(a) Check whether the function f(x) defined as $f(x) = \begin{cases} \dfrac{|x - 3|}{2(x - 3)}, & x < 3 \\ \dfrac{x-6}{6}, & x \ge 3 \end{cases}$ is continuous at $x = 3$ or not.

OR

(b) If $\sqrt{3} (x^2 + y^2) = 4xy$ , then find $\dfrac{dy}{dx}$ at $\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)$ .

Q: (a) Check whether the function f(x) defined as $f(x) = \begin{cases} \dfrac{|x - 3|}{2(x - 3)}, & x < 3 \\ \dfrac{x-6}{6}, & x \ge 3 \end{cases}$ is continuous at $x = 3$ or not. OR (b) If $\sqrt{3} (x^2 + y^2) = 4xy$, then find $\dfrac{dy}{dx}$ at $\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)$.

This question carries an internal choice; either part, fully answered, earns full marks. **Part (a) — Continuity at $x = 3$** - **Left-hand limit ($x < 3$):** since $x - 3 < 0$, $|x - 3| = -(x - 3)$, so $f(x) = -(x - 3)/(2(x - 3)) = -1/2$. Hence $LHL = -1/2$. - **Value and right-hand limit ($x \ge 3$):** $f(x) = (x - 6)/6$, so $f(3) = (3 - 6)/6 = -1/2$ and RHL = −1/2. - Since **$LHL = f(3) = RHL = -1/2$**, the function is **continuous at $x = 3$**. **Part (b) — Implicit differentiation** Differentiating $\sqrt{3}(x^2 + y^2) = 4xy$ gives $dy/dx = (2y - \sqrt{3} x)/(\sqrt{3} y - 2x)$. At $(1/2, \sqrt{3}/2)$: $\text{numerator} = \sqrt{3} - \sqrt{3}/2 = \sqrt{3}/2$, $\text{denominator} = 3/2 - 1 = 1/2$, so **$dy/dx = \sqrt{3}$**.

Q: How does the modulus affect the left-hand limit at $x = 3$?

For $x < 3$, $x - 3$ is negative, so $|x - 3| = -(x - 3)$. The branch $|x - 3|/(2(x - 3))$ therefore simplifies to $-1/2$ for every $x < 3$, making the left-hand limit exactly −1/2.

Q: How do you differentiate $\sqrt{3}(x^2 + y^2) = 4xy$ implicitly?

Differentiate each side with respect to x, treating y as a function of x: the left side gives $\sqrt{3}(2x + 2y \cdot dy/dx)$ and the right side gives $4(y + x \cdot dy/dx)$ by the product rule. Then collect dy/dx terms and solve, giving $dy/dx = (2y - \sqrt{3} x)/(\sqrt{3} y - 2x)$.

Continuity and Differentiability

Official Answer

This question carries an internal choice; either part, fully answered, earns full marks.

Part (a) — Continuity at $x = 3$

Left-hand limit ( $x < 3$ ): since $x - 3 < 0$ , $|x - 3| = -(x - 3)$ , so $f(x) = -(x - 3)/(2(x - 3)) = -1/2$ . Hence $LHL = -1/2$ .
Value and right-hand limit ( $x \ge 3$ ): $f(x) = (x - 6)/6$ , so $f(3) = (3 - 6)/6 = -1/2$ and RHL = −1/2.
Since $LHL = f(3) = RHL = -1/2$ , the function is continuous at $x = 3$ .

Part (b) — Implicit differentiation

Differentiating $\sqrt{3}(x^2 + y^2) = 4xy$ gives $dy/dx = (2y - \sqrt{3} x)/(\sqrt{3} y - 2x)$ . At $(1/2, \sqrt{3}/2)$ : $\text{numerator} = \sqrt{3} - \sqrt{3}/2 = \sqrt{3}/2$ , $\text{denominator} = 3/2 - 1 = 1/2$ , so $dy/dx = \sqrt{3}$ .

continuity at a pointleft-hand and right-hand limitsmodulus functionimplicit differentiationdy/dxpiecewise function

Marking Scheme

1Part (a): 1 mark for correctly evaluating $LHL = -1/2$ (handling $|x - 3| = -(x - 3)$ on $x < 3$ ); 1 mark for showing $f(3) = RHL = -1/2$ and concluding continuity.
2Part (b): 1 mark for correct implicit differentiation giving $dy/dx = (2y - \sqrt{3} x)/(\sqrt{3} y - 2x)$ ; 1 mark for substituting the point to get $dy/dx = \sqrt{3}$ .

Hint

Part (a): on $x < 3$ , replace $|x - 3|$ by $-(x - 3)$ and simplify, then compare LHL, $f(3)$ and RHL. Part (b): differentiate both sides with respect to x (chain rule on $y^2$ , product rule on $4xy$ ), solve for $dy/dx$ , then substitute $(1/2, \sqrt{3}/2)$ .

Quick Oral Answer

For part (a), on $x < 3$ the modulus makes $f(x) = -1/2$ , and the $x \ge 3$ branch gives $f(3) = -1/2$ with the same right limit, so $LHL = f(3) = RHL = -1/2$ and f is continuous. For part (b), implicit differentiation of $\sqrt{3}(x^2 + y^2) = 4xy$ gives $dy/dx = (2y - \sqrt{3} x)/(\sqrt{3} y - 2x)$ , which is $\sqrt{3}$ at $(1/2, \sqrt{3}/2)$ .

Analysis & Explanation

This VSA offers a choice between a continuity check and an implicit-differentiation problem.

Part (a) — concept: A function is continuous at x = a when the left-hand limit, the right-hand limit and the functional value all coincide: $\lim_{x\to a^-} f(x) = f(a) = \lim_{x\to a^+} f(x)$ . The catch here is the modulus branch for $x < 3$ : because x − 3 is negative, $|x - 3| = -(x - 3)$ , which cancels the denominator to a constant $-\frac{1}{2}$ . The $x \ge 3$ branch $\frac{x-6}{6}$ gives the value and the right limit at 3, both $-\frac{1}{2}$ . All three agree, so f is continuous.

Part (b) — concept: When a curve mixes $x^2$ , $y^2$ and $xy$ , y cannot be isolated, so we differentiate implicitly — chain rule on y² (→ $2y \cdot dy/dx$ ) and product rule on $4xy$ (→ $4y + 4x \cdot dy/dx$ ) — then solve for dy/dx and substitute the point. The point $(1/2, \sqrt{3}/2)$ does lie on the curve since $\sqrt{3}(1/4 + 3/4) = \sqrt{3} = 4\cdot(1/2)\cdot(\sqrt{3}/2)$ .

Exam trap: In (a), students forget that |x − 3| flips sign on the $x < 3$ side and wrongly get +1/2, concluding discontinuity; in (b), they forget the product rule on the $4xy$ term.

Common Mistakes

1On the $x < 3$ branch, forgetting that $|x - 3| = -(x - 3)$ , which reverses the sign of the limit (getting $+1/2$ instead of $-1/2$ ) and a wrong conclusion of discontinuity.
2Checking only one side (LHL or RHL) instead of confirming $LHL = f(3) = RHL$ for continuity.
3In part (b), forgetting the product rule on $4xy$ or the chain rule on $y^2$ while differentiating implicitly.

Interesting Facts

The expression $|x - a|/(x - a)$ is the classic 'sign function' building block: it equals $-1$ on one side of $a$ and $+1$ on the other, so such piecewise functions are prime material for testing one-sided limits.

Implicit differentiation lets us find slopes on curves like $\sqrt{3}(x^2 + y^2) = 4xy$ that cannot be rearranged into $y = f(x)$ , which is why it is indispensable for conics and other implicitly-defined relations.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

How does the modulus affect the left-hand limit at $x = 3$ ?

For $x < 3$ , $x - 3$ is negative, so $|x - 3| = -(x - 3)$ . The branch $|x - 3|/(2(x - 3))$ therefore simplifies to $-1/2$ for every $x < 3$ , making the left-hand limit exactly −1/2.

How do you differentiate $\sqrt{3}(x^2 + y^2) = 4xy$ implicitly?

Differentiate each side with respect to x, treating y as a function of x: the left side gives $\sqrt{3}(2x + 2y \cdot dy/dx)$ and the right side gives $4(y + x \cdot dy/dx)$ by the product rule. Then collect dy/dx terms and solve, giving $dy/dx = (2y - \sqrt{3} x)/(\sqrt{3} y - 2x)$ .