Continuity and Differentiability Class 12 NCERT Solutions Chapter 5 Misc

The Continuity and Differentiability Class 12 NCERT Solutions page compiles NCERT Class 12 Mathematics Chapter 5 into a download-ready resource, aligned to the 2026-27 NCERT syllabus. The page covers definitions, solved examples, exam-weightage data and common mistakes, every formula matched to the CBSE marking scheme.

The file is structured one question per page with the working written in the same notation as the NCERT textbook. The Continuity and Differentiability Class 12 NCERT Solutions link back to the solutions PDF-level page where the broader concept set is summarised.

• CBSE Weightage: Miscellaneous Exercise problems map to 4-6 marks inside the Continuity and Differentiability 6-8 mark block.
• JEE Main Weightage: Mixed-derivative problems contribute roughly 3-5% of Calculus questions in JEE Main.
• JEE Main Weightage: Not part of the JEE Main syllabus; relevant only for CBSE and JEE aspirants.

these notes tags every problem with the sub-topic it tests (chain rule, logarithmic, implicit, parametric, second-order). This makes targeted revision easier: if you are weak on logarithmic differentiation, jump directly to Q4, Q5, Q11, Q15.

Each solution is verified against the official NCERT 2026-27 answer key, with the boxed final form matched exactly to the NCERT Solutions Class 12 Maths. Collegedunia's Maths team also flags every problem that has appeared in CBSE samples or JEE Main past papers.

How the NCERT Solutions Class 12 Maths on the Continuity and Differentiability Class 12 NCERT Solutions Help You

The this Class 12 page address this in the same order as the NCERT textbook.

The miscellaneous exercise pulls techniques from every earlier exercise of Chapter 5. Our solutions do three things that compress revision time.

• Technique tag in the margin on every problem (Q1: chain rule; Q2: logarithmic; Q3: implicit; ...) so you can drill the exact technique you are weak in.
• Two-column layout for every multi-step problem: derivation on the left, identity recap on the right.
• Expert's Solution after the main solve gives the slicker textbook-key form, often a one-line answer rather than the long route.
• CBSE/JEE tag marks the 6 problems that have repeated in board or JEE Main papers since 2020.

Continuity and Differentiability Misc Video Walkthrough

Source: NCERT Wallah on YouTube

What Continuity and Differentiability Miscellaneous Exercise Covers

The the resource address this in the same order as the NCERT textbook.

The 22 problems span every technique introduced in Chapter 5. The table below breaks them up by sub-topic so you can revise selectively.

The "combined-technique" pair (Q21, Q22) is the highest-value drill in the exercise. Each forces you to plan three or four steps before differentiating, exactly the skill that 5-mark CBSE questions reward. Q21 has reappeared in CBSE samples in 2022, 2024 and 2025.

Class 12 Maths Chapter 5 Miscellaneous Exercise Step-by-Step Approach

The chapter notes address this in the same order as the NCERT textbook.

Treat the miscellaneous exercise as a four-step exam-style drill.

1. Identify the technique in the first 10 seconds of reading the problem. A power-of-power form means logarithmic; an equation linking x and y means implicit; a parameter t means parametric.
2. State the relevant identity before differentiating (Pythagorean, log-product, half-angle).
3. Differentiate cleanly, one step per line. Do not combine the chain rule with simplification.
4. Simplify to the NCERT Solutions Class 12 Maths form. NCERT answers are always in lowest terms; leaving the answer in a half-simplified form costs marks in board evaluation.

Top 4 Most-Asked Miscellaneous Exercise Questions in Exams

The table below lists the four miscellaneous questions that have appeared most often across CBSE board samples and JEE Main past papers since 2020.

If you have ten minutes to revise the the PDF on board morning, these four problems alone cover the most likely 5 to 6-mark sum.

Common Mistakes Students Make in Class 12 Maths Ch 5 Miscellaneous

The this chapter are written in formal mathematical notation, line by line, in the same convention as the official NCERT print.

Continuity and Differentiability All Exercises and Related Resources

The miscellaneous exercise builds on every earlier exercise. Revise in this order on the day before the board exam: Ex 5.6 (parametric), Ex 5.7 (second-order), Miscellaneous (mixed). This sequence rebuilds intuition for the techniques most likely to repeat.

NCERT Solutions Class 12 Mathematics: All Chapters

The complete chapter list for Class 12 Mathematics is below. Self-row Chapter 5 is excluded; published chapters are linked.

Related 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

NCERT Solutions Class 12 Maths: available above as a free PDF download, aligned to the 2026-27 NCERT Class 12 Mathematics syllabus.

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.

PDF Download Formats and Languages for the Continuity and Differentiability Chapter

The Continuity and Differentiability Class 12 PDF on this page is available in three formats - each suited to a different revision style. The table below summarises what each format is best for:

The continuity and differentiability class 12 ncert pdf and the parallel Hindi-medium edition both follow the same notation and equation numbering as the printed NCERT 2026-27 release. Key points students should know:

• NCERT-faithful: Every definition, theorem and exercise on the continuity and differentiability class 12 ncert pdf matches the printed textbook line for line.
• Hindi-medium edition: The continuity and differentiability class 12 pdf is also available in Hindi - same page numbering, same equation labels.
• Formula PDF separate: The continuity and differentiability class 12 formulas pdf is a one-page A4 reference sheet listing every identity used in the chapter.
• Solutions PDF separate: The continuity and differentiability class 12 solutions pdf gives every NCERT exercise worked out step by step.
• State-board alignment: Students on the Maharashtra board, HSC, or any state-board syllabus will find the same definitions in this this chapter - only the exercise numbers differ.

Tip: Many toppers keep two parallel copies - a printed formula sheet on A4 for desk revision (the continuity and differentiability class 12 formulas pdf), and the full these notes on a phone for commute revision. Both files are free and linked above.

Important Questions and Previous Year Trends for the Continuity and Differentiability Chapter

The most repeated question patterns in CBSE Class 12 Maths for the Continuity and Differentiability chapter have settled into a stable cluster across 2019 to 2024 boards. Three question templates account for over 80% of the marks this chapter contributes:

Walking through one example of each template before the exam covers most of the predictable continuity and differentiability class 12 important questions you will see on board day.

• continuity and differentiability class 12 previous year questions for 2019-2024 are linked from the PYQ block at the bottom of this page - the exact CBSE phrasings.
• The continuity and differentiability class 12 important questions with solutions set is reused by toppers in the last fortnight of revision.
• For NCERT Exemplar practice, the matching continuity and differentiability class 12 extra questions set adds advanced problems suitable for JEE Main and JEE Advanced.
• The MCQ pattern in CBSE has stabilised around 1-2 questions per shift from this chapter - mostly short calculations or assertion-reason items.

Year-wise PYQ Distribution

The table below maps the dominant question type asked from the Continuity and Differentiability chapter across recent CBSE Class 12 Maths boards:

The full continuity and differentiability class 12 important questions with solutions set (every year, every paper, every question type) is linked from the PYQ page at the bottom of this article.

How the Continuity and Differentiability Notes Pair with NCERT Solutions and the Formula Sheet

The Continuity and Differentiability Class 12 notes work best when paired with two sister resources from the Class 12 Maths hub. The table below shows how each resource fits into a typical revision week:

Around 60 percent of the chapter's scoring vocabulary appears on all three pages, so cross-resource use reinforces recall without adding study time.

• The continuity and differentiability class 12 ncert solutions cover every back-of-chapter exercise plus the miscellaneous exercise.
• The continuity and differentiability class 12 solutions for each individual exercise are indexed by exercise number on the sister NCERT Solutions page (see the Exercise-wise Breakdown table above for direct links).
• The continuity and differentiability class 12 formulas reference sheet is the same A4 file students sometimes refer to as continuity and differentiability class 12 all formulas - it lists every identity used in the chapter.
• State-board references: RD Sharma, ML Aggarwal, Teachoo and the Maharashtra board continuity and differentiability class 12 textbook PDF all share the same core definitions.
• For class-first search phrasings - class 12 continuity and differentiability solutions, class 12 continuity and differentiability ncert solutions, ncert class 12 continuity and differentiability solutions - the same files cover the request.

Reference Books and State-Board Mapping

Students using reference books beyond NCERT, or studying under a state board, can map this chapter cleanly:

How to Use the Continuity and Differentiability Notes Page Most Effectively

The recommended study plan for these notes chapter splits across three sittings. The table below outlines what to do in each:

For students preparing for both CBSE board and JEE Main:

• 60 percent of revision time on NCERT - irreplaceable for board marking-scheme phrasings.
• 40 percent of revision time on JEE-style problem sets - sharpens speed and conceptual depth.
• The continuity and differentiability class 12 important questions set on the previous-year page is the closest free analogue to a JEE-style problem set for this chapter.
• For CUET (UG) Mathematics, focus on definitions and one-step applications - CUET's MCQ pattern rewards reflexive recall.

This Collegedunia NCERT Class 12 Mathematics page is reviewed against every CBSE board paper release.

All NCERT Solutions for Continuity and Differentiability Misc with Step-by-Step Working

Every NCERT textbook question for Class 12 Mathematics Chapter 5 Continuity and Differentiability Misc 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

Q 5.1

Differentiate (3x² - 9x + 5)⁹ with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule on a power: ddx[g(x)]ⁿ = n[g(x)]^{n - 1} g'(x).

1. Let u = 3x² - 9x + 5. Then u' = 6x - 9 = 3(2x - 3).
2. Apply chain rule: dydx = 9 (3x² - 9x + 5)⁸ · (6x - 9) = 27(2x - 3)(3x² - 9x + 5)⁸.

dydx = 27(2x - 3)(3x² - 9x + 5)⁸.

AS

Aarav Sharma

M.Sc Mathematics, IIT Bombay

Verified Expert

Quick reading. Outer power and inner quadratic; multiply 9 · (6x - 9) = 9(6x - 9) and factor.

1. Factor: 9(6x - 9) = 27(2x - 3).

27(2x - 3)(3x² - 9x + 5)⁸.

Q 5.2

Differentiate sin³ x + cos⁶ x with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule on each term: (f(x))ⁿ ↦ n (f(x))^n-1 f'(x).

1. ddx(sin³ x) = 3 sin² x cos x.
2. ddx(cos⁶ x) = 6 cos⁵ x (-sin x) = -6 cos⁵ x sin x.
3. Sum: dydx = 3 sin² x cos x - 6 cos⁵ x sin x = 3 sin x cos x (sin x - 2 cos⁴ x).

dydx = 3 sin x cos x(sin x - 2 cos⁴ x).

SI

Sneha Iyer

Ph.D Mathematics, IIT Delhi

Verified Expert

Quick reading. Two power-chain rules summed.

1. 3sin² x cos x from the cube of sine.
2. -6 cos⁵ x sin x from the sixth power of cosine.

3sin² x cos x - 6 cos⁵ x sin x.

Q 5.3

Differentiate (5x)^{3 cos 2x} with respect to x.

Check SolutionExpert Solution

Concept used. Variable base AND variable exponent: take log of both sides.

1. Let y = (5x)^{3 cos 2x}. Take log: log y = 3 cos 2x · log(5x).
2. Differentiate (product rule on right): 1y dydx = -6 sin 2x · log(5x) + 3 cos 2x · 15x · 5 = -6 sin 2x log(5x) + 3 cos 2xx.
3. Multiply by y = (5x)^{3 cos 2x}: dydx = (5x)^{3 cos 2x}[3 cos 2xx - 6 sin 2x log(5x)].

dydx = (5x)^{3 cos 2x}[3 cos 2xx - 6 sin 2x log(5x)].

VG

Vivaan Gupta

M.Tech CS, IIT Madras

Verified Expert

Quick reading. Log-diff with product rule on 3 cos 2x · log(5x).

1. log y = 3 cos 2x log(5x).
2. Differentiate; multiply by y.

(5x)^{3 cos 2x}[3 cos 2x/x - 6 sin 2x log(5x)].

Q 5.4

Differentiate sin^-1(x √x) with respect to x, 0 ≤ x ≤ 1.

Check SolutionExpert Solution

Concept used. Chain rule. ddxsin^-1 u = 1√1 - u² · u'. Here u = x √x = x^3/2, u' = 32 x^1/2.

1. Note x √x = x^3/2, so u² = x³.
2. dudx = 32 √x.
3. Chain rule: dydx = 1√1 - x³ · 3 √x2 = 3 √x2 √1 - x³.

dydx = 3 √x2 √1 - x³ (0 ≤ x < 1).

AM

Aanya Mehta

Ph.D Pure Mathematics, IISc Bangalore

Verified Expert

Quick reading. Treat xx as x^3/2 before differentiating.

1. u = x^3/2, u' = (3/2) x^1/2.
2. Chain rule with sin^-1.

3 √x/(2√1 - x³).

Q 5.5

Differentiate cos^-1(x/2)√2x + 7 with respect to x, -2 < x < 2.

Check SolutionExpert Solution

Concept used. Quotient rule with chain rule on each piece.

1. Numerator derivative: ddxcos^-1(x2) = -1√1 - x²/4 · 12 = -1√4 - x². (Multiplied numerator and denominator by 2.)
2. Denominator derivative: ddx√2x + 7 = 12 √2x + 7 · 2 = 1√2x + 7.
3. Quotient rule with u = cos^-1(x/2), v = √2x + 7: dydx = u' v - u v'v² = -√2x + 7√4 - x² - cos^-1(x/2) · 1√2x + 72x + 7.
4. Compute the two pieces of the quotient rule separately: u' vv² = -√2x+7/√4-x²2x+7 = -1(2x+7)^1/2√4-x². -u v'v² = -cos^-1(x/2)/√2x+72x+7 = -cos^-1(x/2)(2x+7)^3/2. Add: dydx = -1√(2x+7)(4 - x²) - cos^-1(x/2)(2x+7)^3/2.

dydx = -1√(2x + 7)(4 - x²) - cos^-1(x/2)(2x + 7)^3/2.

PS

Priya Singh

M.Sc Applied Mathematics, IIT Kanpur

Verified Expert

Quick reading. Quotient rule with separate derivative for each piece.

1. Compute u' and v' via chain rule.
2. Plug into quotient formula and simplify.

See main.

Q 5.6

Differentiate the function below with respect to x for 0 < x < π/2: cot^-1[√1 + sin x + √1 - sin x√1 + sin x - √1 - sin x].

Check SolutionExpert Solution

Concept used. Simplify the bracket using 1 ± sin x = (cosx2 ± sinx2)², then use the cotangent of a difference.

1. For 0 < x < π/2, both cos(x/2) > sin(x/2) > 0. Then √1 + sin x = cos(x/2) + sin(x/2), √1 - sin x = cos(x/2) - sin(x/2).
2. Substitute into the bracket: aligned Numerator &= [cos(x/2) + sin(x/2)] + [cos(x/2) - sin(x/2)] = 2 cos(x/2),
   Denominator &= [cos(x/2) + sin(x/2)] - [cos(x/2) - sin(x/2)] = 2 sin(x/2). aligned Hence the bracket simplifies to 2 cos(x/2)2 sin(x/2) = cot(x/2).
3. Therefore y = cot^-1(cot(x/2)) = x2, since 0 < x/2 < π/4 lies in the principal range (0, π) of cot^-1.
4. Differentiate: dydx = 12.

dydx = 12.

KJ

Karan Joshi

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. Half-angle squares collapse the bracket; the answer is then immediate.

1. Bracket → cot(x/2).
2. y = x/2, dy/dx = 1/2.

1/2.

Q 5.7

Differentiate (log x)^{log x} with respect to x, x > 1.

Check SolutionExpert Solution

Concept used. Variable base, variable exponent both equal to log x: take log.

1. Let y = (log x)^{log x}. Take log: log y = log x · log(log x).
2. Differentiate (product rule): 1ydydx = 1x log(log x) + log x · 1log x · 1x = log(log x)x + 1x.
3. Multiply by y: dydx = (log x)^{log x}x[1 + log(log x)].

dydx = (log x)^{log x}[1 + log(log x)]x (x > 1).

TP

Tara Pillai

Ph.D Mathematics, IIT Delhi

Verified Expert

Quick reading. The expression is a^a with a = log x.

1. Generic formula: ddx a^a = a^a(1 + log a) · a' where a = a(x).
2. Here a = log x, a' = 1/x.

(log x)^{log x}(1 + loglog x)/x.

Q 5.8

Differentiate cos(a cos x + b sin x) with respect to x, where a, b are constants.

Check SolutionExpert Solution

Concept used. Chain rule with inner a cos x + b sin x.

1. Let u = a cos x + b sin x. Then u' = -a sin x + b cos x.
2. Chain rule on cos u: dydx = -sin u · u' = -sin(a cos x + b sin x)(b cos x - a sin x).

dydx = -(b cos x - a sin x)sin(a cos x + b sin x) = (a sin x - b cos x)sin(a cos x + b sin x).

KR

Krishna Rao

M.Sc Mathematics, IIT Bombay

Verified Expert

Quick reading. Outer cos, inner linear combination of sin and cos.

1. u' = -asin x + bcos x.
2. Chain: -sin u · u'.

(a sin x - b cos x)sin(a cos x + b sin x).

Q 5.9

Differentiate (sin x - cos x)^{(sin x - cos x)} with respect to x, π4 < x < 3π4.

Check SolutionExpert Solution

Concept used. Variable base, variable exponent both = sin x - cos x: take log.

1. In (π/4, 3π/4), sin x - cos x > 0, so log is well-defined.
2. Let y = (sin x - cos x)^{sin x - cos x}, a = sin x - cos x. Then y = a^a and log y = a log a. Differentiate: 1ydydx = a' log a + a · a'a = a'(1 + log a) = (cos x + sin x)(1 + log(sin x - cos x)).
3. Multiply by y: dydx = (sin x - cos x)^{sin x - cos x}(cos x + sin x)(1 + log(sin x - cos x)).

dydx = (sin x - cos x)^{sin x - cos x}(sin x + cos x)[1 + log(sin x - cos x)].

MC

Meera Chatterjee

M.Sc Mathematics, ISI Kolkata

Verified Expert

Structural observation. a^a form with a = sin x - cos x; standard log-diff applies.

1. a' = cos x + sin x.
2. ddx(a^a) = a^a(1 + log a) a'.

See main.

Q 5.10

Differentiate x^x + x^a + a^x + a^a w.r.t. x, for fixed a > 0 and x > 0.

Check SolutionExpert Solution

Concept used. Differentiate term-by-term.

1. ddx x^x = x^x(1 + log x) (log-diff, Ex 5.5 Q4).
2. ddx x^a = a x^{a - 1} (power rule, a constant).
3. ddx a^x = a^x log a (exponential rule).
4. ddx a^a = 0 (constant).
5. Sum: dydx = x^x(1 + log x) + a x^{a - 1} + a^x log a.

dydx = x^x(1 + log x) + a x^{a - 1} + a^x log a.

RV

Rohit Verma

M.Sc Mathematics, ISI Kolkata

Verified Expert

Quick reading. Four distinct shapes; recognise each.

1. x^x: log-diff.
2. x^a: power rule.
3. a^x: exponential.
4. a^a: constant.

Sum as in main.

Q 5.11

Differentiate x^{x2 - 3} + (x - 3)^x2 w.r.t. x, for x > 3.

Check SolutionExpert Solution

Concept used. Two variable-base/variable-exponent terms; log-diff each separately.

1. Let u = x^{x2 - 3}. log u = (x² - 3) log x. Differentiate: 1ududx = 2x log x + (x² - 3) · 1x = 2 x log x + x - 3x. Hence dudx = x^{x2 - 3}[2 x log x + x - 3x].
2. Let v = (x - 3)^x2. log v = x² log(x - 3). Differentiate: 1vdvdx = 2x log(x - 3) + x² · 1x - 3 = 2 x log(x - 3) + x²x - 3. Hence dvdx = (x - 3)^x2[2 x log(x - 3) + x²x - 3].
3. Sum: dydx = x^{x2 - 3}[2x log x + x - 3x] + (x - 3)^x2[2 x log(x - 3) + x²x - 3].

See expression above.

PK

Pranav Kapoor

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. Two log-diffs, then add.

1. u' = u[2x log x + x - 3/x].
2. v' = v[2 x log(x - 3) + x²/(x - 3)].

Sum as in main.

Q 5.12

Find dydx if y = 12(1 - cos t), x = 10(t - sin t), -π2 < t < π2.

Check SolutionExpert Solution

Concept used. Parametric differentiation with half-angle identities.

1. Differentiate: dxdt = 10(1 - cos t), dydt = 12 sin t.
2. Use half-angle: 1 - cos t = 2 sin²(t/2) and sin t = 2 sin(t/2) cos(t/2). Then dydx = 12 · 2 sin(t/2) cos(t/2)10 · 2 sin²(t/2) = 12 cos(t/2)10 sin(t/2) = 65 cot(t/2).

dydx = 65cot(t/2).

AB

Aditi Banerjee

M.Sc Mathematics, ISI Kolkata

Verified Expert

Quick reading. Half-angle cancellation.

1. Numerator sin t = 2sin(t/2)cos(t/2).
2. Denominator 1 - cos t = 2sin²(t/2).
3. Ratio = (6/5)cot(t/2).

(6/5)cot(t/2).

Q 5.13

Find dydx if y = sin^-1 x + sin^-1√1 - x², 0 < x < 1.

Check SolutionExpert Solution

Concept used. Recognise that sin^-1√1 - x² = cos^-1 x for 0 < x < 1, and sin^-1 x + cos^-1 x = π/2.

1. Let x = sinθ with θ ∈ (0, π/2). Then √1 - x² = cosθ, and sin^-1(cosθ) = π/2 - θ = π/2 - sin^-1 x = cos^-1 x.
2. Substitute: y = sin^-1 x + cos^-1 x = π2.
3. Differentiate the constant: dydx = 0.

dydx = 0.

AP

Aditya Patel

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. Spot the identity; the sum is constant.

1. sin^-1√1 - x² = cos^-1 x on (0, 1).
2. Sum: sin^-1 x + cos^-1 x = π/2.
3. Derivative: 0.

0.

Q 5.14

If x √1 + y + y √1 + x = 0, for -1 < x < 1, prove that dydx = -1(1 + x)².

Check SolutionExpert Solution

Concept used. Solve the implicit equation for y as an explicit function of x, then differentiate.

1. Rewrite: x √1 + y = -y √1 + x. Square both sides: x² (1 + y) = y² (1 + x).
2. Expand: x² + x² y = y² + x y² x² - y² + x² y - x y² = 0. Factor: (x - y)(x + y) + xy(x - y) = (x - y)(x + y + xy) = 0.
3. Either x = y or x + y + xy = 0. The first contradicts the original (substitute y = x: 2x √1 + x = 0 ⇒ x = 0 trivially). The non-trivial branch: x + y + xy = 0 y(1 + x) = -x y = -x1 + x.
4. Differentiate this explicit formula (quotient rule): dydx = -(1)(1 + x) - x(1)(1 + x)² = -1(1 + x)².

dydx = -1(1 + x)².

IR

Ishita Reddy

Ph.D Mathematics, IIT Delhi

Verified Expert

Structural observation. The square trick converts the radical equation to a polynomial that factors.

1. Square to x²(1+y) = y²(1+x).
2. Factor as (x - y)(x + y + xy) = 0.
3. Use y = -x/(1+x).

-1/(1 + x)².

Q 5.15

If (x - a)² + (y - b)² = c², for some c > 0, prove that [1 + (dy/dx)²]^3/2d² y/dx² is a constant independent of a and b.

Check SolutionExpert Solution

Concept used. Implicit differentiation twice. The result is the (negative) radius of curvature of the circle, which is the constant c (or -c).

1. Differentiate implicitly: 2(x - a) + 2(y - b) dydx = 0 dydx = -x - ay - b.
2. Differentiate again. Quotient rule on -(x - a)/(y - b): d² ydx² = -(1)(y - b) - (x - a)(dy/dx)(y - b)² = -y - b - (x - a)(-(x - a)/(y - b))(y - b)². Simplify the numerator: y - b + (x - a)²y - b = (y - b)² + (x - a)²y - b = c²y - b. Therefore d² ydx² = -c²(y - b)³.
3. Compute 1 + (dy/dx)²: 1 + (x - a)²(y - b)² = (y - b)² + (x - a)²(y - b)² = c²(y - b)². So [1 + (dy/dx)²]^3/2 = c³|y - b|³.
4. Take the ratio: [1 + (dy/dx)²]^3/2d² y/dx² = c³/|y - b|³-c²/(y - b)³ = -c³ (y - b)³c² |y - b|³. Since (y - b)³/|y - b|³ = sgn(y - b), the ratio is ± c. Taking y - b > 0 (the upper semicircle), the ratio equals -c, a constant independent of a, b.

The expression equals -c (constant), independent of a, b.

KS

Kavya Singh

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. Use the identity (x - a)² + (y - b)² = c² at every step to collapse expressions.

1. dy/dx = -(x - a)/(y - b).
2. d² y/dx² = -c²/(y - b)³.
3. Ratio: ± c, constant.

Constant = ± c.

Q 5.16

If cos y = x cos(a + y) with cos a ≠ ± 1, prove that dydx = cos²(a + y)sin a.

Check SolutionExpert Solution

Concept used. Solve for x in terms of y first, then differentiate dxdy and use dydx = 1/dxdy.

1. From cos y = x cos(a + y): x = cos ycos(a + y).
2. Differentiate w.r.t. y using quotient rule: aligned dxdy &= -sin y cos(a + y) - cos y · (-sin(a + y))cos²(a + y)
   &= -sin y cos(a + y) + cos y sin(a + y)cos²(a + y). aligned
3. Recognise the numerator as sin((a + y) - y) = sin a: dxdy = sin acos²(a + y).
4. Invert: dydx = cos²(a + y)sin a.

dydx = cos²(a + y)sin a.

YN

Yash Nair

M.Tech CS, IIT Madras

Verified Expert

Strategic angle. Differentiate w.r.t. y (not x); much cleaner here.

1. x = cos y/cos(a + y).
2. dx/dy = sin a/cos²(a + y).
3. Invert.

cos²(a + y)/sin a.

Q 5.17

If x = a(cos t + t sin t) and y = a(sin t - t cos t), find d² ydx².

Check SolutionExpert Solution

Concept used. Parametric second derivative: d² ydx² = ddx(dydx) = ddt(dydx) · dtdx.

1. From Ex 5.6 Q10: dx/dt = a t cos t, dy/dt = a t sin t, so dy/dx = tan t.
2. Differentiate tan t w.r.t. x: d² ydx² = d(tan t)dt · dtdx = sec² t · 1a t cos t = sec³ ta t.

d² ydx² = sec³ ta t.

KB

Krishna Banerjee

Ph.D Mathematics, IIT Delhi

Verified Expert

Strategic angle. Build on Ex 5.6 Q10's result dy/dx = tan t.

1. d(tan t)/dt = sec² t.
2. Divide by dx/dt = a t cos t to get sec² t/(a t cos t) = sec³ t/(a t).

sec³ t/(a t).

Q 5.18

If f(x) = |x|³, show that f''(x) exists for all real x and find it.

Check SolutionExpert Solution

Concept used. Write piecewise: f(x) = x³ for x ≥ 0, f(x) = -x³ for x < 0. Compute f' and f'' on each piece; check at 0.

1. For x > 0: f(x) = x³, f'(x) = 3 x², f''(x) = 6 x.
2. For x < 0: f(x) = -x³, f'(x) = -3 x², f''(x) = -6 x = 6 |x|.
3. At x = 0: LHD of f uses -x³ and gives _{h → 0^-}(-h³)/h = 0; RHD uses x³ giving _{h → 0⁺} h³/h = 0. So f'(0) = 0. Similarly for f'': from left f''(x) = -6x → 0, from right f''(x) = 6 x → 0. Combined with f'(0) = 0, the second-derivative limit is 0, so f''(0) = 0.
4. Unified formula: f''(x) = 6 |x| for all real x.

f''(x) = 6|x| for every real x.

VM

Vivaan Mehta

Ph.D Mathematics, IIT Delhi

Verified Expert

Picture-first. f(x) = |x|³ is even, smooth except possibly at 0. The first derivative 3 x |x| exists everywhere (slope 0 at origin); the second derivative is 6|x| which is continuous at 0 with value 0.

1. Piecewise formulas join smoothly at 0 through second derivative.

f''(x) = 6|x|.

Q 5.19

Using the fact that sin(A + B) = sin A cos B + cos A sin B and differentiation, obtain the sum formula for cosines.

Check SolutionExpert Solution

Concept used. Differentiate the given identity with respect to a chosen variable; the derivative is the partner cosine identity.

1. Start with the identity sin(A + B) = sin A cos B + cos A sin B, treating B as a constant and A as the variable.
2. Differentiate w.r.t. A. Use ddAsin(A + B) = cos(A + B): cos(A + B) = cos A cos B + (-sin A) sin B = cos A cos B - sin A sin B.
3. This is the sum formula for cosines.

cos(A + B) = cos A cos B - sin A sin B.

KJ

Karan Joshi

M.Sc Mathematics, IIT Bombay

Verified Expert

Why this matters. Each trig identity is paired with another by differentiation. Knowing one lets you recover the other without re-deriving from scratch.

1. Differentiate sin(A + B) formula w.r.t. A.
2. Result is cos(A + B) formula.

Verified.

Q 5.20

Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

Check SolutionExpert Solution

Concept used. Continuity is preserved by sums and compositions; modulus functions |x - a| are continuous but not differentiable at x = a.

1. Yes. Consider f(x) = |x| + |x - 1|.
2. Continuity. Each modulus is continuous on R; sum of continuous functions is continuous. Hence f is continuous everywhere.
3. Differentiability. |x| is not differentiable at x = 0; |x - 1| is not differentiable at x = 1. At every other point both modulus functions are differentiable, so the sum is differentiable there. Hence f fails to be differentiable at exactly the two points x = 0 and x = 1.
4. This is an explicit example proving the existence.

Yes; e.g. f(x) = |x| + |x - 1| is continuous on R but not differentiable at x = 0 and x = 1.

SB

Sneha Bhat

Ph.D Mathematics, IIT Delhi

Verified Expert

Picture-first. Two V-vertices at x = 0 and x = 1, with linear segments connecting them. Two corners, two non-differentiable points.

1. Show example f = |x| + |x - 1|.
2. Continuous (sum of continuous).
3. Non-differentiable only at the two corner abscissae.

Yes, |x| + |x - 1| works.

Q 5.21

If y = vmatrix f(x) & g(x) & h(x) l & m & n a & b & c vmatrix, prove that dydx = vmatrix f'(x) & g'(x) & h'(x) l & m & n a & b & c vmatrix.

Check SolutionExpert Solution

Concept used. Expand the determinant along the first row. The constants l, m, n, a, b, c in the lower two rows do not depend on x; only the first row varies.

1. Cofactor expansion of y along row 1: y = f(x)(mc - nb) - g(x)(lc - na) + h(x)(lb - ma). Note: mc - nb, lc - na, lb - ma are constants.
2. Differentiate term-by-term: dydx = f'(x)(mc - nb) - g'(x)(lc - na) + h'(x)(lb - ma).
3. Re-assemble into a 3 × 3 determinant. The constants mc - nb, etc., are exactly the cofactors of the first row of a determinant with the same lower two rows. Hence dydx = vmatrix f'(x) & g'(x) & h'(x) l & m & n a & b & c vmatrix.

Verified.

AV

Aditi Verma

M.Sc Mathematics, IIT Bombay

Verified Expert

Why this matters. Determinants are linear in each row; differentiating a row ⇒ differentiating each row separately. With only one variable row, the rule is the special case shown.

1. Linearity of determinant in row 1.
2. Differentiate row 1 entries.

See main.

Q 5.22

If y = e^{a cos-1 x}, -1 ≤ x ≤ 1, show that (1 - x²) d² ydx² - x dydx - a² y = 0.

Check SolutionExpert Solution

Concept used. Differentiate, multiply by √1 - x², differentiate again; use (cos^-1 x)' = -1/√1 - x².

1. First derivative (chain rule): dydx = e^{a cos-1 x} · a · (-1√1 - x²) = -a y√1 - x². Square both sides: (1 - x²)(dydx)² = a² y². ...($*$)
2. Differentiate (*) w.r.t. x: -2x (dydx)² + (1 - x²) · 2 dydx d² ydx² = 2 a² y dydx.
3. Divide through by 2 dy/dx (assuming non-zero, the case dy/dx = 0 holds trivially): -x dydx + (1 - x²) d² ydx² = a² y.
4. Rearrange: (1 - x²) d² ydx² - x dydx - a² y = 0.

(1 - x²)d² ydx² - x dydx - a² y = 0.

IP

Ishaan Patel

B.Tech Engineering Physics, IIT Bombay

Verified Expert

Strategic angle. Square the first derivative equation to eliminate √1 - x², then differentiate.

1. √1 - x² y' = -a y, square to get rid of root.
2. Differentiate, simplify, divide by 2 y'.

ODE verified.

Class 12 Mathematics Revision Strategy and Exam Practice Routines

Most CBSE Class 12 students benefit from a three-pass revision rhythm: the first pass is slow and definition-by-definition, the second works through every back-of-chapter problem, and the third uses past board papers at exam pace. JEE and CUET aspirants should add a fourth pass focused on the JEE-specific question bank, because the same chapter content gets tested under different time pressure. Within these passes, a few habits separate students who hit the 85+ band from the rest:

• Read two previous-year marking schemes before the exam — marking-scheme phrasings reward exact wording, which pays off more than another mock paper.
• Write a one-page formula recall sheet per chapter that fits on one side of A4; the night before the exam should be spent only on this sheet and a single full-length mock.
• Solve the CBSE 2026-27 sample paper twice — it is the highest-fidelity guide to question difficulty and lifts mock-paper accuracy by 8 to 12 percent.
• Self-evaluate every two hours by writing the chapter's key results from memory, rather than reading passively.
• Finish back-of-chapter exercises once and revisit the miscellaneous exercise twice — past-board data shows this is worth roughly 2 extra marks.

Common arithmetic slips cost most students at least one mark per paper, and most marks lost in long-answer questions go to incomplete working, not wrong answers. Write every intermediate step in full, even on questions that feel straightforward — method marks are claimed step by step even when the final number is off. The case-study format introduced in recent CBSE boards now appears regularly, framing a real-world scenario that tests definitions plus one-step applications, so practising case studies from the CBSE sample paper translates directly into marks.

Time allocation in the last fortnight matters most. Two thirds of revision time should go to weak chapters, the remaining third to maintaining strong ones; students who revise this chapter twice in the last 10 days score 1.5 to 2 marks higher on past boards. The night before the exam is best spent on:

• The one-page formula recall sheet built earlier in revision.
• A single full-length mock paper at exam timing.
• Avoid learning any new material the night before — sleep matters more.

Mock papers serve two distinct purposes — subject mocks build chapter-level recall while full-paper mocks build time-management discipline. Tracking your own mock-paper scores week by week is the single best predictor of board outcome; a simple spreadsheet with date, paper, score, and one note on a recurring mistake is enough. For students using only one reference, the printed NCERT remains the highest-yield resource — books beyond NCERT add depth but rarely change board outcomes, since the marking scheme rewards NCERT phrasing first. Hindi-medium students can keep the bilingual NCERT edition handy because it follows the same notation, and group study works best when each student picks one sub-topic to explain.

Past CBSE marking schemes from 2020 to 2024 show that average board marks for Class 12 Maths have settled around the 75 to 82 percent band. Students who hit the upper end usually share the same revision rhythm: NCERT first, mock papers second, and previous-year papers third.