These NCERT Solutions cover Exercise 5.6 of Class 12 Maths Chapter 5 Continuity and Differentiability in full, one question per page, in the same notation as the textbook. The free PDF is right below.
- CBSE Weightage: Exercise 5.6 problems map to 3-4 marks inside the Continuity and Differentiability 6-8 mark block.
- JEE Main Weightage: Parametric differentiation contributes roughly 2-3% of the Calculus questions.
Every solved sum in these notes follows the standard parametric rule dydx = dy/dtdx/dt , with the intermediate dx/dt and dy/dt shown line by line. Trigonometric simplifications (double-angle, half-angle, Pythagorean identities) are written before differentiating, so the algebra never gets compressed.
Every answer is cross-checked against the official NCERT key and the 2026-27 textbook. Where a simplified single-fraction answer is expected (Q7, Q9), the final boxed expression matches the NCERT reference verbatim.
What Exercise 5.6 Covers: The Parametric Differentiation Map
The the resource address this in the same order as the NCERT textbook.
Exercise 5.6 is dedicated entirely to differentiating functions where both x and y are expressed through a third variable, the parameter t. Direct elimination is awkward (sometimes impossible) so the chain-rule shortcut becomes the default tool.
| Q No. | Parametric form | Technique used | Answer type |
|---|---|---|---|
| 1 | x = 2at2, y = at4 | Power rule | Polynomial in t |
| 2 | x = acosθ, y = bcosθ | Trig derivative | Constant ratio b/a |
| 3 | x = sin t, y = cos 2t | Chain + double-angle | Linear in sin t |
| 4 | x = 4t, y = 4/t | Power rule | -1/t2 |
| 5 | x = cosθ - cos 2θ, y = sinθ - sin 2θ | Sum-to-product | Single trig ratio |
| 6 | x = a(θ - sinθ), y = a(1 + cosθ) | Half-angle identity | -cot(θ/2) |
| 7 | x = sin3 t√cos 2t, y = cos3 t√cos 2t | Quotient + chain | Negative tangent product |
| 8 | x = acos t + (t/2), y = asin t | Log derivative + chain | tan t |
| 9 | x = asecθ, y = btanθ | Secant-tangent | bacscθ |
| 10 | x = a(cosθ + θ), y = a(sinθ - θ) | Product rule | tanθ |
| 11 | Show dydx = -yx for x = √asin-1t, y = √acos-1t | Logarithmic differentiation | Proof type |
Three problems (Q6, Q8, Q10) are repeat favourites in CBSE boards and the school pre-board circuit. Q11 is a proof-style sum, which means full marks depend on the structure of the working, not just the final answer.
Continuity and Differentiability Ex 5 6 Video Walkthrough
Source: NCERT Wallah on YouTube
How the Continuity and Differentiability Class 12 NCERT Solutions on the Continuity and Differentiability Class 12 NCERT Solutions Help You
The chapter notes address this in the same order as the NCERT textbook.
Parametric differentiation looks small but the algebra is where students leak marks. Our solutions show every t-derivative on a separate line, every trigonometric simplification before differentiating, and the final ratio in lowest terms.
- Two-column layout for each problem: dx/dt on the left, dy/dt on the right, ratio at the bottom.
- Identity reminders printed in the margin (e.g. 1-cosθ = 2sin2(θ/2) ) so you do not have to flip back to Chapter 3.
- Sign-tracking callouts on Q6, Q7, Q10 where most students drop a negative.
- Expert's Solution after each main solve, giving the slicker route (often half-angle substitution) that examiners reward.
Continuity and Differentiability Exercise 5.6 Step-by-Step Approach
The the PDF address this in the same order as the NCERT textbook.
The parametric workflow is identical for every problem in this exercise. Internalise this four-step routine and you can clear Exercise 5.6 in under 25 minutes.
- Read both equations and decide whether trigonometric simplification is needed BEFORE differentiating (Q3, Q5, Q6 reward this).
- Compute dx/dt (or dx/dθ ) and keep the result un-simplified at first.
- Compute dy/dt separately. Do not combine yet.
- Form the ratio dy/dtdx/dt and simplify using the identity that fits the structure of the expression.
For proof-style Q11, you also need a fifth step: substitute back the original expressions for x and y to land on -y/x .
Exam Relevance of Class 12 Maths Chapter 5 Exercise 5.6
Parametric differentiation almost never appears as a standalone question. It shows up inside larger sums - tangent and normal questions in Chapter 6, area-under-curve sums in Chapter 8, even differential-equation modelling in Chapter 9. The table below maps the last five CBSE board sittings.
| Year | Marks from parametric form | Linked chapter |
|---|---|---|
| 2025 | 3 | Chapter 6 (tangent slope) |
| 2024 | 4 | Chapter 5 standalone |
| 2023 | 3 | Chapter 5 standalone |
| 2022 | - | - |
| 2021 | 4 | Chapter 6 (rate of change) |
Across the last five sittings, parametric differentiation has carried 3-4 marks on average, with one zero-weight year (2022). Treat it as a near-guaranteed 3-mark question for CBSE 2026.
Common Mistakes Students Make in Exercise 5.6
The this chapter are written in formal mathematical notation, line by line, in the same convention as the official NCERT print.
- Combining dx/dt and dy/dt too early. If you simplify before forming the ratio, sign errors creep in.
- Forgetting θ is the variable, not x. Several students differentiate as if dx replaces dθ .
- Dropping the half-angle substitution on Q6. Without 1+cosθ = 2cos2(θ/2) the answer never collapses to -cot(θ/2) .
- Forgetting to simplify on Q9. The expected final answer is bacscθ , not bsec2θasecθ .
NCERT Solutions Class 12 Mathematics: All Chapters
The full set of solutions for the 2026-27 NCERT syllabus is listed below. Solutions for chapters in active publication (Ch 1, 4, 5, 6 ...) are linked; the remaining chapters are being staged on the same template.
| Chapter | NCERT Solutions |
|---|---|
| Chapter 1 | Relations and Functions NCERT Solutions |
| Chapter 2 | Inverse Trigonometric Functions NCERT Solutions |
| Chapter 3 | Matrices NCERT Solutions |
| Chapter 4 | Determinants NCERT Solutions |
| Chapter 6 | Application of Derivatives NCERT Solutions |
| Chapter 7 | Integrals NCERT Solutions |
| Chapter 8 | Application of Integrals NCERT Solutions |
| Chapter 9 | Differential Equations NCERT Solutions |
| Chapter 10 | Vector Algebra NCERT Solutions |
| Chapter 11 | Three Dimensional Geometry NCERT Solutions |
| Chapter 12 | Linear Programming NCERT Solutions |
| Chapter 13 | Probability NCERT Solutions |
Other Resources for Continuity and Differentiability
- NCERT Notes Class 12 Maths Chapter 5
- Formula Sheet Class 12 Maths Chapter 5
- Handwritten Notes Class 12 Maths Chapter 5
- Exemplar Solutions Class 12 Maths Chapter 5
these notes: available above as a free PDF download, aligned to the 2026-27 NCERT Class 12 Mathematics syllabus.
Exercise-wise Breakdown of the Continuity and Differentiability Chapter
The Continuity and Differentiability chapter splits into 7 numbered exercises plus a Miscellaneous Exercise. The table below maps every exercise to the specific concept it tests, so students can plan revision per exercise and click straight into the worked solutions.
| Exercise | Topic Tested |
|---|---|
| Exercise 5.1 | Continuity at a point and on an interval |
| Exercise 5.2 | Algebra of continuous functions |
| Exercise 5.3 | Differentiability and chain rule |
| Exercise 5.4 | Derivatives of inverse trigonometric functions |
| Exercise 5.5 | Logarithmic differentiation |
| Exercise 5.6 | Parametric and implicit differentiation |
| Exercise 5.7 | Second-order derivatives; Rolle's and Mean Value Theorem |
| Miscellaneous Exercise | Mixed continuity and differentiability problems |
All NCERT Solutions for Continuity and Differentiability Ex 5.6 with Step-by-Step Working
Every NCERT textbook question for Class 12 Mathematics Chapter 5 Continuity and Differentiability Ex 5.6 is listed below with its full Solution and Expert Solution hidden inside collapsible tabs. Click Check Solution to reveal the step-by-step working; click Expert Solution for the expanded explanation.
Questions
Find dydx if x = 2at2, y = at4.
Find dydx if x = a cosθ, y = b cosθ.
Find dydx if x = sin t, y = cos 2t.
Find dydx if x = 4t, y = 4t.
Find dydx if x = cosθ - cos 2θ, y = sinθ - sin 2θ.
Find dydx if x = a(θ - sinθ), y = a(1 + cosθ).
Find dydx if x = sin3 t√cos 2t, y = cos3 t√cos 2t.
Find dydx if x = a(cos t + t2), y = a sin t.
Find dydx if x = a secθ, y = b tanθ.
Find dydx if x = a(cosθ + θ sinθ), y = a(sinθ - θ cosθ).
If x = asin-1 t, y = acos-1 t, show that dydx = -yx.
Student Feedback - Continuity and Differentiability Exercise 5.6 (Collegedunia Survey, March 2026):
- 73% of Class 12 students surveyed rated this chapter as one of the higher-weightage units in board prep.
- The average student lost 1.2 marks per question by skipping an intermediate parametric step.
- Toppers reported that writing the formula recall sheet for this chapter added 1-2 marks on the long-answer question.
Continuity and Differentiability Class 12 NCERT Solutions - Frequently Asked Questions
Ques. How many questions are there in Exercise 5.6 of Class 12 Maths Chapter 5?
Ans. Exercise 5.6 of NCERT Class 12 Maths Chapter 5 has 11 questions, all on parametric differentiation, including one proof-style problem (Q11).
Ques. What is the parametric form of differentiation used in Exercise 5.6?
Ans. Parametric differentiation uses the rule dydx = dy/dtdx/dt , where both x and y are expressed in terms of a third variable t. This avoids the need to eliminate the parameter, which is often algebraically expensive.
Ques. Is Exercise 5.6 important for CBSE Class 12 Maths Boards?
Ans. Yes. Parametric differentiation has carried 3 to 4 marks in four of the last five CBSE Class 12 Mathematics board exams, either standalone or inside Application of Derivatives sums.
Ques. What is the answer to Question 6 of Exercise 5.6?
Ans. For x = a(θ - sinθ) and y = a(1 + cosθ) , the final answer is dydx = -cot(θ/2) , obtained after applying the half-angle identities 1+cosθ = 2cos2(θ/2) and 1-cosθ = 2sin2(θ/2) .
Ques. Does Exercise 5.6 appear in JEE Main?
Ans. Yes. Parametric differentiation is part of the JEE Main Calculus block and contributes roughly 2 to 3 percent of the section weightage. It also feeds into JEE Main Application of Derivatives sums.
Ques. How is Exercise 5.6 different from Exercise 5.5?
Ans. Exercise 5.5 covers logarithmic and implicit differentiation, where one equation links x and y directly. Exercise 5.6 uses a third variable (parameter), and the derivative is computed as a ratio of two separate t-derivatives.
Comments