CBSE Board Examination 2026 · Section B · 2 marks

Q21: Continuity of a piecewise function at $x = 3$ ⁠, or $\frac{dy}{dx}$ on $\sqrt3(x^2 + y^2) = 4xy$

Board 2026Ch 5 · Continuity and Differentiability

The question

(a) Check whether function $f(x)$ defined as

f(x) = \begin{cases} \dfrac{|x - 3|}{2(x - 3)}, & x < 3 \\[6pt] \dfrac{x - 6}{6}, & x \ge 3 \end{cases}

is continuous at $x = 3$ or not?

(b) If $\sqrt3(x^2 + y^2) = 4xy$ ⁠, then find $\dfrac{dy}{dx}$ at $\left(\dfrac12, \dfrac{\sqrt3}{2}\right)$ ⁠.

Exam answer

(a)

For $x < 3$ ⁠, $|x - 3| = -(x - 3)$ ⁠, so

Why?

The modulus of a number is the number itself when it is positive, and the number with its sign changed when it is negative:

|a| = \begin{cases} a, & a \ge 0 \\ -a, & a < 0 \end{cases}

For $x < 3$ ⁠, the number $x - 3$ is negative, so $|x - 3| = -(x - 3)$ ⁠.

In the next line we may cancel $x - 3$ ⁠, because $x - 3 \ne 0$ when $x < 3$ ⁠. So to the left of 3 the function is the constant $-\frac12$ ⁠.

f(x) = \frac{-(x - 3)}{2(x - 3)} = -\frac12

f(3) = \frac{3 - 6}{6} = -\frac12

Why?

The second rule is for $x \ge 3$ ⁠, and that includes $x = 3$ ⁠. So $f(3)$ comes from $\frac{x - 6}{6}$ ⁠.

\begin{aligned} f(3) &= \frac{3 - 6}{6} = \frac{-3}{6} \\ &= -\frac{3 \div 3}{6 \div 3} = -\frac12 \end{aligned}

\begin{aligned} \text{LHL} &= \lim_{x \to 3^-} f(x) \\ &= \lim_{x \to 3^-}\left(-\frac12\right) = -\frac12 \end{aligned}

\begin{aligned} \text{RHL} &= \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+}\frac{x - 6}{6} \\ &= \frac{3 - 6}{6} = -\frac12 \end{aligned}

Why?

The left-hand limit uses values just below 3, so it uses the first rule. That rule is the constant $-\frac12$ ⁠, and the limit of a constant is the constant.

The right-hand limit uses values just above 3, so it uses the second rule. $\frac{x - 6}{6}$ is a polynomial, so we can put $x = 3$ straight in.

LHL = RHL = $f(3)$ ⁠, hence $f(x)$ is continuous at $x = 3$ ⁠.

Why?

$f$ is continuous at $x = a$ when both one-sided limits exist and equal the value there:

\lim_{x \to a^-}f = \lim_{x \to a^+}f = f(a)

Watch out: $\frac{|x - 3|}{2(x - 3)}$ looks as if it might jump at 3, but for $x < 3$ it is always $-\frac12$ ⁠, which is exactly $f(3)$ ⁠.

(b)

Differentiating both sides w.r.t. $x$ ⁠,

\sqrt3\left(2x + 2y\frac{dy}{dx}\right) = 4\left(x\frac{dy}{dx} + y\right)

Why?

$y$ is not given as a formula in $x$ ⁠, so we differentiate the equation as it stands (implicit differentiation), treating $y$ as a function of $x$ ⁠.

By the chain rule, and by the product rule for $xy$ ⁠:

\begin{aligned} \frac{d}{dx}(y^2) &= 2y\frac{dy}{dx} \\ \frac{d}{dx}(xy) &= x\frac{dy}{dx} + y \end{aligned}

So, side by side:

\begin{aligned} &\frac{d}{dx}\big(\sqrt3(x^2 + y^2)\big) \\ &= \sqrt3\left(2x + 2y\frac{dy}{dx}\right) \\[4pt] &\frac{d}{dx}(4xy) \\ &= 4\left(x\frac{dy}{dx} + y\right) \end{aligned}

2\sqrt3x + 2\sqrt3y\frac{dy}{dx} = 4x\frac{dy}{dx} + 4y

\begin{aligned} (2\sqrt3y - 4x)\frac{dy}{dx} &= 4y - 2\sqrt3x \\ \frac{dy}{dx} &= \frac{2y - \sqrt3x}{\sqrt3y - 2x} \end{aligned}

Why?

First open the brackets: $\sqrt3 \cdot 2x = 2\sqrt3x$ ⁠, $\sqrt3 \cdot 2y\frac{dy}{dx} = 2\sqrt3y\frac{dy}{dx}$ ⁠, and on the right $4 \cdot x\frac{dy}{dx} = 4x\frac{dy}{dx}$ ⁠, $4 \cdot y = 4y$ ⁠.

Collect the terms with $\frac{dy}{dx}$ on the left and the rest on the right, then take $\frac{dy}{dx}$ out as a common factor.

\begin{aligned} &2\sqrt3y\frac{dy}{dx} - 4x\frac{dy}{dx} \\ &= 4y - 2\sqrt3x \\[4pt] &(2\sqrt3y - 4x)\frac{dy}{dx} \\ &= 4y - 2\sqrt3x \\[4pt] &\frac{dy}{dx} = \frac{4y - 2\sqrt3x}{2\sqrt3y - 4x} \end{aligned}

Both the numerator and the denominator have a factor 2, which cancels.

\begin{aligned} \frac{dy}{dx} &= \frac{2(2y - \sqrt3x)}{2(\sqrt3y - 2x)} \\ &= \frac{2y - \sqrt3x}{\sqrt3y - 2x} \end{aligned}

At $\left(\frac12, \frac{\sqrt3}{2}\right)$ ⁠,

\begin{aligned} \frac{dy}{dx} &= \frac{2\cdot\frac{\sqrt3}{2} - \sqrt3\cdot\frac12}{\sqrt3\cdot\frac{\sqrt3}{2} - 2\cdot\frac12} \\ &= \frac{\frac{\sqrt3}{2}}{\frac32 - 1} = \frac{\frac{\sqrt3}{2}}{\frac12} \end{aligned}

Why?

The point lies on the curve: both sides of the equation equal $\sqrt3$ there.

\begin{aligned} &\sqrt3\left(\left(\tfrac12\right)^2 + \left(\tfrac{\sqrt3}{2}\right)^2\right) \\ &= \sqrt3\left(\tfrac14 + \tfrac34\right) = \sqrt3(1) = \sqrt3 \\[4pt] &4\left(\tfrac12\right)\left(\tfrac{\sqrt3}{2}\right) = \frac{4\sqrt3}{4} = \sqrt3 \end{aligned}

Put $x = \frac12$ ⁠, $y = \frac{\sqrt3}{2}$ ⁠. On the top, $\sqrt3 - \frac{\sqrt3}{2} = \frac{\sqrt3}{2}$ ⁠. On the bottom, $\sqrt3\cdot\sqrt3 = 3$ ⁠, so the first term is $\frac32$ ⁠.

\begin{aligned} \text{Top: } &2\cdot\tfrac{\sqrt3}{2} = \sqrt3 \\ &\sqrt3\cdot\tfrac12 = \tfrac{\sqrt3}{2} \\ &\sqrt3 - \tfrac{\sqrt3}{2} = \tfrac{\sqrt3}{2} \\[4pt] \text{Bottom: } &\sqrt3\cdot\tfrac{\sqrt3}{2} = \tfrac{3}{2} \\ &2\cdot\tfrac12 = 1 \\ &\tfrac32 - 1 = \tfrac12 \end{aligned}

Dividing by $\frac12$ is the same as multiplying by 2.

\frac{\frac{\sqrt3}{2}}{\frac12} = \frac{\sqrt3}{2} \times 2 = \sqrt3

$\therefore\ \dfrac{dy}{dx} = \sqrt3$ at $\left(\dfrac12, \dfrac{\sqrt3}{2}\right)$ ⁠.

Where this comes from

Chapter 5, Continuity and Differentiability: Continuity at a point; implicit differentiation.

Question type: (a) Is $f$ continuous at the joint? Compare three numbers; (b) Implicit

How to spot it: click to see

(a) Type 1. Is $f$ continuous at the joint? Compare three numbers

Spot it: $f$ is given in pieces, and you are asked where it is continuous, or where it breaks.

Here: Left of 3, $|x - 3| = -(x - 3)$ ⁠, so the first piece is just $-\frac12$ ⁠. The piece $\frac{x - 6}{6}$ gives RHL and $f(3)$ ⁠, both $-\frac12$ ⁠: all three agree, so it is continuous.

(b) Type 5. Implicit: $x$ and $y$ mixed in one equation

Spot it: an equation such as $x^2 + xy + y^2 = 100$ that you cannot easily solve for $y$ ⁠.

Here: $\sqrt3(x^2 + y^2) = 4xy$ mixes $x$ and $y$ ⁠: differentiate both sides, with the product rule on $4xy$ ⁠, collect the $\frac{dy}{dx}$ terms, then put in $\left(\frac12, \frac{\sqrt3}{2}\right)$ ⁠.

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Continuity and Differentiability in past papers: question 1 of 35 · see them all

Q21: Continuity of a piecewise function at x=3x = 3x=3⁠, or dydx\frac{dy}{dx}dxdy​ on 3(x2+y2)=4xy\sqrt3(x^2 + y^2) = 4xy3​(x2+y2)=4xy

Q21: Continuity of a piecewise function at $x = 3$ ⁠, or $\frac{dy}{dx}$ on $\sqrt3(x^2 + y^2) = 4xy$