Question 212 marksVery Short AnswerSection B

(a) Check whether the function f(x) defined as f(x)={∣x−3∣2(x−3),x<3x−66,x≥3f(x) = \begin{cases} \dfrac{|x - 3|}{2(x - 3)}, & x < 3 \\ \dfrac{x-6}{6}, & x \ge 3 \end{cases} is continuous at x=3x = 3 or not.


OR


(b) If 3(x2+y2)=4xy\sqrt{3} (x^2 + y^2) = 4xy, then find dydx\dfrac{dy}{dx} at (12,32)\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right).

Continuity and Differentiability
Continuity and Differentiability

Official answer

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


Part (a) — Continuity at x=3x = 3


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

Part (b) — Implicit differentiation


Differentiating 3(x2+y2)=4xy\sqrt{3}(x^2 + y^2) = 4xy gives dy/dx=(2y−3x)/(3y−2x)dy/dx = (2y - \sqrt{3} x)/(\sqrt{3} y - 2x). At (1/2,3/2)(1/2, \sqrt{3}/2): numerator=3−3/2=3/2\text{numerator} = \sqrt{3} - \sqrt{3}/2 = \sqrt{3}/2, denominator=3/2−1=1/2\text{denominator} = 3/2 - 1 = 1/2, so dy/dx=3dy/dx = \sqrt{3}.

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

Marking scheme

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

Hint

Part (a): on x<3x < 3, replace ∣x−3∣|x - 3| by −(x−3)-(x - 3) and simplify, then compare LHL, f(3)f(3) and RHL. Part (b): differentiate both sides with respect to x (chain rule on y2y^2, product rule on 4xy4xy), solve for dy/dxdy/dx, then substitute (1/2,3/2)(1/2, \sqrt{3}/2).

Quick oral answer

For part (a), on x<3x < 3 the modulus makes f(x)=−1/2f(x) = -1/2, and the x≥3x \ge 3 branch gives f(3)=−1/2f(3) = -1/2 with the same right limit, so LHL=f(3)=RHL=−1/2LHL = f(3) = RHL = -1/2 and f is continuous. For part (b), implicit differentiation of 3(x2+y2)=4xy\sqrt{3}(x^2 + y^2) = 4xy gives dy/dx=(2y−3x)/(3y−2x)dy/dx = (2y - \sqrt{3} x)/(\sqrt{3} y - 2x), which is 3\sqrt{3} at (1/2,3/2)(1/2, \sqrt{3}/2).

Common mistakes

  1. 1.On the x<3x < 3 branch, forgetting that ∣x−3∣=−(x−3)|x - 3| = -(x - 3), which reverses the sign of the limit (getting +1/2+1/2 instead of −1/2-1/2) and a wrong conclusion of discontinuity.
  2. 2.Checking only one side (LHL or RHL) instead of confirming LHL=f(3)=RHLLHL = f(3) = RHL for continuity.
  3. 3.In part (b), forgetting the product rule on 4xy4xy or the chain rule on y2y^2 while differentiating implicitly.

Analysis and 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→a−f(x)=f(a)=lim⁡x→a+f(x)\lim_{x\to a^-} f(x) = f(a) = \lim_{x\to a^+} f(x). The catch here is the modulus branch for x<3x < 3: because x − 3 is negative, ∣x−3∣=−(x−3)|x - 3| = -(x - 3), which cancels the denominator to a constant −12-\frac{1}{2}. The x≥3x \ge 3 branch x−66\frac{x-6}{6} gives the value and the right limit at 3, both −12-\frac{1}{2}. All three agree, so f is continuous.


Part (b) — concept: When a curve mixes x2x^2, y2y^2 and xyxy, y cannot be isolated, so we differentiate implicitly — chain rule on y² (→ 2y⋅dy/dx2y \cdot dy/dx) and product rule on 4xy4xy (→ 4y+4x⋅dy/dx4y + 4x \cdot dy/dx) — then solve for dy/dx and substitute the point. The point (1/2,3/2)(1/2, \sqrt{3}/2) does lie on the curve since 3(1/4+3/4)=3=4⋅(1/2)⋅(3/2)\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<3x < 3 side and wrongly get +1/2, concluding discontinuity; in (b), they forget the product rule on the 4xy4xy term.

Interesting facts

The expression ∣x−a∣/(x−a)|x - a|/(x - a) is the classic 'sign function' building block: it equals −1-1 on one side of aa and +1+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 3(x2+y2)=4xy\sqrt{3}(x^2 + y^2) = 4xy that cannot be rearranged into y=f(x)y = f(x), which is why it is indispensable for conics and other implicitly-defined relations.

Question details

Q21 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Continuity and Differentiability
Topic
Continuity and Differentiability
Bloom's level
apply
Syllabus topic
Unit 2: Continuity and Differentiability — Continuity of a Function at a Point / Implicit Differentiation
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

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

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

How do you differentiate 3(x2+y2)=4xy\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 3(2x+2y⋅dy/dx)\sqrt{3}(2x + 2y \cdot dy/dx) and the right side gives 4(y+x⋅dy/dx)4(y + x \cdot dy/dx) by the product rule. Then collect dy/dx terms and solve, giving dy/dx=(2y−3x)/(3y−2x)dy/dx = (2y - \sqrt{3} x)/(\sqrt{3} y - 2x).