Class 12 Mathematics - NCERT Mathematics Part I
NCERT Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability
Chapter 5 covers continuity of functions, differentiability, chain rule, implicit differentiation, exponential and logarithmic differentiation, parametric differentiation, second order derivatives, and the Mean Value Theorem (Rolle's Theorem and Lagrange's Mean Value Theorem). This is part of the Calculus unit which carries 35 marks in the CBSE Class 12 board exam 2027 - the highest weighted unit. Continuity and Differentiability alone typically contributes about 8-10 marks, making it one of the most important chapters for scoring.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 5 Continuity and Differentiability has 8 exercises plus a Miscellaneous Exercise. Exercise 5.1 (34 Qs) covers continuity of functions. Exercise 5.2 (10 Qs) covers chain rule differentiation. Exercise 5.3 (15 Qs) covers implicit differentiation. Exercise 5.4 (10 Qs) covers exponential and logarithmic function derivatives. Exercise 5.5 (18 Qs) covers logarithmic differentiation. Exercise 5.6 (11 Qs) covers parametric differentiation. Exercise 5.7 (17 Qs) covers second order derivatives. Exercise 5.8 (6 Qs) covers Mean Value Theorem. The Miscellaneous Exercise has 23 questions. Total 144 questions.
Table of Contents
5.1Exercise 5.1 - Continuity of Functions
10 key questions shown below covering continuity proofs, piecewise functions, and discontinuity detection. The remaining 24 questions follow similar patterns.
Q1Prove that the function f(x) = 5x − 3 is continuous at x = 0, x = −3, and x = 5.
A function f is continuous at x = a if lim(x→a) f(x) = f(a).
At x = 0: f(0) = 5(0) − 3 = −3. lim(x→0)(5x − 3) = −3 = f(0). Continuous. ✓
At x = −3: f(−3) = 5(−3) − 3 = −18. lim(x→−3)(5x − 3) = −18 = f(−3). Continuous. ✓
At x = 5: f(5) = 5(5) − 3 = 22. lim(x→5)(5x − 3) = 22 = f(5). Continuous. ✓
Since f(x) = 5x − 3 is a polynomial, it is continuous everywhere on ℝ.
Q2Examine the continuity of the function f(x) = 2x² − 1 at x = 3.
f(3) = 2(3)² − 1 = 2(9) − 1 = 18 − 1 = 17.
lim(x→3)(2x² − 1) = 2(9) − 1 = 17.
Since lim(x→3) f(x) = 17 = f(3), f(x) = 2x² − 1 is continuous at x = 3. ✓
As a polynomial, f is continuous at every point in ℝ.
Q3Examine the following functions for continuity: (a) f(x) = x − 5 (b) f(x) = 1/(x − 5), x ≠ 5 (c) f(x) = (x² − 25)/(x + 5), x ≠ −5 (d) f(x) = |x − 5|
(a) f(x) = x − 5: Polynomial function, continuous for all x ∈ ℝ. ✓
(b) f(x) = 1/(x − 5), x ≠ 5: For any c ≠ 5: lim(x→c) 1/(x−5) = 1/(c−5) = f(c). At x = 5, f is not defined. Continuous for all x ∈ ℝ − {5}. ✓
(c) f(x) = (x² − 25)/(x + 5), x ≠ −5: Simplify: (x−5)(x+5)/(x+5) = x − 5 for x ≠ −5. This is a polynomial. Continuous for all x ∈ ℝ − {−5}. ✓
(d) f(x) = |x − 5|: Write as {5−x if x < 5, x−5 if x ≥ 5}.
At x = 5: f(5) = 0. LHL: lim(x→5⁻)(5−x) = 0. RHL: lim(x→5⁺)(x−5) = 0.
LHL = RHL = f(5) = 0. For x ≠ 5, f is a polynomial on each piece. Continuous for all x ∈ ℝ. ✓
Q5Is the function f defined by f(x) = |x| continuous at x = 0?
f(x) = |x| = {−x if x < 0; x if x ≥ 0}.
f(0) = 0. LHL: lim(x→0⁻)(−x) = 0. RHL: lim(x→0⁺)(x) = 0.
Since LHL = RHL = 0 = f(0), f(x) = |x| is continuous at x = 0. ✓
Note: |x| is continuous everywhere on ℝ, but not differentiable at x = 0 (sharp corner).
Q6Discuss the continuity of the function f defined by: f(x) = 5, if x ≤ 1 f(x) = x, if x > 1
At x = 1 (boundary point):
f(1) = 5 (since 1 ≤ 1, use first piece).
LHL: lim(x→1⁻) f(x) = 5.
RHL: lim(x→1⁺) f(x) = lim(x→1⁺) x = 1.
Since LHL = 5 ≠ 1 = RHL, the limit does not exist. f is discontinuous at x = 1. ✗
For x < 1: f(x) = 5 (constant), continuous. For x > 1: f(x) = x (polynomial), continuous.
The only point of discontinuity is x = 1.
Q10Discuss the continuity of the function f defined by: f(x) = (x − |x|)/x, if x ≠ 0 f(0) = 0
Simplify f(x) for x ≠ 0:
For x > 0: |x| = x, so f(x) = (x − x)/x = 0.
For x < 0: |x| = −x, so f(x) = (x + x)/x = 2.
So f(x) = {2 if x < 0; 0 if x ≥ 0}.
At x = 0: f(0) = 0. LHL: lim(x→0⁻) 2 = 2. RHL: lim(x→0⁺) 0 = 0.
Since LHL = 2 ≠ 0 = RHL, f is discontinuous at x = 0. ✗
For x < 0 and x > 0, f is constant, hence continuous. Only point of discontinuity: x = 0.
Q12Find all points of discontinuity of f, where f is defined by: f(x) = x³ − 3, if x ≤ 2 f(x) = x² + 1, if x > 2
At x = 2 (boundary point):
f(2) = (2)³ − 3 = 5. LHL: lim(x→2⁻)(x³ − 3) = 5. RHL: lim(x→2⁺)(x² + 1) = 5.
Since LHL = RHL = 5 = f(2), f is continuous at x = 2. ✓
For x < 2: polynomial, continuous. For x > 2: polynomial, continuous.
f is continuous everywhere. No points of discontinuity. ✓
Q17Find all points of discontinuity of f, where f is defined by: f(x) = 1/(x + 1), if x ≥ 0 f(x) = x², if x < 0
At x = 0 (boundary point):
f(0) = 1/(0 + 1) = 1. LHL: lim(x→0⁻) x² = 0. RHL: lim(x→0⁺) 1/(x+1) = 1.
Since LHL = 0 ≠ 1 = RHL, f is discontinuous at x = 0. ✗
For x < 0: polynomial, continuous. For x > 0: rational with denominator > 0, continuous.
Only point of discontinuity: x = 0.
Q22Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
cos x: Continuous for all x ∈ ℝ (follows from limit identities lim(h→0) sin h/h = 1 and lim(h→0) cos h = 1).
cosec x = 1/sin x: Continuous everywhere except where sin x = 0, i.e., x = nπ (n ∈ ℤ). Domain of continuity: ℝ − {nπ}. ✓
sec x = 1/cos x: Continuous everywhere except where cos x = 0, i.e., x = (2n+1)π/2 (n ∈ ℤ). Domain of continuity: ℝ − {(2n+1)π/2}. ✓
cot x = cos x/sin x: Continuous everywhere except where sin x = 0, i.e., x = nπ (n ∈ ℤ). Domain of continuity: ℝ − {nπ}. ✓
Q34The function f(x) = [x] (greatest integer function) is continuous at: (A) 4 (B) −2 (C) 1 (D) 1.5
[x] is discontinuous at every integer (LHL ≠ RHL at integer points).
(A) x = 4: LHL = 3, RHL = 4. Discontinuous. ✗
(B) x = −2: LHL = −3, RHL = −2. Discontinuous. ✗
(C) x = 1: LHL = 0, RHL = 1. Discontinuous. ✗
(D) x = 1.5: f(1.5) = 1. For x near 1.5, [x] = 1. lim(x→1.5)[x] = 1 = f(1.5). Continuous. ✓
Answer: (D) 1.5
5.2Exercise 5.2 - Differentiation (Chain Rule)
Q1Differentiate the function sin(x² + 5) with respect to x.
Let y = sin(x² + 5). By chain rule: dy/dx = cos(x² + 5) × 2x = 2x cos(x² + 5).
Outer: sin → cos, Inner: (x² + 5) → 2x.
Q2Differentiate the function cos(sin x) with respect to x.
Let y = cos(sin x). By chain rule:
dy/dx = −sin(sin x) × cos x = −cos x × sin(sin x).
Q3Differentiate the function sin(ax + b) with respect to x.
Let y = sin(ax + b). dy/dx = cos(ax + b) × a = a cos(ax + b).
Q4Differentiate sec(tan(√x)) with respect to x.
Let y = sec(tan(√x)). Three nested functions - apply chain rule step by step:
Let u = √x, v = tan(√x), y = sec(tan(√x)).
dy/dv = sec(tan√x) tan(tan√x), dv/du = sec²(√x), du/dx = 1/(2√x).
dy/dx = sec(tan√x) × tan(tan√x) × sec²(√x) × 1/(2√x)
= [sec(tan√x) × tan(tan√x) × sec²(√x)] / (2√x)
Q5Differentiate sin(ax + b) / cos(cx + d) with respect to x.
Let y = sin(ax + b) / cos(cx + d).
Using the quotient rule: d/dx(u/v) = (v × du/dx − u × dv/dx) / v².
Here u = sin(ax + b), v = cos(cx + d).
du/dx = cos(ax + b) × a = a cos(ax + b) [chain rule]
dv/dx = −sin(cx + d) × c = −c sin(cx + d) [chain rule]
dy/dx = [cos(cx + d) × a cos(ax + b) − sin(ax + b) × (−c sin(cx + d))] / cos²(cx + d)
= [a cos(ax + b) cos(cx + d) + c sin(ax + b) sin(cx + d)] / cos²(cx + d)
= [a cos(ax + b) cos(cx + d) + c sin(ax + b) sin(cx + d)] / cos²(cx + d)
Q6Differentiate cos(x³) × sin²(x⁵) with respect to x.
Let y = cos(x³) × sin²(x⁵).
Using the product rule: d/dx(uv) = u(dv/dx) + v(du/dx).
Here u = cos(x³), v = sin²(x⁵).
du/dx: d/dx[cos(x³)] = −sin(x³) × 3x² = −3x² sin(x³) [chain rule]
dv/dx: d/dx[sin²(x⁵)] = 2 sin(x⁵) × d/dx[sin(x⁵)] [chain rule on outer]
= 2 sin(x⁵) × cos(x⁵) × 5x⁴ [chain rule on inner]
= 10x⁴ sin(x⁵) cos(x⁵)
dy/dx = cos(x³) × [10x⁴ sin(x⁵) cos(x⁵)] + sin²(x⁵) × [−3x² sin(x³)]
= 10x⁴ cos(x³) sin(x⁵) cos(x⁵) − 3x² sin(x³) sin²(x⁵)
Q7Differentiate 2√(cot(x²)) with respect to x.
Let y = 2√(cot(x²)) = 2[cot(x²)]^(1/2).
Using the chain rule, differentiate step by step:
Step 1: dy/dx = 2 × (1/2) × [cot(x²)]^(−1/2) × d/dx[cot(x²)]
= [cot(x²)]^(−1/2) × d/dx[cot(x²)]
= 1/√(cot(x²)) × d/dx[cot(x²)]
Step 2: d/dx[cot(x²)] = −cosec²(x²) × d/dx(x²) = −cosec²(x²) × 2x = −2x cosec²(x²)
Step 3: Combine:
dy/dx = [1/√(cot(x²))] × (−2x cosec²(x²))
= −2x cosec²(x²) / √(cot(x²))
Alternatively: = −2x cosec²(x²) × [cot(x²)]^(−1/2)
Q8Differentiate cos(√x) with respect to x.
y = cos(√x). By chain rule:
dy/dx = −sin(√x) × 1/(2√x) = −sin(√x) / (2√x).
Q9Prove that the function f given by f(x) = |x − 1|, x ∈ ℝ, is not differentiable at x = 1.
f(x) = |x − 1| = {1 − x if x < 1; x − 1 if x ≥ 1}.
LHD at x = 1: lim(h→0⁻) [f(1+h) − f(1)]/h = lim(h→0⁻) |h|/h = lim(h→0⁻)(−h)/h = −1.
RHD at x = 1: lim(h→0⁺) [f(1+h) − f(1)]/h = lim(h→0⁺) |h|/h = lim(h→0⁺) h/h = 1.
Since LHD (−1) ≠ RHD (1), f is not differentiable at x = 1. ∎
Note: f is continuous at x = 1, showing continuity does NOT imply differentiability.
Q10Differentiate e^(sin⁻¹x) with respect to x.
y = e^(sin⁻¹x). By chain rule: dy/dx = e^(sin⁻¹x) × 1/√(1 − x²) = e^(sin⁻¹x) / √(1 − x²), for |x| < 1.
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If f(x) = {kx + 1, if x ≤ 5; 3x − 5, if x > 5} is continuous at x = 5, find the value of k.
For continuity at x = 5: lim(x→5⁻) f(x) = lim(x→5⁺) f(x) = f(5).
f(5) = k(5) + 1 = 5k + 1 (using first piece since 5 ≤ 5).
RHL: lim(x→5⁺) (3x − 5) = 15 − 5 = 10.
For continuity: 5k + 1 = 10 → 5k = 9 → k = 9/5.
If y = (sin⁻¹x)², prove that (1 − x²)y₂ − xy₁ − 2 = 0.
y = (sin⁻¹x)². Differentiating: y₁ = 2(sin⁻¹x) × 1/√(1−x²) = 2sin⁻¹x/√(1−x²).
So y₁√(1−x²) = 2sin⁻¹x. Squaring: y₁²(1−x²) = 4(sin⁻¹x)² = 4y.
Differentiate both sides: 2y₁y₂(1−x²) + y₁²(−2x) = 4y₁.
Divide by 2y₁ (y₁ ≠ 0): y₂(1−x²) − xy₁ = 2.
Rearranging: (1−x²)y₂ − xy₁ − 2 = 0. Hence proved. ∎
Differentiate xˣ + x^(cos x) with respect to x.
Let y = u + v where u = xˣ and v = x^(cos x).
For u = xˣ: ln u = x ln x. Differentiate: (1/u)(du/dx) = ln x + x(1/x) = ln x + 1.
du/dx = xˣ(ln x + 1).
For v = x^(cos x): ln v = cos x × ln x. Differentiate: (1/v)(dv/dx) = −sin x × ln x + cos x × (1/x).
dv/dx = x^(cos x) × [cos x/x − sin x × ln x].
dy/dx = xˣ(1 + ln x) + x^(cos x)[cos x/x − sin x × ln x].
If x = a(cos t + t sin t) and y = a(sin t − t cos t), find d²y/dx².
dx/dt = a(−sin t + sin t + t cos t) = at cos t.
dy/dt = a(cos t − cos t + t sin t) = at sin t.
dy/dx = (dy/dt)/(dx/dt) = (at sin t)/(at cos t) = tan t.
d/dt(dy/dx) = d/dt(tan t) = sec²t.
d²y/dx² = [d/dt(dy/dx)]/(dx/dt) = sec²t/(at cos t) = 1/(at cos³t).
d²y/dx² = sec³t/(at).
Differentiate sin⁻¹(2x√(1−x²)) with respect to x.
Let y = sin⁻¹(2x√(1−x²)). Put x = sin θ, so θ = sin⁻¹x.
2x√(1−x²) = 2 sin θ √(1−sin²θ) = 2 sin θ cos θ = sin 2θ.
y = sin⁻¹(sin 2θ) = 2θ = 2 sin⁻¹x.
dy/dx = 2 × 1/√(1−x²) = 2/√(1−x²).
If y = eˣ sin x, prove that d²y/dx² − 2(dy/dx) + 2y = 0.
y = eˣ sin x.
y₁ = eˣ sin x + eˣ cos x = eˣ(sin x + cos x).
y₂ = eˣ(sin x + cos x) + eˣ(cos x − sin x) = eˣ(2 cos x).
Now: y₂ − 2y₁ + 2y = 2eˣ cos x − 2eˣ(sin x + cos x) + 2eˣ sin x
= 2eˣ cos x − 2eˣ sin x − 2eˣ cos x + 2eˣ sin x = 0. Hence proved. ∎
Find the value of k for which f(x) = {kx², if x ≤ 2; 3, if x > 2} is continuous at x = 2.
For continuity at x = 2: lim(x→2⁻) f(x) = lim(x→2⁺) f(x) = f(2).
f(2) = k(2)² = 4k (using first piece).
RHL: lim(x→2⁺) f(x) = 3.
For continuity: 4k = 3 → k = 3/4.
Differentiate (log x)^(cos x) with respect to x.
Let y = (log x)^(cos x). Taking log: ln y = cos x × ln(log x).
Differentiating: (1/y)(dy/dx) = −sin x × ln(log x) + cos x × [1/(log x)] × (1/x).
dy/dx = (log x)^(cos x) × [−sin x × ln(log x) + cos x/(x log x)].
dy/dx = (log x)^(cos x) × [cos x/(x log x) − sin x × ln(log x)].
If y = 3 cos(log x) + 4 sin(log x), show that x²y₂ + xy₁ + y = 0.
y = 3 cos(log x) + 4 sin(log x).
y₁ = −3 sin(log x) × (1/x) + 4 cos(log x) × (1/x) = [−3 sin(log x) + 4 cos(log x)]/x.
xy₁ = −3 sin(log x) + 4 cos(log x).
Differentiating xy₁: xy₂ + y₁ = [−3 cos(log x) × (1/x) − 4 sin(log x) × (1/x)].
x(xy₂ + y₁) = −3 cos(log x) − 4 sin(log x) = −y.
x²y₂ + xy₁ = −y.
x²y₂ + xy₁ + y = 0. Hence proved. ∎
If y = (tan⁻¹x)², show that (x² + 1)²y₂ + 2x(x² + 1)y₁ = 2.
y = (tan⁻¹x)². y₁ = 2tan⁻¹x × 1/(1+x²) = 2tan⁻¹x/(1+x²).
So (1+x²)y₁ = 2tan⁻¹x. Differentiate: (1+x²)y₂ + 2xy₁ = 2/(1+x²).
Multiply by (1+x²): (1+x²)²y₂ + 2x(1+x²)y₁ = 2.
(x²+1)²y₂ + 2x(x²+1)y₁ = 2. Hence proved. ∎
Verify Rolle's theorem for f(x) = x² + 2x − 8 on [−4, 2].
Check conditions:
(i) f(x) = x² + 2x − 8 is a polynomial, so continuous on [−4, 2]. ✓
(ii) f(x) is differentiable on (−4, 2). ✓
(iii) f(−4) = 16 − 8 − 8 = 0, f(2) = 4 + 4 − 8 = 0. f(−4) = f(2) = 0. ✓
All conditions satisfied. By Rolle's theorem, ∃ c ∈ (−4, 2) such that f'(c) = 0.
f'(x) = 2x + 2. Setting f'(c) = 0: 2c + 2 = 0 → c = −1 ∈ (−4, 2). ✓
Rolle's theorem is verified.
If y = sin(log x), prove that x²y₂ + xy₁ + y = 0.
y = sin(log x). y₁ = cos(log x) × (1/x) = cos(log x)/x.
xy₁ = cos(log x). Differentiate: xy₂ + y₁ = −sin(log x) × (1/x) = −sin(log x)/x.
Multiply by x: x²y₂ + xy₁ = −sin(log x) = −y.
x²y₂ + xy₁ + y = 0. Hence proved. ∎
Differentiate (sin x)ˣ + sin⁻¹(√x) with respect to x.
Let y = u + v where u = (sin x)ˣ and v = sin⁻¹(√x).
For u = (sin x)ˣ: ln u = x ln(sin x). Differentiate: (1/u)(du/dx) = ln(sin x) + x × (cos x/sin x) = ln(sin x) + x cot x.
du/dx = (sin x)ˣ[ln(sin x) + x cot x].
For v = sin⁻¹(√x): dv/dx = 1/√(1−x) × 1/(2√x) = 1/(2√(x−x²)).
dy/dx = (sin x)ˣ[ln(sin x) + x cot x] + 1/(2√(x−x²)).
Verify LMVT for f(x) = x³ − 5x² − 3x on [1, 3].
f(x) = x³ − 5x² − 3x is a polynomial, so continuous on [1, 3] and differentiable on (1, 3).
f(1) = 1 − 5 − 3 = −7. f(3) = 27 − 45 − 9 = −27.
[f(3) − f(1)]/(3 − 1) = (−27 − (−7))/2 = −20/2 = −10.
f'(x) = 3x² − 10x − 3. Set f'(c) = −10:
3c² − 10c − 3 = −10 → 3c² − 10c + 7 = 0 → (3c − 7)(c − 1) = 0.
c = 7/3 or c = 1. Since c must be in (1, 3): c = 7/3 ∈ (1, 3). ✓
LMVT is verified.
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): Every differentiable function is continuous.
Reason (R): If a function f is differentiable at x = a, then lim(x→a) f(x) = f(a).
Answer: (A)
Both are TRUE and the reason explains the assertion. Differentiability at a point implies the function must be continuous at that point, because the limit used in the derivative definition requires the function value to equal the limit. This is a fundamental theorem.
AR2
Assertion (A): Every continuous function is differentiable.
Reason (R): The function f(x) = |x| is continuous at x = 0 but not differentiable at x = 0.
Answer: (D)
Assertion is FALSE: Continuity does NOT imply differentiability. A function can be continuous but have a sharp corner (like |x| at x = 0) where the derivative does not exist. Reason is TRUE: |x| is indeed continuous but not differentiable at x = 0, which serves as a counterexample.
AR3
Assertion (A): The function f(x) = [x] (greatest integer function) is discontinuous at all integer points.
Reason (R): At any integer n, lim(x→n⁻) [x] = n − 1 and lim(x→n⁺) [x] = n, so the left hand limit ≠ right hand limit.
Answer: (A)
Both are TRUE and the reason explains the assertion. At every integer n, the greatest integer function has a jump discontinuity because the left limit (n−1) does not equal the right limit (n). At non-integer points, the function is constant in a neighborhood and hence continuous.
AR4
Assertion (A): If f(x) = sin x, then f satisfies Rolle's theorem on [0, π].
Reason (R): sin 0 = 0 = sin π, sin x is continuous on [0, π], and differentiable on (0, π).
Answer: (A)
Both are TRUE and the reason explains the assertion. All three conditions of Rolle's theorem are satisfied: (1) sin x is continuous on [0, π], (2) sin x is differentiable on (0, π), (3) f(0) = f(π) = 0. Indeed, f'(c) = cos c = 0 gives c = π/2 ∈ (0, π).
AR5
Assertion (A): d/dx(eˣ) = eˣ for all x ∈ ℝ.
Reason (R): The exponential function eˣ is its own derivative.
Answer: (A)
Both are TRUE and the reason explains the assertion. The function eˣ is unique in that it equals its own derivative: d/dx(eˣ) = eˣ. This follows from the definition of e as the base of the natural logarithm and the series expansion of eˣ.
AR6
Assertion (A): If f(x) = |x − 2|, then f'(2) exists and equals 0.
Reason (R): f(x) = |x − 2| has a minimum at x = 2.
Answer: (D)
Assertion is FALSE: f(x) = |x − 2| is NOT differentiable at x = 2. The LHD = −1 and RHD = 1, which are not equal, so the derivative does not exist at x = 2. Reason is TRUE: f(x) = |x − 2| does have a minimum value of 0 at x = 2. But having a minimum does not guarantee differentiability (this only works for smooth functions).
Case Study Questions
A scientist models the temperature T (in °C) of a chemical reaction as a function of time t (in hours) using the piecewise function: T(t) = {t² + 2t + 1, if 0 ≤ t ≤ 3; at + b, if 3 < t ≤ 6; 25, if t > 6} For the model to be physically meaningful, the temperature function must be continuous (no sudden jumps in temperature).
(i) Find T(3) using the first piece of the function. (1 mark)
T(3) = (3)² + 2(3) + 1 = 9 + 6 + 1 = 16°C.
(ii) Find the values of a and b such that T(t) is continuous at t = 3 and t = 6. (2 marks)
At t = 3: For continuity, lim(t→3⁺) T(t) = T(3).
3a + b = 16 ... (1)
At t = 6: For continuity, lim(t→6⁻) T(t) = lim(t→6⁺) T(t).
6a + b = 25 ... (2)
Subtract (1) from (2): 3a = 9 → a = 3.
Substitute in (1): 9 + b = 16 → b = 7.
So T(t) = 3t + 7 for 3 < t ≤ 6.
(iii)(a) What is the rate of change of temperature at t = 2? (2 marks)
For 0 ≤ t ≤ 3: T(t) = t² + 2t + 1. T'(t) = 2t + 2.
At t = 2: T'(2) = 2(2) + 2 = 6°C per hour.
The temperature is increasing at a rate of 6°C/hour at t = 2.
(iii)(b) Is the rate of change of temperature continuous at t = 3? Justify your answer. (2 marks)
Left derivative at t = 3: T'(t) = 2t + 2, so T'(3⁻) = 2(3) + 2 = 8.
Right derivative at t = 3: T(t) = 3t + 7, so T'(3⁺) = 3.
Since 8 ≠ 3, the derivative is not continuous at t = 3.
The temperature function is continuous at t = 3, but its rate of change has a sudden jump from 8°C/hour to 3°C/hour. This means the function is continuous but not differentiable at t = 3.
An engineer designs a section of a roller coaster track using parametric equations: x = a(θ − sin θ) and y = a(1 − cos θ) where θ is the parameter (angle in radians) and a is a positive constant representing the radius of the generating circle. This is known as a cycloid curve. The engineer needs to compute the slope and curvature at various points to ensure safety.
(i) Find dx/dθ and dy/dθ. (1 mark)
x = a(θ − sin θ) → dx/dθ = a(1 − cos θ).
y = a(1 − cos θ) → dy/dθ = a sin θ.
dx/dθ = a(1 − cos θ), dy/dθ = a sin θ.
(ii) Find the slope dy/dx in terms of θ. (2 marks)
dy/dx = (dy/dθ) / (dx/dθ) = (a sin θ) / (a(1 − cos θ)) = sin θ / (1 − cos θ).
Using half-angle formulas: sin θ = 2 sin(θ/2) cos(θ/2), 1 − cos θ = 2 sin²(θ/2).
dy/dx = [2 sin(θ/2) cos(θ/2)] / [2 sin²(θ/2)] = cos(θ/2) / sin(θ/2) = cot(θ/2).
(iii)(a) Find the slope of the track at θ = π/2. (2 marks)
dy/dx at θ = π/2: cot(π/4) = 1.
The slope is 1, meaning the track makes a 45° angle with the horizontal at this point.
(iii)(b) Find d²y/dx² in terms of θ. (2 marks)
dy/dx = cot(θ/2). d/dθ(dy/dx) = d/dθ[cot(θ/2)] = −cosec²(θ/2) × (1/2) = −cosec²(θ/2)/2.
d²y/dx² = [d/dθ(dy/dx)] / (dx/dθ) = [−cosec²(θ/2)/2] / [a(1 − cos θ)]
= [−cosec²(θ/2)/2] / [a × 2sin²(θ/2)]
= −cosec²(θ/2) / [4a sin²(θ/2)]
= −1 / [4a sin⁴(θ/2)]
= −cosec⁴(θ/2) / (4a).
MCQ Practice
MCQ1. The function f(x) = [x] (greatest integer function) is continuous at:
(A) x = 1
(B) x = −1
(C) x = 1.5
(D) x = 2
Answer: (C) - The greatest integer function [x] is discontinuous at all integer values and continuous at all non-integer values. Since 1.5 is not an integer, **f is continuous at x = 1.5.**
MCQ2. If f(x) = x|x|, then f'(0) is:
(A) 0
(B) 1
(C) −1
(D) does not exist
Answer: (A) - f(x) = x|x| = {−x², x < 0; x², x ≥ 0}. f'(x) = {−2x, x < 0; 2x, x ≥ 0}. LHD at 0: lim(x→0⁻)(−2x) = 0. RHD at 0: lim(x→0⁺)(2x) = 0. Since LHD = RHD = 0, **f'(0) = 0.**
MCQ3. d/dx(e^(log x)) equals:
(A) eˣ
(B) 1
(C) 1/x
(D) x
Answer: (B) - e^(log x) = e^(ln x) = x (since e and ln are inverse functions). So d/dx(e^(log x)) = d/dx(x) = **1.**
MCQ4. If y = log(log x), then dy/dx is:
(A) 1/x
(B) 1/(x log x)
(C) log x/x
(D) 1/log x
Answer: (B) - y = log(log x). By chain rule: dy/dx = 1/(log x) × d/dx(log x) = 1/(log x) × 1/x = **1/(x log x).**
MCQ5. If f(x) = |cos x|, then f'(π/4) is:
(A) 1/√2
(B) −1/√2
(C) √2
(D) −√2
Answer: (B) - At x = π/4, cos x = 1/√2 > 0, so |cos x| = cos x in a neighborhood of π/4. Thus f'(π/4) = −sin(π/4) = **−1/√2.**
MCQ6. If y = sin⁻¹(2x/(1 + x²)), then dy/dx is:
(A) 1/(1 + x²)
(B) 2/(1 + x²)
(C) −2/(1 + x²)
(D) 2x/(1 + x²)
Answer: (B) - Put x = tan θ, then 2x/(1+x²) = 2tan θ/(1+tan²θ) = sin 2θ. So y = sin⁻¹(sin 2θ) = 2θ = 2tan⁻¹x. dy/dx = 2 × 1/(1+x²) = **2/(1+x²).**
MCQ7. The derivative of eˣ³ with respect to x is:
(A) eˣ³
(B) 3x²eˣ³
(C) eˣ³ × 3x
(D) x³eˣ³
Answer: (B) - Using chain rule: d/dx(eˣ³) = eˣ³ × d/dx(x³) = eˣ³ × 3x² = **3x²eˣ³.**
MCQ8. If f(x) = {sin x/x, if x ≠ 0; k, if x = 0} is continuous at x = 0, then k equals:
(A) 0
(B) 1
(C) −1
(D) 1/2
Answer: (B) - For continuity at x = 0: k = f(0) = lim(x→0) sin x/x = **1** (standard limit). So k = 1.
MCQ9. If x = at² and y = 2at, then d²y/dx² is:
(A) −1/(2at³)
(B) 1/t²
(C) −1/(at²)
(D) 0
Answer: (A) - dx/dt = 2at, dy/dt = 2a. dy/dx = 2a/(2at) = 1/t. d/dt(1/t) = −1/t². d²y/dx² = (−1/t²)/(2at) = **−1/(2at³).**
MCQ10. Rolle's theorem is applicable to f(x) = x² − 4x + 3 on:
(A) [0, 3]
(B) [1, 3]
(C) [0, 4]
(D) [−1, 1]
Answer: (B) - f(x) = x² − 4x + 3 = (x−1)(x−3). f(1) = 0 and f(3) = 0. So f(1) = f(3) = 0. The function is a polynomial (continuous and differentiable everywhere). All three conditions of Rolle's theorem are satisfied on **[1, 3].**
MCQ11. If y = xˣ, then dy/dx is:
(A) xˣ(1 + log x)
(B) xˣ × log x
(C) xˣ⁻¹ × x
(D) xˣ + x
Answer: (A) - y = xˣ. Take log: ln y = x ln x. Differentiate: (1/y)(dy/dx) = ln x + x × (1/x) = ln x + 1 = 1 + ln x. So dy/dx = xˣ(1 + ln x) = **xˣ(1 + log x).**
MCQ12. If f(x) is continuous on [a, b] and differentiable on (a, b), then by LMVT there exists c ∈ (a, b) such that f'(c) equals:
(A) f(b) − f(a)
(B) [f(b) − f(a)]/(b − a)
(C) [f(a) − f(b)]/(b − a)
(D) f(a) + f(b)
Answer: (B) - Lagrange's Mean Value Theorem states that f'(c) = **[f(b) − f(a)]/(b − a).** This is the slope of the secant line joining (a, f(a)) and (b, f(b)).
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
Find the derivative of eˢⁱⁿˣ with respect to x.
d/dx(eˢⁱⁿˣ) = eˢⁱⁿˣ × cos x = **cos x × eˢⁱⁿˣ**.
Is f(x) = |x| differentiable at x = 0?
No. LHD = −1 and RHD = 1 at x = 0. Since LHD ≠ RHD, f(x) = |x| is **not differentiable at x = 0**, though it is continuous there.
If y = log(sin x), find dy/dx.
dy/dx = (1/sin x) × cos x = cos x/sin x = **cot x**.
Write the derivative of tan⁻¹(eˣ) with respect to x.
d/dx[tan⁻¹(eˣ)] = 1/(1 + (eˣ)²) × eˣ = **eˣ/(1 + e²ˣ)**.
If y = 5ˣ, find dy/dx.
d/dx(5ˣ) = 5ˣ × ln 5 = **5ˣ log 5**.
2-3 Mark Questions
If f(x) = {(x² − 1)/(x − 1), if x ≠ 1; k, if x = 1} is continuous at x = 1, find k.
lim(x→1) (x²−1)/(x−1) = lim(x→1) (x+1)(x−1)/(x−1) = lim(x→1) (x+1) = 2. For continuity: k = 2. k = 2.
Differentiate sin⁻¹(2x√(1−x²)) with respect to x.
Put x = sin θ. 2x√(1−x²) = 2sinθ cosθ = sin2θ. y = sin⁻¹(sin2θ) = 2θ = 2sin⁻¹x. dy/dx = 2/√(1−x²).
If x = a cos³θ and y = a sin³θ, find dy/dx.
dx/dθ = −3a cos²θ sinθ. dy/dθ = 3a sin²θ cosθ. dy/dx = (3a sin²θ cosθ)/(−3a cos²θ sinθ) = −sinθ/cosθ = −tan θ.
Verify Rolle's theorem for f(x) = sin x on [0, π].
f is continuous on [0,π], differentiable on (0,π), f(0) = 0 = f(π). f'(x) = cos x. f'(c) = 0 → cos c = 0 → c = π/2 ∈ (0,π). Rolle's theorem verified with c = π/2.
3-4 Mark Questions
Differentiate (x + 1/x)ˣ + x^(1+1/x) with respect to x.
Let u = (x + 1/x)ˣ. ln u = x ln(x + 1/x). du/dx = (x + 1/x)ˣ[ln(x + 1/x) + x(1 − 1/x²)/(x + 1/x)]. Let v = x^(1+1/x). ln v = (1+1/x)ln x. dv/dx = x^(1+1/x)[(1+1/x)/x + ln x × (−1/x²)]. dy/dx = du/dx + dv/dx.
If y = (sin x)^(tan x), find dy/dx.
ln y = tan x × ln(sin x). (1/y)(dy/dx) = sec²x × ln(sin x) + tan x × (cos x/sin x) = sec²x ln(sin x) + 1. dy/dx = (sin x)^(tan x) × [sec²x × ln(sin x) + 1].
If y = eˣ(sin x + cos x), prove that d²y/dx² − 2(dy/dx) + 2y = 0.
y₁ = eˣ(sinx + cosx) + eˣ(cosx − sinx) = 2eˣcosx. y₂ = 2eˣcosx − 2eˣsinx = 2eˣ(cosx − sinx). y₂ − 2y₁ + 2y = 2eˣ(cosx−sinx) − 4eˣcosx + 2eˣ(sinx+cosx) = 2eˣcosx − 2eˣsinx − 4eˣcosx + 2eˣsinx + 2eˣcosx = 0. Hence proved.
Key Concepts
Important Formulas
Study Plan - Continuity and Differentiability
Continuity of Functions (Exercise 5.1)
Read Sections 5.1-5.2. Understand the three conditions for continuity and algebra of continuous functions. Complete 15-20 questions from Exercise 5.1 covering polynomial continuity, piecewise functions, and finding points of discontinuity. Focus on problems with LHL/RHL comparison.
Chain Rule and Implicit Differentiation (Exercises 5.2, 5.3)
Read Sections 5.3-5.4. Master the chain rule for composite functions and implicit differentiation. Complete Exercise 5.2 (all 10 Qs) and Exercise 5.3 (all 15 Qs). Practice nested chain rule applications (3+ layers) and implicit equations like x² + y² = r².
Exponential, Logarithmic, and Parametric Differentiation (Exercises 5.4, 5.5, 5.6)
Read Sections 5.5-5.7. Learn derivatives of eˣ, ln x, aˣ, and logarithmic differentiation for [f(x)]^g(x) type functions. Complete Exercises 5.4 (10 Qs), 5.5 (18 Qs), and 5.6 (11 Qs). These are the highest-scoring topics in board exams.
Second Order Derivatives and Mean Value Theorem (Exercises 5.7, 5.8)
Read Sections 5.8-5.9. Master second order derivatives for both explicit and parametric functions. Understand and practice Rolle's Theorem and LMVT. Complete Exercise 5.7 (17 Qs) and Exercise 5.8 (6 Qs). Focus on the 'prove that' pattern for second derivatives.
Miscellaneous Exercise, PYQs, and Revision
Complete the Miscellaneous Exercise (23 Qs). Solve PYQs from 2019-2026. Revise all formulas, especially chain rule, logarithmic differentiation, and parametric formulas. Practice MCQs, assertion-reason, and case study questions. Time yourself: aim to solve a 5-mark logarithmic differentiation problem in under 8 minutes.
Common Mistakes to Avoid
- ✗Forgetting to apply the chain rule when differentiating composite functions. For example, d/dx(sin(3x)) = 3cos(3x), NOT cos(3x). Always multiply by the derivative of the inner function.
- ✗Checking only one-sided limit for piecewise continuity. At boundary points of piecewise functions, you must check BOTH the left hand limit (LHL) and the right hand limit (RHL) separately, and compare them with f(a).
- ✗Confusing continuity with differentiability. Every differentiable function is continuous, but NOT every continuous function is differentiable. The classic example is f(x) = |x| at x = 0: continuous but not differentiable.
- ✗Wrong application of logarithmic differentiation: forgetting to multiply by y at the end. After finding (1/y)(dy/dx), you must multiply both sides by y to get dy/dx. Students often leave the answer as (1/y)(dy/dx).
- ✗Incorrect second derivatives in parametric form. For d²y/dx², the formula is d²y/dx² = [d/dt(dy/dx)]/(dx/dt), NOT d²y/dt² ÷ d²x/dt². This is a very common error.
- ✗Not verifying all three conditions of Rolle's theorem or LMVT before applying them. You must explicitly check: (1) continuity on [a,b], (2) differentiability on (a,b), and for Rolle's: (3) f(a) = f(b). Missing any condition means the theorem cannot be applied.
Tips for Board Exam
- ✓For finding k in piecewise continuity problems: equate the left hand limit and right hand limit at the boundary point. This gives you the equation to solve for k. Always state the continuity condition explicitly before solving.
- ✓Use substitution to simplify inverse trigonometric differentiation: put x = sin θ for expressions with √(1−x²), x = tan θ for expressions with 1/(1+x²), x = cos θ for expressions involving √(1−x²). This converts complex expressions to simple double-angle forms.
- ✓For logarithmic differentiation of [f(x)]^g(x): take ln of both sides FIRST, then use the formula ln y = g(x) × ln f(x). Differentiate using product rule on the right side. This is much simpler than direct differentiation.
- ✓In second order derivative problems, look for patterns: if you can express xy₁ or (1+x²)y₁ in a simple form, differentiate that expression instead of computing y₂ from scratch. This saves significant time in the exam.
- ✓Rolle's theorem and LMVT verification questions follow a fixed template: (1) State and verify all conditions, (2) Find f'(x), (3) Set f'(c) = 0 (Rolle's) or f'(c) = [f(b)−f(a)]/(b−a) (LMVT), (4) Solve for c and verify c is in (a,b). Practice this template until it becomes automatic.
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Start Continuity and Differentiability - Choose Your ModeAbout Continuity and Differentiability - Class 12 Mathematics
Chapter 5 Continuity and Differentiability is the most important chapter in the CBSE Class 12 Mathematics syllabus from a marks perspective, belonging to the Calculus unit which carries 35 marks - the highest weightage of any unit. The chapter begins with the concept of continuity, establishing that a function is continuous at a point when its limit equals its value at that point. Students learn to test continuity for polynomial, rational, trigonometric, and piecewise functions, with special emphasis on finding constants that make piecewise functions continuous. The chapter then introduces differentiability, establishing the crucial relationship that every differentiable function is continuous but not vice versa (|x| at x = 0 being the classic counterexample). The chain rule for composite functions, implicit differentiation, and derivatives of exponential and logarithmic functions form the technical core of the chapter.
The second half covers advanced differentiation techniques that are heavily tested in board exams: logarithmic differentiation for functions of the form [f(x)]^g(x), parametric differentiation using dy/dx = (dy/dt)/(dx/dt), and second order derivatives including the parametric formula d²y/dx² = [d/dt(dy/dx)]/(dx/dt). The chapter concludes with two fundamental theorems of calculus - Rolle's Theorem and Lagrange's Mean Value Theorem (LMVT) - which guarantee the existence of specific tangent lines under given conditions. For the CBSE 2027 board exam, students should prioritize logarithmic differentiation (5-mark question appears frequently), second order derivative proofs like showing x²y₂ + xy₁ + y = 0, finding k for continuity of piecewise functions (2-mark question in almost every paper), and MVT verification problems. With 144 questions across 8 exercises and a Miscellaneous Exercise, thorough practice of this chapter is essential for maximizing marks in the Calculus unit.
Frequently Asked Questions
How many exercises are in NCERT Class 12 Maths Chapter 5 Continuity and Differentiability?
There are 8 exercises (5.1 to 5.8) plus a Miscellaneous Exercise, totalling 144 questions. Exercise 5.1 has 34 questions on continuity, Exercise 5.2 has 10 questions on chain rule, Exercise 5.3 has 15 on implicit differentiation, Exercise 5.4 has 10 on exponential/logarithmic derivatives, Exercise 5.5 has 18 on logarithmic differentiation, Exercise 5.6 has 11 on parametric derivatives, Exercise 5.7 has 17 on second order derivatives, Exercise 5.8 has 6 on MVT, and the Miscellaneous Exercise has 23 questions.
How many marks does Continuity and Differentiability carry in the CBSE board exam?
Continuity and Differentiability is part of the Calculus unit (35 marks total - the highest weighted unit). Chapter 5 alone typically contributes about 8-10 marks, making it one of the most important chapters. Expect 1 MCQ (1 mark), 1 short answer on continuity or chain rule (2-3 marks), and 1 long answer on logarithmic differentiation or second order derivatives (5 marks).
What is the difference between continuity and differentiability?
Continuity at a point means lim(x→a) f(x) = f(a) (no breaks or jumps in the graph). Differentiability at a point means the derivative exists (the graph has a well-defined tangent line). Key relationship: Every differentiable function is continuous, but a continuous function may NOT be differentiable (e.g., f(x) = |x| at x = 0 has a sharp corner - continuous but not differentiable).
What is the chain rule and when do I use it?
The chain rule states: d/dx[f(g(x))] = f'(g(x)) × g'(x). Use it whenever you differentiate a composite function (a function inside another function). For example, d/dx[sin(x²)] = cos(x²) × 2x. The rule says: differentiate the outer function, keep the inner function unchanged, then multiply by the derivative of the inner function.
When do I use logarithmic differentiation?
Use logarithmic differentiation when: (1) The function is of the form [f(x)]^g(x) (variable base and variable exponent), like xˣ or (sin x)^(cos x). (2) The function involves complex products or quotients that simplify with logarithms. The method: take ln of both sides, use log properties to simplify, then differentiate using implicit differentiation.
What is Rolle's Theorem and how is it different from LMVT?
Rolle's Theorem: If f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then ∃ c ∈ (a,b) with f'(c) = 0 (horizontal tangent). LMVT (Lagrange's Mean Value Theorem): If f is continuous on [a,b] and differentiable on (a,b), then ∃ c ∈ (a,b) with f'(c) = [f(b)−f(a)]/(b−a) (tangent parallel to secant). Rolle's is a special case of LMVT where f(a) = f(b), making the right side zero.
What is the formula for second derivative in parametric form?
If x = f(t) and y = g(t), then: First derivative: dy/dx = (dy/dt)/(dx/dt). Second derivative: d²y/dx² = [d/dt(dy/dx)] / (dx/dt). Important: Do NOT use d²y/dt² ÷ d²x/dt². You must first find dy/dx as a function of t, then differentiate that with respect to t, and finally divide by dx/dt.
Is NCERT enough for Continuity and Differentiability in the CBSE board exam?
Yes, NCERT is sufficient for the board exam. All questions are based on NCERT patterns. Prioritize: Exercise 5.1 (continuity with piecewise functions), Exercise 5.5 (logarithmic differentiation - appears frequently as a 5-mark question), Exercise 5.7 (second order derivatives), and Exercise 5.8 (MVT). Also practice the Miscellaneous Exercise and previous year questions. For competitive exams (JEE), you may need supplementary materials for advanced applications.