Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f:R\to R$ be defined as $f(x)=\{\begin{array}{ll}2\sin (-\frac{\pi }{2}),& \text{if }x<-1\\ a{x}^2+x+b,& \text{if }-1\le x\le 1\\ \sin (\pi x),& \text{if }x>1\end{array}$ If $f(x)$ is continuous on $R$, then $a+b$ equals :

Imagine you are an architect designing a bridge that spans across the entire real number line. This bridge is a piecewise construction, and for it to be safe—for it to be continuous—there can be no gaps where the sections meet.

If the road ends at height $y=-2$ and the next section begins at $y=5$, you have a dangerous jump. Today, we are going to ensure our mathematical bridge is perfectly connected by enforcing continuity at the boundaries $x=-1$ and $x=1$.

Where the Road Meets the Curve

Our function $f(x)$ is defined in three parts. Let's look at the first boundary, $x=-1$.

To the left of this point, our bridge is a constant:

Using the identity $\sin (-\theta )=-\sin (\theta )$, our expression becomes $-2\sin \left(\frac{\pi }{2}\right)$. Since $\sin \left(\frac{\pi }{2}\right)=1$, the value is simply $-2$. This is our Left-Hand Limit (LHL) at $x=-1$.

Now, consider the middle section: $f(x)=a{x}^2+x+b$. As we approach $x=-1$ from the right, we must ensure the parabola meets our horizontal line at exactly the same height.

Substituting $x=-1$ into the quadratic, we get:

For the bridge to be continuous, these two heights must be identical:

Rearranging this, we find our first crucial relationship:

The Wavy Connection

Now, let's travel to the other side of our bridge, at $x=1$. To the right of this point, our function is a sine wave: $f(x)=\sin (\pi x)$.

As we approach $x=1$ from the right, we evaluate the limit:

This is our Right-Hand Limit (RHL) at $x=1$.

Returning to our middle section, we approach $x=1$ from the left. Substituting $x=1$ into our parabola $a{x}^2+x+b$, we get:

For the bridge to be continuous, the height of the parabola must match the height of the sine wave:

If we shift the $1$ to the other side, we get:

Look at what we have discovered! Both boundary conditions, despite being at different points and involving different functions, lead us to the exact same conclusion: $a+b=-1$.

It is as if the math itself is whispering the answer to us. We did not need to find $a$ or $b$ individually; the structure of the problem was designed to reveal the sum $a+b$ directly.

By respecting the rules of continuity—ensuring the LHL equals the RHL at every junction—we have successfully built our bridge. The value of $a+b$ is $-1$, and our bridge is perfectly, beautifully continuous.