Continuity and Differentiability Class 12 NCERT Solutions Chapter 5 Ex 5.2

The Continuity and Differentiability Class 12 NCERT Solutions given here cover Exercise 5.2 of Class 12 Mathematics Chapter 5 Continuity and Differentiability in full. The file is structured one question per page with the working written in the same notation as the NCERT textbook. The chapter notes link back to the solutions PDF-level page where the broader concept set is summarised.

• CBSE Weightage: 8-10 marks (Continuity and Differentiability + Applications of Derivatives, Calculus unit total 35 marks)
• JEE Main Weightage: 6-8% of the Mathematics section, with 2-3 questions on chain rule and composite differentiation
• Exercise 5.2 problem count: 10 questions covering chain rule on composite trig functions, inverse trig differentiation, and proving differentiability

These Collegedunia NCERT Solutions for Exercise 5.2 are written by experienced Class 12 Mathematics teachers, follow the latest 2026-27 NCERT print, and lay out each chain-rule unwrap, intermediate u = g(x) substitution, and final derivative on a separate line so you can copy the layout verbatim into a CBSE board script.

Every question shows the formula recall, the substitution, and the boxed final derivative.

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.2

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

Exercise 5.2 trains you to differentiate composite functions of the form y = f(g(x)) using the chain rule dydx = f'(g(x)) · g'(x) .

Most questions chain two or three layers (a polynomial inside a trig function inside a square root), and the Collegedunia solution for each one breaks the layering into clearly named substitution variables so you never lose track of which dudx you are computing.

The two-layer composites in the middle of Exercise 5.2 are the highest-yield CBSE practice in these notes, and the Collegedunia Expert Solution writes out the u and v substitution table so a student can audit the calculation in under 60 seconds.

Continuity and Differentiability Ex 5 2 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 this Class 12 page address this in the same order as the NCERT textbook.

Continuity and Differentiability Exercise 5.2 is graded as moderate-to-hard in the 2026-27 NCERT print, primarily because students rush the second chain-rule pass and drop a factor. The Collegedunia solution PDF for this exercise is built around the following promises:

• Every chain-rule layer printed on its own line with the substituted variable named, so the differentiation sequence reads top-to-bottom without ambiguity.
• Formula recall printed at the top of any question that needs a non-trivial derivative identity ( ddxsin^-1x = 1√1-x² , ddxtan^-1x = 11+x² ).
• Expert Solution block after every main solution that re-derives the answer using the first-principle limit definition, giving you the option to verify by an independent method.
• Tip callouts at common mistake points: forgetting the inner derivative, sign error on cos^-1 derivatives, dropping the modulus in ddx|x| .

The this resource Class 12 Mathematics Exercise 5.2 solutions are aligned line-by-line with the official 2026-27 NCERT textbook reprint, so the question numbering matches your printed copy exactly.

Important Concepts Covered in Exercise 5.2

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

Exercise 5.2 stays inside the four concept boxes below. Reviewing them before attempting the questions cuts solving time by roughly half.

The multi-layer composite cluster (Q3, 5, 6, 9) is where the this resource Expert Solution earns its place: each substitution is written in a 3-column table so the reader can trace exactly which derivative belongs to which layer without re-reading the calculation.

Exam Relevance of Continuity and Differentiability Exercise 5.2

Continuity and Differentiability is part of the Calculus unit, which itself carries 35 marks in the CBSE Class 12 Mathematics paper. Exercise 5.2 contributes the bulk of the 2-mark and 4-mark chain-rule differentiation prompts. A short snapshot of recent appearances:

The chain rule is the workhorse identity for every later exercise in this chapter, so failure to learn Ex 5.2 directly costs marks in Ex 5.3 (implicit), Ex 5.4 (exponential / log), and Ex 5.5 (logarithmic differentiation) as well.

Common Mistakes Students Make in Exercise 5.2

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

Three predictable error patterns account for roughly 80% of the marks lost on this exercise in CBSE board scripts. The this resource solutions highlight each one inline.

• Forgetting the inner derivative after differentiating the outer function. Students often write ddxsin(2x) = cos(2x) and lose the factor of 2.
• Sign error on cos^-1 and cot^-1 derivatives. Both carry a negative sign that is easy to drop when the outer function is also being unwrapped.
• Skipping the 1 - x² or 1 + x² denominator for inverse trig results, especially when a chain rule sits on top.

Related Resources for Class 12 Maths Chapter 5

NCERT Solutions for Class 12 Maths: All Chapters

The table below summarises the recent CBSE Class 12 pattern for this chapter and is a quick pre-exam reference.

the PDF: 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.

All NCERT Solutions for Continuity and Differentiability Ex 5.2 with Step-by-Step Working

Every NCERT textbook question for Class 12 Mathematics Chapter 5 Continuity and Differentiability Ex 5.2 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 sin(x² + 5) with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule. If y = f(g(x)), then dydx = f'(g(x)) · g'(x). Here f(u) = sin u with f'(u) = cos u, and g(x) = x² + 5 with g'(x) = 2x.

1. Set u = x² + 5 so that y = sin u.
2. Differentiate the outer: dydu = cos u = cos(x² + 5).
3. Differentiate the inner: dudx = ddx(x² + 5) = 2x + 0 = 2x.
4. Multiply by the chain rule: dydx = dydu · dudx = cos(x² + 5) · 2x = 2x cos(x² + 5).

dydx = 2x cos(x² + 5).

AS

Aarav Sharma

M.Sc Mathematics, IIT Bombay

Verified Expert

Quick reading. Outer is sine; inner is the polynomial x² + 5 with derivative 2x.

1. ddxsin() = cos() · ddx.
2. Apply with = x² + 5, derivative 2x.

2x cos(x² + 5).

Q 5.2

Differentiate cos(sin x) with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule applied to a cosine-of-sine composite. ddxcos u = -sin u · dudx; here u = sin x, so dudx = cos x.

1. Let u = sin x, y = cos u.
2. dydu = -sin u = -sin(sin x).
3. dudx = cos x.
4. Chain rule: dydx = -sin(sin x) · cos x.

dydx = -cos x · sin(sin x).

SI

Sneha Iyer

Ph.D Mathematics, IIT Delhi

Verified Expert

Structural observation. Two trig layers; differentiate each in order.

1. Outer derivative: -sin() at = sin x.
2. Inner derivative: cos x.
3. Product: -sin(sin x) cos x.

-cos x sin(sin x).

Q 5.3

Differentiate sin(ax + b) with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule with outer sin and inner linear function ax + b.

1. Let u = ax + b so that y = sin u.
2. dydu = cos u = cos(ax + b).
3. dudx = a.
4. Chain rule: dydx = cos(ax + b) · a = a cos(ax + b).

dydx = a cos(ax + b).

VG

Vivaan Gupta

M.Tech CS, IIT Madras

Verified Expert

Quick reading. The chain rule scales the cosine by the constant a.

1. Derivative of sin(linear) = a cos(same linear).

a cos(ax + b).

Q 5.4

Differentiate sec(tan(√x)) with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule applied to a composite of three functions: sec, tan, √·. The derivative of sec u is sec u tan u; the derivative of tan v is sec² v; the derivative of √x is 12√x.

1. Let w = √x, v = tan w, y = sec v. Then y = sec(tan(√x)).
2. dydv = sec v tan v = sec(tan√x) tan(tan√x).
3. dvdw = sec² w = sec²(√x).
4. dwdx = 12√x.
5. Chain rule (three layers): dydx = dydv · dvdw · dwdx = sec(tan√x) · tan(tan√x) · sec²(√x) · 12√x.

dydx = sec(tan√x) · tan(tan√x) · sec² √x2√x.

AM

Aanya Mehta

Ph.D Pure Mathematics, IISc Bangalore

Verified Expert

Strategic angle. Three-layer chain: sec → tan → √ .

1. Layer 1: sec()' = sec tan .
2. Layer 2: tan()' = sec² .
3. Layer 3: √x' = 1/(2√x).
4. Multiply all three with = tan√x, √x, x respectively.

sec(tan√x) tan(tan√x) sec²√x2√x.

Q 5.5

Differentiate sin(ax + b)cos(cx + d) with respect to x.

Check SolutionExpert Solution

Concept used. Quotient rule. If y = u/v then dydx = u'v - uv'v². Combined with the chain rule for u = sin(ax + b) and v = cos(cx + d).

1. Let u = sin(ax + b) and v = cos(cx + d).
2. Derivatives of numerator and denominator (chain rule): u' = a cos(ax + b), v' = -c sin(cx + d).
3. Quotient rule: dydx = u'v - uv'v² = a cos(ax + b) cos(cx + d) - sin(ax + b)(-c sin(cx + d))cos²(cx + d).
4. Simplify the numerator: = a cos(ax + b) cos(cx + d) + c sin(ax + b) sin(cx + d)cos²(cx + d).

dydx = a cos(ax + b) cos(cx + d) + c sin(ax + b) sin(cx + d)cos²(cx + d).

PS

Priya Singh

M.Sc Applied Mathematics, IIT Kanpur

Verified Expert

Quick reading. Quotient rule with chain-rule numerator/denominator derivatives.

1. u = sin(ax+b), u' = acos(ax+b).
2. v = cos(cx+d), v' = -csin(cx+d).
3. (u/v)' = (u'v - uv')/v² as written.

acos(ax+b)cos(cx+d) + csin(ax+b)sin(cx+d)cos²(cx+d).

Q 5.6

Differentiate cos x³ · sin²(x⁵) with respect to x.

Check SolutionExpert Solution

Concept used. Product rule: (uv)' = u'v + uv'. Each factor needs the chain rule.

1. Let u = cos(x³) and v = sin²(x⁵) = (sin(x⁵))².
2. Derivative of u: chain rule, outer cos, inner x³. u' = -sin(x³) · ddx(x³) = -3x² sin(x³).
3. Derivative of v: write v = (sin(x⁵))². Chain rule three layers (square, sine, x⁵): v' = 2 sin(x⁵) · cos(x⁵) · 5 x⁴ = 10 x⁴ sin(x⁵) cos(x⁵).
4. Product rule: dydx = u'v + uv' = -3x² sin(x³) sin²(x⁵) + 10 x⁴ cos(x³) sin(x⁵) cos(x⁵).

dydx = -3x² sin(x³) sin²(x⁵) + 10 x⁴ cos(x³) sin(x⁵) cos(x⁵).

KJ

Karan Joshi

M.Sc Mathematics, IIT Bombay

Verified Expert

Structural observation. Product of two trig composites; differentiate by parts then layer-by-layer.

1. u' = -3x² sin(x³).
2. v = sin²(x⁵), v' = 2 sin(x⁵) cos(x⁵) · 5 x⁴ using sin(2θ) = 2 sinθ cosθ this equals 5 x⁴ sin(2 x⁵).
3. Sum: -3x² sin(x³) sin²(x⁵) + 5 x⁴ cos(x³) sin(2 x⁵).

Equivalent form: -3x² sin(x³) sin²(x⁵) + 5 x⁴ cos(x³) sin(2 x⁵).

Q 5.7

Differentiate 2 √cot(x²) with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule three layers: outer √·, middle cot, inner x². The derivative of √u is 12√u; the derivative of cot v is -csc² v.

1. Let w = x², v = cot w, y = 2 √v.
2. dydv = 2 · 12√v = 1√v = 1√cot(x²).
3. dvdw = -csc² w = -csc²(x²).
4. dwdx = 2x.
5. Chain rule: dydx = 1√cot(x²) · (-csc²(x²)) · 2x = -2x csc²(x²)√cot(x²).

dydx = -2x csc²(x²)√cot(x²).

PK

Pranav Kapoor

M.Sc Mathematics, IIT Bombay

Verified Expert

Quick reading. Square-root strips the 2 at the front; cot gives a -csc²; x² gives 2x.

1. Derivative of 2√u is u'/√u.
2. Here u = cot(x²), u' = -csc²(x²) · 2x.
3. Combine to -2x csc²(x²)/√cot(x²).

-2x csc²(x²)/√cot(x²).

Q 5.8

Differentiate cos(√x) with respect to x.

Check SolutionExpert Solution

Concept used. Chain rule with outer cos and inner √x. ddx√x = 12√x.

1. Let u = √x, y = cos u.
2. dydu = -sin u = -sin√x.
3. dudx = 12√x.
4. Chain rule: dydx = -sin√x · 12√x = -sin√x2√x.

dydx = -sin√x2√x.

DB

Diya Bhat

Ph.D Mathematics, IIT Delhi

Verified Expert

Quick reading. Two-layer composite; product of -sin√x and 1/(2√x).

1. Outer: -sin√x.
2. Inner: 1/(2√x).
3. Product: -sin√x/(2√x).

-sin√x/(2√x).

Q 5.9

Prove that the function f given by f(x) = |x - 1|, x ∈ R, is not differentiable at x = 1.

Check SolutionExpert Solution

Concept used. f is differentiable at c iff _{h → 0^-} f(c+h) - f(c)h _{h → 0⁺} f(c+h) - f(c)h both exist (finite) and are equal. We compute LHD and RHD at c = 1.

1. f(1) = |1 - 1| = 0. For h near 0: f(1 + h) = |1 + h - 1| = |h|.
2. LHD at 1. For h → 0^-, h < 0 so |h| = -h: _{h → 0^-} f(1+h) - f(1)h = _{h → 0^-} |h| - 0h = _{h → 0^-} -hh = -1.
3. RHD at 1. For h → 0⁺, h > 0 so |h| = h: _{h → 0⁺} f(1+h) - f(1)h = _{h → 0⁺} hh = 1.
4. LHD = -1 but RHD = 1, so the two one-sided derivatives are unequal. Hence f is not differentiable at x = 1.

f(x) = |x - 1| is not differentiable at x = 1 (corner with LHD = -1 and RHD = 1).

YN

Yash Nair

M.Tech CS, IIT Madras

Verified Expert

Picture-first. The graph of y = |x - 1| is a V-shape with the vertex at (1, 0); slope -1 to the left of 1 and +1 to the right.

1. Left-side slope -1, right-side slope +1.
2. Slopes unequal ⇒ derivative does not exist at x = 1.

Not differentiable at x = 1.

Q 5.10

Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3, is not differentiable at x = 1 and x = 2.

Check SolutionExpert Solution

Concept used. A function that is discontinuous at c cannot be differentiable at c (Theorem 3 of the chapter: every differentiable function is continuous). We show f is discontinuous at x = 1 and x = 2; this rules out differentiability automatically. (Alternatively, the LHD/RHD limits are computed and shown unequal.)

1. At x = 1. f(1) = [1] = 1. For h → 0^- (so 1 + h ∈ (0, 1)), [1 + h] = 0: _{h → 0^-} f(1+h) - f(1)h = _{h → 0^-} 0 - 1h = _{h → 0^-} -1h = +∞. The LHD is not finite, hence f is not differentiable at x = 1. (One can also note: f is discontinuous at x = 1, hence not differentiable there.)
2. At x = 2. f(2) = [2] = 2. For h → 0^- (so 1 + h < 2 but > 1), [2 + h] = 1: _{h → 0^-} f(2+h) - f(2)h = _{h → 0^-} 1 - 2h = _{h → 0^-} -1h = +∞. Again the LHD diverges; f is not differentiable at x = 2.
3. In both cases the function jumps by 1 as x crosses the integer, so the difference quotient (f(c+h)-f(c))/h blows up. Hence differentiability fails.

f(x) = [x] is not differentiable at x = 1 and x = 2 (jump discontinuities at both points).

AB

Aditi Banerjee

M.Sc Mathematics, ISI Kolkata

Verified Expert

Strategic angle. A discontinuity automatically kills differentiability. So we just verify the jump.

1. LHL of [x] at x = 1 is 0; value is 1. Discontinuous; not differentiable.
2. LHL of [x] at x = 2 is 1; value is 2. Discontinuous; not differentiable.

Not differentiable at x = 1, 2.

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.