Q.Examine the differentiability of $f$, where $f$ is defined by $f(x) = \begin{cases} x[x], & 0 \le x < 2 \\ (x - 1)x, & 2 \le x < 3 \end{cases}$ at $x = 2$ (here $[x]$ denotes the greatest integer function).

We need to decide whether $f$ is continuous and/or differentiable at $x=1$. The function is defined piecewise:

The key idea: continuity checks whether the graph has a break; differentiability checks whether it has a sharp corner. At a piecewise boundary, we must compare the left-hand and right-hand limits (for continuity) and the left-hand and right-hand derivatives (for differentiability).

1. Check continuity at $x=1$

   • Left-hand limit: as $x \to 1^-$, we use $f(x)=x$, so $\lim_{x \to 1^-} f(x) = 1$.
   • Right-hand limit: as $x \to 1^+$, we use $f(x)=2x-1$, so $\lim_{x \to 1^+} f(x) = 2(1)-1 = 1$.
   • Function value: $f(1) = 1$ (since $x=1$ falls in the first piece). Since left limit = right limit = $f(1)$, the function is continuous at $x=1$.

2. Check differentiability at $x=1$

   Differentiability requires that the derivative from the left equals the derivative from the right.

   • Left-hand derivative: for $x \le 1$, $f(x)=x$, so $f'(x)=1$. Thus the left-hand derivative at $x=1$ is $1$.
   • Right-hand derivative: for $x > 1$, $f(x)=2x-1$, so $f'(x)=2$. Thus the right-hand derivative at $x=1$ is $2$. …

Concept: If f' is continuous everywhere then f must be differentiable everywhere, hence continuous; both f and f' must match at the junction x = -1.

f(x) = b x^2 + a x + 4 , x >= -1

f(x) = a x^2 + b , x < -1

Continuity at x = -1:

b(1) + a(-1) + 4 = a(1) + b

b - a + 4 = a + b

4 = 2a => a = 2

Derivative matching at x = -1:

Right branch: f'(x) = 2bx + a -> at x = -1: -2b + a

Left branch: f'(x) = 2ax -> at x = -1: -2a …

The key idea is that continuity at a point means the function's value equals its limit there. Here, $f(0)$ is given as $\frac12$, so we need to find $K$ such that $\lim_{x\to 0} f(x) = \frac12$.

The expression $\frac{1-\cos Kx}{x\sin x}$ is a classic $0/0$ form at $x=0$. The numerator involves $\cos Kx$, and the denominator has $x\sin x$. The standard trick is to use the small-angle approximations or the known limit $\lim_{\theta\to 0}\frac{1-\cos\theta}{\theta^2} = \frac12$. This lets us rewrite the numerator in terms of $(Kx)^2$, and the denominator in terms of $x^2$, so the ratio becomes a constant times $K^2$.

Let's work through it step by step.

1. Set up the continuity condition. For $f$ to be continuous at $x=0$, we require

So we need

2. Rewrite the numerator using a standard limit. Multiply numerator and denominator by $(Kx)^2$ in a clever way:

This separates the limit into a product of two known limits.

3. Evaluate the first factor. As $x\to 0$, let $\theta = Kx$. Then

This is a fundamental limit you should remember.

4. Evaluate the second factor.

As $x\to 0$, $\frac{\sin x}{x} \to 1$, so $\frac{x}{\sin x} \to 1$. Hence

5. Combine the two limits. Since both limits exist, the product is the product of the limits: $$$$