes at $x = -1$. This is the critical moment where most students stumble. We cannot use a single integral; we must split our journey into two parts.

### Phase 3: The Integral Setup

To find the area, we use the fundamental principle:
$$	ext{Area} = \int (y_{upper} - y_{lower}) dx$$

Our first region spans from $x = -2$ to $x = -1$. Here, the upper curve is the green parabola $y = -\sqrt{-\frac{x}{2}}$, and the lower curve is the blue parabola $y = x^2 + 2x - 1$.

The integral is:
$$I_1 = \int_{-2}^{-1} \left(-\sqrt{-\frac{x}{2}} - (x^2 + 2x - 1)ight) dx$$

Our second region spans from $x = -1$ to $x = -1/8$. Here, the upper curve remains the green parabola, but the lower curve is now the straight line $y = 2x$.

The integral is:
$$I_2 = \int_{-1}^{-1/8} \left(-\sqrt{-\frac{x}{2}} - 2xight) dx$$

### Phase 4: The Grand Finale

Now, we perform the integration. For the term $-\sqrt{-\frac{x}{2}}$, which is $-\frac{1}{\sqrt{2}}(-x)^{1/2}$, the integral is $\frac{\sqrt{2}}{3}(-x)^{3/2}$.

Evaluating $I_1$ gives us $\frac{\sqrt{2}}{3} + \frac{1}{3}$. Evaluating $I_2$ gives us $\frac{193}{192} - \frac{\sqrt{2}}{3}$.

Look closely at these two results. The term $\frac{\sqrt{2}}{3}$ appears in both, but with opposite signs. As we add $I_1$ and $I_2$ to find the total area, these irrational terms vanish into thin air!

This is the elegance of mathematics. We are left with:
$$\frac{1}{3} + \frac{193}{192}$$

Converting to a common denominator of $192$, we get:
$$\frac{64 + 193}{192} = \frac{257}{192}$$

We have conquered the problem. We navigated the continuity, visualized the geometry, respected the boundaries, and arrived at the solution. The final area is **$\frac{257}{192}$**.

Visualized Solution

Continuity at $x = 1$

Continuity at $x = - 1$

Solving for $a$ and $b$

Defining $f (x)$ for the 3rd Quadrant

The Parabola $x = - 2 y^{2}$

Intersection at $x = - \frac{1}{8}$

Intersection at $x = - 2$

Setting up the Integral

Evaluating Part 1

Evaluating Part 2

Final Summation

The Sigma Insight: Area Bounded by Curves

The Detective Story of Continuity and Curves

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

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

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

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Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) for the 3rd Quadrant

The Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−2y2

Intersection at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−81​

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