Pattern 1 — Constants That Make a Function Continuous or Differentiable

JEE Main's favourite question from this chapter: a piecewise function with unknown constants, and the demand that it be continuous — or the stricter demand that it be differentiable — at the joint.

Continuity at the joint $x = c$: equate three things — LHL, RHL, $f(c)$. One equation per unknown usually.

Differentiability at the joint: two conditions — first continuity (always check it; differentiability presumes it), then equality of the one-sided derivatives. Two equations, so typically two unknowns.

Worked template: $f(x) = \begin{cases} ax^2 + 1, & x \leq 1 \\ x + a, & x > 1 \end{cases}$ differentiable at $x = 1$: continuity gives $a + 1 = 1 + a$ (automatic here); matching slopes gives $2ax\big\vert_{x=1} = 1$, so $a = \frac{1}{2}$.

When the piece at the joint is a limit-form like $\frac{\sin 5x}{3x}$ or $\frac{1 - \cos 2x}{x^2}$, the constant is just the standard limit: $\lim_{x\to 0}\frac{\sin kx}{x} = k$, $\lim_{x\to 0}\frac{1 - \cos kx}{x^2} = \frac{k^2}{2}$, $\lim_{x\to 0}\frac{e^{kx} - 1}{x} = k$, $\lim_{x\to 0}\frac{\log(1 + kx)}{x} = k$.

Pattern 2 — Counting Kinks

Every $\vert x - a \vert$ contributes a corner at $a$ and smoothness everywhere else, and sums of moduli stack their corners. So:

1. $\vert x - 1 \vert + \vert x - 3 \vert$: non-differentiable at exactly $x = 1$ and $x = 3$.
2. $\vert x \vert + \vert x - 1 \vert + \vert x + 1 \vert$: exactly three bad points, $-1, 0, 1$.
3. $\vert \sin x \vert$: corners wherever $\sin x = 0$ with nonzero slope — every multiple of $\pi$.

General kink principle: $\vert f(x) \vert$ picks up a corner at each simple zero of $f$ (where $f = 0$ but $f' \neq 0$); at a zero of even multiplicity — like $x^2$ at $0$ — the graph touches smoothly and no corner forms. And one exotic specimen worth knowing: $f(x) = x^2\sin\frac{1}{x}$ (with $f(0) = 0$) is differentiable even at $0$ — the squeeze gives $f'(0) = 0$ — showing differentiability at a point can coexist with wild oscillation nearby.

Pattern 3 — The Substitution Dictionary

Before differentiating any inverse-trig expression, try to collapse it:

Each collapse turns a quotient-rule marathon into differentiating $2\theta$ or $3\theta$ — watch the interval, which decides the sign.

Pattern 4 — Differentiating One Function w.r.t. Another

"$\text{Differentiate } u \text{ w.r.t. } v$" means $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}$ — compute both $x$-derivatives and divide. E.g. the derivative of $e^x$ w.r.t. $\sqrt{x}$ is $\dfrac{e^x}{1/(2\sqrt{x})} = 2\sqrt{x}\,e^x$.

Worked Examples

Example 1 — A limit-form continuity constant

Find $k$ so that $f(x) = \begin{cases} \dfrac{1 - \cos 2x}{x^2}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $0$.

Step 1 — the standard limit: $1 - \cos 2x = 2\sin^2 x$, so $\dfrac{1 - \cos 2x}{x^2} = 2\left(\dfrac{\sin x}{x}\right)^2 \to 2$.

Answer: $k = 2$ — continuity at a removable point just means "set the value equal to the limit".

Example 2 — A differentiability constant

Find $a$ so that $f(x) = \begin{cases} ax^2 + 1, & x \leq 1 \\ x + a, & x > 1 \end{cases}$ is differentiable at $x = 1$.

Step 1 — continuity first: LHL $= a + 1$, RHL $= 1 + a$ — equal for every $a$; no constraint yet.

Step 2 — match the one-sided derivatives: left slope $2ax\big\vert_{1} = 2a$; right slope $1$.

Answer: $2a = 1$, so $a = \dfrac{1}{2}$ — the differentiability demand adds the equation that continuity alone could not supply.

Example 3 — Two constants across three pieces

Find $a$ and $b$ so that $f(x) = \begin{cases} 1, & x \leq 3 \\ ax + b, & 3 < x < 5 \\ 7, & x \geq 5 \end{cases}$ is continuous.

Step 1 — joint at $3$: $3a + b = 1$.

Step 2 — joint at $5$: $5a + b = 7$.

Step 3 — solve the pair: subtracting, $2a = 6$, so $a = 3$ and $b = 1 - 9 = -8$.

Answer: $a = 3$, $b = -8$ — the middle line must hit both flat pieces exactly.

Example 4 — Counting kinks in a modulus sum

At how many points is $f(x) = \vert x - 1 \vert + \vert x - 3 \vert$ not differentiable?

Step 1 — one corner per modulus: corners can only occur at $x = 1$ and $x = 3$.

Step 2 — check they survive: near $x = 1$ the second term equals the linear function $3 - x$, which is smooth — so the corner of the first term is undisturbed; symmetrically at $x = 3$.

Answer: exactly $2$ points — continuous everywhere, corners at $1$ and $3$.

Example 5 — The oscillating exception

Show that $f(x) = \begin{cases} x^2\sin\frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ is differentiable at $0$.

Step 1 — the difference quotient: $\dfrac{f(h) - f(0)}{h} = h\sin\dfrac{1}{h}$.

Step 2 — squeeze: $\left\vert h\sin\frac{1}{h} \right\vert \leq \vert h \vert \to 0$.

Answer: $f'(0) = 0$ exists — the $x^2$ factor tames the infinite oscillation of $\sin\frac{1}{x}$. (With $x\sin\frac{1}{x}$ instead, the quotient becomes $\sin\frac{1}{h}$, which has no limit — not differentiable. The power of $x$ decides.)

Example 6 — Dictionary collapse with a sine substitution

Differentiate $\sin^{-1}\left(2x\sqrt{1 - x^2}\right)$ for $\vert x \vert < \dfrac{1}{\sqrt{2}}$.

Step 1 — substitute $x = \sin\theta$: $2x\sqrt{1 - x^2} = 2\sin\theta\cos\theta = \sin 2\theta$.

Step 2 — collapse: $y = 2\theta = 2\sin^{-1} x$ on this interval.

Step 3 — differentiate: $\dfrac{2}{\sqrt{1 - x^2}}$.

Answer: $\dfrac{2}{\sqrt{1 - x^2}}$ — the interval $\vert x \vert < \frac{1}{\sqrt 2}$ keeps $2\theta$ inside the principal branch, which is what licenses the collapse.

Example 7 — A quarter-turn collapse

Differentiate $\tan^{-1}\left(\dfrac{\cos x}{1 + \sin x}\right)$ w.r.t. $x$.

Step 1 — rewrite the fraction: with $\cos x = \sin\left(\frac{\pi}{2} - x\right)$ and $1 + \sin x = 1 + \cos\left(\frac{\pi}{2} - x\right)$, the half-angle identity gives $$\frac{\cos x}{1 + \sin x} = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)$$

Step 2 — collapse and differentiate: $y = \frac{\pi}{4} - \frac{x}{2}$, so $y' = -\frac{1}{2}$.

Answer: $-\dfrac{1}{2}$ — constant, with no quotient rule anywhere in sight.

Example 8 — Relative differentiation

Differentiate $e^x$ with respect to $\sqrt{x}$.

Step 1 — both $x$-rates: $\frac{d}{dx}e^x = e^x$ and $\frac{d}{dx}\sqrt{x} = \frac{1}{2\sqrt{x}}$.

Step 2 — divide: $\dfrac{e^x}{1/(2\sqrt{x})} = 2\sqrt{x}\,e^x$.

Answer: $2\sqrt{x}\,e^x$ — the ratio-of-rates recipe, every time.