Q.Find the value of $k$ so that the function $f$ is continuous at the indicated point: $f(x) = \begin{cases} \dfrac{1 - \cos kx}{x \sin x}, & x \ne 0 \\ \dfrac{1}{2}, & x = 0 \end{cases}$ at $x = 0$.

The core idea here is that direct substitution gives $\frac{0}{0}$, so we need to manipulate the expression algebraically. The presence of $\sqrt{2}\cos x - 1$ suggests rationalizing by multiplying numerator and denominator by the conjugate $\sqrt{2}\cos x + 1$, and the $\cot x - 1$ in the denominator can be rewritten in terms of $\sin x$ and $\cos x$ to reveal cancellations.

Let’s work through it step by step.

1. Check the form at $x = \frac{\pi}{4}$

   $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$, so $\sqrt{2}\cos\frac{\pi}{4} = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1$, making the numerator $1 - 1 = 0$.

   $\cot\frac{\pi}{4} = 1$, so the denominator is $1 - 1 = 0$.

   This is a $\frac{0}{0}$ indeterminate form, so we proceed with algebraic manipulation.

2. Rationalize the numerator

   Multiply numerator and denominator by $\sqrt{2}\cos x + 1$:

because $(\sqrt{2}\cos x)^2 - 1^2 = 2\cos^2 x - 1$.

3. Simplify $2\cos^2 x - 1$

   Recall the double-angle identity: $\cos 2x = 2\cos^2 x - 1$. So the numerator becomes $\cos 2x$.

4. Rewrite the denominator in terms of sine and cosine

   $\cot x - 1 = \frac{\cos x}{\sin x} - 1 = \frac{\cos x - \sin x}{\sin x}$.

   So the expression is now:

5. Use another identity for $\cos 2x$ $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$. This is perfect because it cancels the $(\cos x - \sin x)$ factor in the denominator:

provided $\cos x \neq \sin x$ (which holds near $x = \pi/4$ except at the point itself). …

Concept & Intuition

This function is a classic example of a piecewise-defined function that behaves like the sign of $x$ for all nonzero inputs. For $x>0$, $|x|/x = x/x = 1$; for $x<0$, $|x|/x = (-x)/x = -1$. At $x=0$, the function is defined separately as $0$. The key question is: does the function have a limit as $x$ approaches $0$? Because the left-hand and right-hand limits are different, the limit does not exist, so the function cannot be continuous at $0$. Everywhere else, the function is constant ($1$ or $-1$), so it is continuous there.

Step-by-step reasoning

1. Simplify the expression for $x \neq 0$

   For $x > 0$, $|x| = x$, so $f(x) = x/x = 1$.

   For $x < 0$, $|x| = -x$, so $f(x) = (-x)/x = -1$.

   Thus, for all $x \neq 0$, $f(x)$ is either $1$ (if $x>0$) or $-1$ (if $x<0$).

2. Check continuity at $x=0$

   A function is continuous at a point if the left-hand limit, right-hand limit, and the function value at that point all agree.

   • Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 1 = 1$.
   • Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-1) = -1$.
   • Function value: $f(0) = 0$. Since $1 \neq -1 \neq 0$, the limit does not exist, so $f$ is discontinuous at $x=0$.

3. Check continuity for $x > 0$ (any positive number)

   For any $a > 0$, there is an interval around $a$ that stays positive (e.g., $(a/2, 2a)$). On that interval, $f(x) = 1$ (constant). A constant function is continuous everywhere. So $f$ is continuous at every $x > 0$.

4. Check continuity for $x < 0$ (any negative number)

   Similarly, for any $b < 0$, there is an interval around $b$ that stays negative. On that interval, $f(x) = -1$ (constant). So $f$ is continuous at every $x < 0$. …

Rolle’s theorem is a special case of the Mean Value Theorem. It says: if a function $f$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c$ in $(a, b)$ such that $f'(c) = 0$.

To check where the theorem is not applicable, we test each condition — continuity, differentiability, and equal endpoints — for every option. The moment any one condition fails, Rolle’s theorem does not apply.

Let’s go through each case.

1. Option (A): $f(x) = x^2 - 4x + 5$ on $[1, 3]$

   This is a polynomial — continuous and differentiable everywhere.

   Check endpoints: $f(1) = 1 - 4 + 5 = 2$, $f(3) = 9 - 12 + 5 = 2$. So $f(1) = f(3)$.

   All conditions satisfied. Rolle’s theorem applies.

2. Option (B): $f(x) = x^2 - x$ on $[0, 1]$

   Again a polynomial — continuous and differentiable everywhere.

   Endpoints: $f(0) = 0$, $f(1) = 1 - 1 = 0$. So $f(0) = f(1)$.

   All conditions satisfied. Rolle’s theorem applies.

3. Option (C): $f(x) = |x|$ on $[-2, 2]$

   This is the absolute value function. It is continuous everywhere, including at $x = 0$.

   Endpoints: $f(-2) = 2$, $f(2) = 2$ — equal.

   But is it differentiable on $(-2, 2)$? No — at $x = 0$, the graph has a sharp corner. The left-hand derivative is $-1$, the right-hand derivative is $+1$, so $f$ is not differentiable at $x = 0$, which lies inside the open interval.

   Since differentiability fails, Rolle’s theorem does not apply. …

Step 1 — Right-hand limit

For $x \geq 2$, $f(x) = x^2-1$, so

Step 2 — Left-hand limit

For $x < 2$, $f(x) = x+1$, so

$\lim_{x \to 2^-} f(x) = 2+1 = 3$ …