Class 12 Maths NCERT Solutions

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Continuity and Differentiability Class 12 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 5, Continuity and Differentiability — 121 questions from Ex 5.1 to Ex 5.8, each worked through step by step in the CBSE marking pattern. Continuity at a point, differentiability, chain rule, implicit and logarithmic differentiation, and parametric and second derivatives.

Class:12Subject:MathsChapter:5

5 Key Formulas

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 12 Maths Chapter 5?

Chapter 5 carries 8 exercise questions, numbered Ex 5.1 to Ex 5.8. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Table of Contents9 sections

Revision & Practice

Chapter Overview

Continuity and differentiability is where Class 12 calculus begins. A function is continuous at a point when its left and right limits meet the function value there, and differentiable when the left- and right-hand derivatives exist and agree. Every differentiable function is continuous, but the converse fails at corners, cusps and jumps — the two ideas below separate those cases question by question. Work each one with a pencil first; the solutions keep the exam-pattern working short.

Board pattern

Marks are awarded for the test, not the verdict. Always write the left-hand limit, the right-hand limit and the function value for continuity, and the left- and right-hand derivatives for differentiability. State the conclusion in one line: "LHL = RHL = f(a), so f is continuous at x = a". A bare "continuous" scores nothing.

$Ex 5.1 — Continuity at a point, piecewise and composite functions (34 Q)$
$Ex 5.2 — Derivatives of composite functions, chain rule (10 Q)$
$Ex 5.3 — Implicit and inverse-trigonometric derivatives (15 Q)$
$Ex 5.4 — Derivatives of exponential and logarithmic functions (10 Q)$
$Ex 5.5 — Logarithmic differentiation (18 Q)$
$Ex 5.6 — Derivatives of parametric functions (11 Q)$
$Ex 5.7 — Second order derivatives (17 Q)$
$Ex 5.8 — Rolle’s and Lagrange’s Mean Value Theorems (6 Q)$
$Misc — Mixed problems on the whole chapter (21 Q)$

Exercise 5.1 — Continuity

34Exercise questions

Step-by-step solution

1f is a polynomial, hence continuous everywhere; at each point LHL = RHL = f(a). At x = 0, f(0) = −3; at x = −3, f(−3) = −18; at x = 5, f(5) = 22.

Final answer

Continuous at all three points.

Final answer

Continuous at x = 3, since f(3) = 17 = LHL = RHL.

Final answer

All four are continuous on their domains. (a) polynomial; (b) rational, continuous for x ≠ 5; (c) equals x − 5 for x ≠ −5, a removable discontinuity at x = −5; (d) |x − 5| is continuous everywhere.

Final answer

f(n) = nⁿ; as a polynomial, f is continuous at every real x, in particular at x = n.

Step-by-step solution

1At x = 0: LHL = 0, RHL = 0, f(0) = 0 — continuous.
2At x = 1: LHL = 1, RHL = 5 — discontinuous.
3At x = 2: LHL = 5, RHL = 5, f(2) = 5 — continuous.

Final answer

Continuous at x = 0 and x = 2; discontinuous at x = 1, since LHL = 1 but RHL = 5.

Final answer

Discontinuous at x = 2 (LHL = 7, RHL = 1).

Step-by-step solution

1At x = −3: LHL = −6, RHL = f(−3) = 6 — discontinuous.
2At x = 3: LHL = |3|+3 = 6, RHL = 20 — discontinuous.

Final answer

Discontinuous at x = −3 and x = 3.

Final answer

No point of discontinuity: |x| is continuous everywhere, including x = 0 where the limit is 0 = f(0).

Final answer

Discontinuous at x = 0, since LHL = −1 and RHL = 1.

Final answer

Continuous at x = 1 (LHL = 2 = RHL = f(1) = 2); no discontinuity.

Final answer

Continuous at x = 2: LHL = 8 − 3 = 5, RHL = 4 + 1 = 5, f(2) = 5. No discontinuity.

Final answer

Discontinuous at x = 1: LHL = 0 but RHL = 1.

Final answer

Discontinuous at x = 1: LHL = 6 but RHL = −4.

Final answer

Discontinuous at x = 1 and x = 3 (jumps of 1); continuous everywhere else.

Final answer

Continuous at x = 0 (both limits 0); discontinuous at x = 1, where LHL = 0 but RHL = 4.

Final answer

Continuous everywhere: at x = −1 both limits equal −2; at x = 1 both equal 2.

Step-by-step solution

1LHL = 3a + 1 and f(3) = 3a + 1; RHL = 3b + 3.
2Set equal: 3a + 1 = 3b + 3 ⇒ 3a − 3b = 2.

Final answer

a - b = t f r a c 23 (t e x t e q u i v a l e n tl y 3 a - 3 b = 2)

Step-by-step solution

1At x = 0: LHL = λ(0 − 0) = 0, RHL = 1; 0 ≠ 1 for every λ.
2At x = 1: f is given by 4x + 1 near x = 1, a polynomial — continuous.

Final answer

No value of λ makes f continuous at x = 0, because LHL = 0 while RHL = 1. At x = 1, f(x) = 4x + 1 is continuous.

Step-by-step solution

1At an integer n: g(n) = n − n = 0.
2LHL = n − (n − 1) = 1, RHL = n − n = 0. Since LHL ≠ RHL, g is discontinuous at n.

Final answer

g is discontinuous at every integer and continuous at every non-integer.

Final answer

Yes. f(π) = π² + 5; since x² and sin x are continuous, f is continuous at x = π.

Final answer

All three are continuous on ℝ: sums, differences and products of the continuous functions sin x and cos x.

Final answer

cos x is continuous on ℝ. cosec x is continuous on ℝ − {nπ}; sec x on ℝ − {(2n+1)π/2}; cot x on ℝ − {nπ}, n ∈ ℤ.

Final answer

No point of discontinuity. At x = 0, LHL = lim (sin x)/x = 1, RHL = f(0) = 1, so f is continuous at 0.

Final answer

f is continuous everywhere. At x = 0, |x² sin(1/x)| ≤ x² → 0, so the limit is 0 = f(0).

Final answer

Discontinuous at x = 0: f(0) = 1, but LHL = RHL = sin 0 − cos 0 = −1.

Step-by-step solution

1Put x = π/2 + h: k cos x/(π − 2x) = k(−sin h)/(−2h) → k/2 as h → 0.
2Set k/2 = 3 ⇒ k = 6.

Final answer

k = 6

Step-by-step solution

1LHL = 4k, RHL = 3, f(2) = 4k; set 4k = 3.

Final answer

k = df r a c 34

Step-by-step solution

1LHL = kπ + 1, RHL = cos π = −1; set kπ + 1 = −1.

Final answer

k = - df r a c 2 p i

Step-by-step solution

1LHL = 5k + 1, RHL = 10; set 5k + 1 = 10.

Final answer

k = df r a c 95

Step-by-step solution

1At x = 2: 2a + b = 5.
2At x = 10: 10a + b = 21.
3Subtract: 8a = 16 ⇒ a = 2, then b = 1.

Final answer

a = 2, b = 1

Final answer

cos is continuous and x² is continuous; a composition of continuous functions is continuous.

Final answer

cos x is continuous, and the modulus function is continuous; a composition of continuous functions is continuous.

Final answer

sin|x| = (sin ∘ |·|)(x). Both sin x and |x| are continuous, so the composite sin|x| is continuous on ℝ.

Final answer

Discontinuous at x = 0 and x = −1, where the two modulus terms change slope and the derivative jumps.

Exercise 5.2 — Derivatives of Composite Functions

10Exercise questions

Final answer

2 x cos (x^{2} + 5)

Final answer

- cos x, s in (s in x)

Final answer

a cos (a x + b)

Final answer

df r a c se c^{2} s q r t x, sec (t an s q r t x), t an (t an s q r t x) 2 s q r t x

Final answer

a cos (a x + b) sec (c x + d) + c s in (a x + b) t an (c x + d) sec (c x + d)

Final answer

- 3 x^{2} s in x^{3}, s i n^{2} (x^{5}) + 10 x^{4} cos x^{3}, s in (x^{5}) cos (x^{5})

Final answer

df r a c - 2 s q r t 2, x s in (x^{2}) s q r t s in 2 x^{2}

Final answer

- df r a c s in s q r t x 2 s q r t x

Step-by-step solution

1LHD = lim (f(1+h) − f(1))/h = lim (−h)/h = −1 as h → 0⁻.
2RHD = lim (f(1+h) − f(1))/h = lim h/h = 1 as h → 0⁺.

Final answer

LHD = −1 and RHD = 1 at x = 1; since they differ, f is not differentiable at x = 1.

Step-by-step solution

1At x = 1: LHD = lim [1+h]−[1])/h = −1/h → ∞; RHD = 0.
2At x = 2: LHD = (1 − 2)/h = −1/h → ∞; RHD = 0.

Final answer

Not differentiable at x = 1 and x = 2: one-sided derivatives are infinite/0 and unequal.

Exercise 5.3 — Implicit and Inverse Trigonometric Derivatives

15Exercise questions

Final answer

df r a c d y d x = df r a c cos x - 2 3

Step-by-step solution

12 + 3y' = cos y · y'.
2y'(3 − cos y) = −2.

Final answer

df r a c d y d x = df r a c 2 cosy - 3

Final answer

df r a c d y d x = df r a c - a 2 b y + s in y

Final answer

df r a c d y d x = df r a c se c^{2} x - y x + 2 y - 1

Final answer

df r a c d y d x = - df r a c 2 x + y x + 2 y

Final answer

df r a c d y d x = - df r a c 3 x^{2} + 2 x y + y^{2} x^{2} + 2 x y + 3 y^{2}

Final answer

df r a c d y d x = df r a c y s in (x y) s in 2 y - x s in (x y)

Final answer

df r a c d y d x = df r a c s in 2 x s in 2 y

Step-by-step solution

1Put x = tan θ; then 2x/(1+x²) = sin 2θ, so y = 2θ = 2 tan⁻¹x.

Final answer

df r a c d y d x = df r a c 2 1 + x^{2}

Step-by-step solution

1Put x = tan θ; then the argument is tan 3θ, so y = 3θ = 3 tan⁻¹x.

Final answer

df r a c d y d x = df r a c 3 1 + x^{2}

Step-by-step solution

1Put x = tan θ; (1−x²)/(1+x²) = cos 2θ, so y = 2θ = 2 tan⁻¹x.

Final answer

df r a c d y d x = df r a c 2 1 + x^{2}

Step-by-step solution

1Put x = tan θ; (1−x²)/(1+x²) = cos 2θ, and sin⁻¹(cos 2θ) = 2θ.

Final answer

df r a c d y d x = df r a c 2 1 + x^{2}

Step-by-step solution

1Put x = tan θ; 2x/(1+x²) = sin 2θ, so y = π/2 − 2θ.

Final answer

df r a c d y d x = df r a c - 2 1 + x^{2}

Step-by-step solution

1Put x = sin θ; 2x√(1−x²) = sin 2θ, so y = 2θ = 2 sin⁻¹x.

Final answer

df r a c d y d x = df r a c 2 s q r t 1 - x^{2}

Step-by-step solution

1Put x = cos θ; 1/(2x²−1) = 1/cos 2θ = sec 2θ, so y = 2θ = 2 cos⁻¹x.

Final answer

df r a c d y d x = df r a c - 2 s q r t 1 - x^{2}

Exercise 5.4 — Exponential and Logarithmic Derivatives

10Exercise questions

Final answer

df r a c e^{x} (s in x - cos x) s i n^{2} x

Final answer

df r a c e^{s i n^{- 1} x} s q r t 1 - x^{2}

Final answer

3 x^{2} e^{x^{3}}

Final answer

df r a c - e^{- x} cos (t a n^{- 1} e^{- x}) 1 + e^{- 2 x}

Final answer

- e^{x} t an (e^{x})

Final answer

e^{x} + 2 x e^{x^{2}} + 3 x^{2} e^{x^{3}} + 4 x^{3} e^{x^{4}} + 5 x^{4} e^{x^{5}}

Final answer

df r a c e^{s q r t x /2} 4 s q r t x

Final answer

df r a c 1 x l o g x

Final answer

df r a c - s in x l o g x - cos x x (l o g x)^{2}

Final answer

- l e f t (df r a c 1 x + e^{x} r i g h t) s in (l o g x + e^{x})

Exercise 5.5 — Logarithmic Differentiation

18Exercise questions

Final answer

df r a c d y d x = - cos x cos 2 x cos 3 x, (t an x + 2 t an 2 x + 3 t an 3 x)

Step-by-step solution

1log y = ½[log(x−1)+log(x−2)−log(x−3)−log(x−4)−log(x−5)].

Final answer

df r a c d y d x = df r a c 12 y l e f t [df r a c 1 x - 1 + df r a c 1 x - 2 - df r a c 1 x - 3 - df r a c 1 x - 4 - df r a c 1 x - 5 r i g h t]

Final answer

df r a c d y d x = (l o g x)^{cos x} l e f t [- s in x, l o g (l o g x) + df r a c cos x x l o g x r i g h t]

Final answer

df r a c d y d x = x^{x} (1 + l o g x) - 2^{s in x} (cos x) l o g 2

Final answer

df r a c d y d x = y l e f t [df r a c 2 x + 3 + df r a c 3 x + 4 + df r a c 4 x + 5 r i g h t]

Final answer

df r a c d y d x = l e f t (x + df r a c 1 x r i g h t)^{x} l e f t [l o g! l e f t (x + df r a c 1 x r i g h t) + df r a c x^{2} - 1 x^{2} + 1 r i g h t] + x^{1 + f r a c 1 x} l e f t [df r a c x + 1 - l o g x x^{2} r i g h t]

Final answer

df r a c d y d x = (l o g x)^{x} l e f t [l o g (l o g x) + df r a c 1 l o g x r i g h t] + x^{l o g x} c d o t df r a c 2 l o g x x

Final answer

df r a c d y d x = (s in x)^{x} l e f t [l o g (s in x) + x co t x r i g h t] + df r a c 1 2 s q r t x - x^{2}

Final answer

df r a c d y d x = x^{s in x} l e f t [cos x l o g x + df r a c s in x x r i g h t] + (s in x)^{cos x} l e f t [cos x co t x - s in x l o g (s in x) r i g h t]

Final answer

df r a c d y d x = x^{x cos x} l e f t [cos x l o g x - x s in x l o g x + cos x r i g h t] - df r a c 4 x (x^{2} - 1)^{2}

Final answer

df r a c d y d x = (x cos x)^{x} l e f t [l o g (x cos x) + 1 - x t an x r i g h t] + (x s in x)^{f r a c 1 x} l e f t [df r a c x co t x + 1 - l o g (x s in x) x^{2} r i g h t]

Final answer

df r a c d y d x = - df r a c x^{y} l e f t (f r a c y x r i g h t) + y^{x} l o g y x^{y} l o g x + y^{x} l e f t (f r a c x y r i g h t)

Step-by-step solution

1Take logs: x log y = y log x.
2Differentiate: log y + x y'/y = y' log x + y/x.

Final answer

df r a c d y d x = df r a c f r a c y x - l o g y f r a c x y - l o g x

Final answer

df r a c d y d x = df r a c l o g (cosy) + y t an x l o g (cos x) + x t an y

Step-by-step solution

1log x + log y = x − y.
2Differentiate: 1/x + y'/y = 1 − y'.

Final answer

df r a c d y d x = df r a c y (x - 1) x (1 + y)

Step-by-step solution

1log f = log(1+x)+log(1+x²)+log(1+x⁴)+log(1+x⁸).
2f'/f = 1/(1+x)+2x/(1+x²)+4x³/(1+x⁴)+8x⁷/(1+x⁸).
3At x = 1, f(1) = 16 and the bracket equals 7/2, so f'(1) = 56.

Final answer

f^{'} (1) = 56

Step-by-step solution

1Product rule: (2x−5)(x³+7x+9)+(x²−5x+8)(3x²+7).
2Expanding and log-differentiation both give the same polynomial.

Final answer

df r a c d y d x = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11

Step-by-step solution

1Repeated product rule: treat uv as one factor, then expand.
2Log-differentiation: log(uvw) = log u + log v + log w; differentiate and multiply by uvw.

Final answer

Both methods give u′vw + uv′w + uvw′, so the formula holds.

Exercise 5.6 — Derivatives of Parametric Functions

11Exercise questions

Step-by-step solution

1dx/dt = 4at, dy/dt = 4at³; divide.

Final answer

df r a c d y d x = t^{2}

Final answer

df r a c d y d x = - df r a c b a co t t h e t a

Final answer

df r a c d y d x = - 4 s in t

Final answer

df r a c d y d x = - df r a c 1 t^{2}

Final answer

df r a c d y d x = df r a c cos t h e t a - 2 cos 2 t h e t a 2 s in 2 t h e t a - s in t h e t a

Final answer

df r a c d y d x = - co t df r a c t h e t a 2

Final answer

df r a c d y d x = co tt, df r a c 1 - 2 cos 2 t 1 + 2 cos 2 t

Step-by-step solution

1dx/dt = a(−sin t + 1/sin t) = a cos²t/sin t; dy/dt = a cos t.

Final answer

df r a c d y d x = t an t

Final answer

df r a c d y d x = df r a c b a, t e x t cosec, t h e t a

Final answer

df r a c d y d x = t an t h e t a

Step-by-step solution

1(1/x)dx/dt = (1/2)log a /√(1−t²).
2(1/y)dy/dt = −(1/2)log a /√(1−t²).
3Ratio: (dy/dt)/(dx/dt) = −y/x.

Final answer

Differentiating log x and log y and dividing gives dy/dx = −y/x.

Exercise 5.7 — Second Order Derivatives

17Exercise questions

Final answer

df r a c d^{2} y d x^{2} = 2

Final answer

df r a c d^{2} y d x^{2} = 380 x^{18}

Final answer

df r a c d^{2} y d x^{2} = - 2 s in x - x cos x

Final answer

df r a c d^{2} y d x^{2} = - df r a c 1 x^{2}

Final answer

df r a c d^{2} y d x^{2} = 6 x l o g x + 5 x

Final answer

df r a c d^{2} y d x^{2} = 2 e^{x} (5 cos 5 x - 12 s in 5 x)

Final answer

df r a c d^{2} y d x^{2} = 9 e^{6 x} (3 cos 3 x - 4 s in 3 x)

Final answer

df r a c d^{2} y d x^{2} = df r a c - 2 x (1 + x^{2})^{2}

Final answer

df r a c d^{2} y d x^{2} = df r a c - (l o g x + 1) x^{2} (l o g x)^{2}

Final answer

df r a c d^{2} y d x^{2} = df r a c - [s in (l o g x) + cos (l o g x)] x^{2}

Final answer

y″ = −5cos x + 3sin x = −y, so y″ + y = 0.

Step-by-step solution

1y′ = −(1−x²)^{−1/2}, y″ = −x(1−x²)^{−3/2}.
2With x = cos y and √(1−x²) = sin y, y″ = −cos y/sin³y.

Final answer

df r a c d^{2} y d x^{2} = - co t y, t e x t cosec^{2} y

Final answer

x y₁ = −3sin(log x) + 4cos(log x); differentiating and substituting gives x²y₂ + xy₁ + y = 0.

Final answer

y′ = mAe^{mx}+nBe^{nx}, y″ = m²Ae^{mx}+n²Be^{nx}; substitution using mn(Ae^{mx}+Be^{nx}) gives 0.

Final answer

y″ = 500·49 e^{7x} + 600·49 e^{−7x} = 49y.

Final answer

From e^y = 1/(x+1), y = −log(x+1), so y′ = −1/(x+1) and y″ = 1/(x+1)² = (y′)².

Final answer

y₁ = 2 tan⁻¹x/(1+x²); differentiating and clearing (1+x²)² gives the identity.

Exercise 5.8 — Rolle's and Mean Value Theorems

6Exercise questions

Step-by-step solution

1f(−4) = 16 − 8 − 8 = 0 and f(2) = 4 + 4 − 8 = 0.
2f′(−1) = 0.

Final answer

f is continuous on [−4,2] and differentiable on (−4,2); f(−4) = f(2) = 0, so there is c with f′(c) = 0. f′(x) = 2x + 2 = 0 gives c = −1 ∈ (−4,2).

Step-by-step solution

1(i) [x] is discontinuous at 5,6,7,8 in [5,9]; f(5) = 5 ≠ 9 = f(9).
2(ii) discontinuous at −1,0,1; f(−2) = −2 ≠ 2 = f(2).
3(iii) continuity and differentiability hold, but endpoint values differ.

Final answer

Rolle's theorem applies to none. (i) and (ii): the greatest integer function is not continuous at the integers inside the interval. (iii): f is continuous and differentiable but f(1) = 0 ≠ f(2) = 3. Hence the converse of Rolle's theorem is not true.

Final answer

Suppose f(−5) = f(5); then by Rolle's theorem there is c ∈ (−5,5) with f′(c) = 0, contradicting the hypothesis. Hence f(−5) ≠ f(5).

Step-by-step solution

1f(1) = −6, f(4) = −3; slope = (−3+6)/3 = 1.

Final answer

Slope (f(4)−f(1))/3 = 1; f′(c) = 2c − 4 = 1 gives c = 5/2 ∈ (1,4).

Step-by-step solution

1f(1) = −7, f(3) = −27; slope = (−27+7)/2 = −10.
23c² − 10c − 3 = −10 ⇒ 3c² − 10c + 7 = 0 ⇒ (3c−7)(c−1) = 0.
3c = 7/3 lies in (1,3).

Final answer

c = df r a c 73

Step-by-step solution

1(iii) f(1) = 0, f(2) = 3; slope = 3.

Final answer

MVT applies only to (iii). (i) and (ii) are not continuous (nor differentiable) throughout the interval. For (iii), f′ obeys (f(2)−f(1))/1 = 3; f′(c) = 2c = 3 gives c = 3/2 ∈ (1,2).

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Product rule

\frac{d}{d x} uv = u^{'} v + u v^{'}

Quotient rule

\frac{d}{d x} (\frac{u}{v}) = \frac{u ^{'} v - u v ^{'}}{v ^{2}}

Chain rule

\frac{d y}{d x} = \frac{d y}{d u} u^{'} (x)

Parametric derivative

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

Second derivative

\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (\frac{d y}{d x})

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

A function can be continuous but not differentiable, but differentiability always implies continuity — so a discontinuity proves the derivative does not exist.
For implicit functions differentiate both sides and keep y² terms as products using the chain rule, never as 2y.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Maths Chapter 5 (Continuity and Differentiability)?

There are 8 exercise questions in this chapter, numbered Ex 5.1 to Ex 5.8. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Continuity and Differentiability Class 12 Maths?

The formulas this chapter's questions actually turn on are: Product rule, Quotient rule, Chain rule, Parametric derivative, Second derivative. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Continuity and Differentiability important for JEE Main?

Very important — the largest chapter in Class 12 Maths, a major board unit, and the foundation of every later calculus chapter and of JEE Main differentiation.

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