Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f(x)=\{\begin{array}{ll}-x-\frac{\pi }{2},& x\le -\frac{\pi }{2}\\ -\cos x,& -\frac{\pi }{2}<x\le 0\\ x-1,& 0<x\le 1\\ \ln x,& x>1\end{array}$

Welcome, fellow explorers of the calculus realm! Today, we are not just solving a problem; we are embarking on a journey of discovery.

We have a piecewise function, $f(x)$, which is like a patchwork quilt of different mathematical behaviors. Our mission is to investigate the 'seams' of this quilt—the boundary points where the function changes its identity—and determine if the graph is connected (continuous) and smooth (differentiable).

Imagine you are walking along a path. At $x=-\frac{\pi }{2}$, the path changes from a linear function, $f(x)=-x-\frac{\pi }{2}$, to a trigonometric one, $f(x)=-\cos x$.

For the path to be continuous, the two segments must meet at the exact same spot. We calculate the Left Hand Limit (LHL) as we approach from the left:

Now, we check the Right Hand Limit (RHL) from the right:

Since both limits are zero, the path is perfectly connected. The function is continuous at $x=-\frac{\pi }{2}$.

Next, we arrive at $x=0$. Here, the function switches from $-\cos x$ to $x-1$.

To check for differentiability, we need to see if the slope of the tangent line is the same from both sides. The derivative of $-\cos x$ is $\sin x$, and at $x=0$:

This means the slope from the left is horizontal. However, the derivative of $x-1$ is $1$. The slope from the right is $1$.

Because $0 eq 1$, the graph has a sharp corner or a 'kink' at $x=0$. It is continuous, but not differentiable.

Now, let's move to $x=1$. First, we check continuity: the left branch $x-1$ gives $1-1=0$, and the right branch $\ln x$ gives $\ln (1)=0$. It is continuous!

Now, for the smoothness: the derivative of $x-1$ is $1$. The derivative of $\ln x$ is $\frac{1}{x}$, and at $x=1$:

Since the slopes match perfectly, the line transitions into the logarithmic curve without a single bump. It is differentiable!

Finally, we encounter the point $x=-\frac{3}{2}$, which is $-1.5$. A common mistake is to treat every point as a boundary.

But look closely: $-\frac{\pi }{2}$ is approximately $-1.57$. Since $-1.5>-1.57$, this point lies strictly inside the interval $(-\frac{\pi }{2},0)$.

In this region, the function is just $-\cos x$. Since cosine is a smooth, differentiable function, any point inside this interval is also smooth. No limits needed here!

We have successfully navigated the landscape of this function. We found continuity at the first boundary, a sharp corner at the second, a smooth transition at the third, and a perfectly differentiable interior point at the end.

Every option provided in the problem is correct. Keep this detective mindset, and you will master any function that comes your way!