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Continuity and Differentiability: Class 12 Maths Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A civil engineer models the vertical profile of a road bridge. The height h(x) = √(4x + 9) metres at horizontal distance x metres (x ≥ 0) from the start of the bridge is used to verify a structural identity involving derivatives.

A civil engineer is modelling the vertical profile of a road bridge. The height (in metres) of the bridge deck above ground at a horizontal distance x metres from the start is given by

h(x) = √(4x + 9), x ≥ 0.

The engineer needs to verify a structural identity involving the rate of change of height and its acceleration (second derivative) to ensure the design is consistent.

(i) Find dh/dx. [1 mark]
(ii) Find d²h/dx². [1 mark]
(iii) Show that h · (d²h/dx²) + (dh/dx)² = 0, and interpret what this identity tells the engineer about the curvature formula for this bridge profile. [2 marks]

Show answer
Part (i) — Find dh/dx. [1 mark]

Given h(x) = √(4x + 9) = (4x + 9)<super>1/2</super>.

Differentiating w.r.t. x,

dh/dx = (1/2)(4x + 9)<super>−1/2</super> · 4

∴ dh/dx = 2/√(4x + 9)


Part (ii) — Find d<super>2</super>h/dx<super>2</super>. [1 mark]

Differentiating dh/dx = 2(4x + 9)<super>−1/2</super> w.r.t. x,

d<super>2</super>h/dx<super>2</super> = 2 · (−1/2)(4x + 9)<super>−3/2</super> · 4

∴ d<super>2</super>h/dx<super>2</super> = −4/(4x + 9)<super>3/2</super>


Part (iii) — Show that h · (d<super>2</super>h/dx<super>2</super>) + (dh/dx)<super>2</super> = 0. [2 marks]

L.H.S. = h · (d<super>2</super>h/dx<super>2</super>) + (dh/dx)<super>2</super>

Substituting the results from parts (i) and (ii),

= √(4x + 9) · [−4/(4x + 9)<super>3/2</super>] + [2/√(4x + 9)]<super>2</super>

= −4(4x + 9)<super>1/2</super>/(4x + 9)<super>3/2</super> + 4/(4x + 9)

= −4/(4x + 9) + 4/(4x + 9)

= 0

∴ L.H.S. = R.H.S. Hence proved.

Interpretation: The identity h · (d<super>2</super>h/dx<super>2</super>) + (dh/dx)<super>2</super> = 0 is equivalent to d<super>2</super>(h<super>2</super>)/dx<super>2</super> = 2[(dh/dx)<super>2</super> + h·(d<super>2</super>h/dx<super>2</super>)] being governed solely by a constant term (since h<super>2</super> = 4x + 9 is linear in x, its second derivative is zero). This tells the engineer that the bridge profile h(x) = √(4x + 9) has no acceleration in h<super>2</super>, confirming the design follows a smooth, uniformly-varying curvature with no abrupt structural inflection.
Q2Case-based4 marks

A civil engineer models the deflection of a beam using piecewise and implicit functions to study continuity and slope behaviour at critical points along the beam's length.

A civil engineer is analysing the stress distribution along a beam. The deflection function of the beam is modelled as:

f(x) = { (x² sin(1/x) + 3x, x ≠ 0
{ 0, x = 0

(i) The engineer first checks whether the deflection function is continuous at x = 0. Verify whether f(x) is continuous at x = 0. [1 mark]

(ii) Next, the engineer needs to confirm differentiability at x = 0 to ensure no abrupt change in slope. Check whether f(x) is differentiable at x = 0. [1 mark]

(iii) For a different section of the beam, the deflection satisfies the implicit relation:

x² + y² − 2x − 4y + 3 = 0

The engineer needs to find dy/dx at the point (2, 1) to determine the slope of the deflection curve at that point. Find dy/dx at (2, 1).

OR

For yet another section, the deflection is given by y = (cos x)^(sin x), x ∈ (0, π/2). Find dy/dx. [2 marks]

Show answer
(i) For continuity at x = 0, we check whether lim(x→0) f(x) = f(0).

Now, f(0) = 0.

For x ≠ 0:
|f(x) − 0| = |x² sin(1/x) + 3x| ≤ |x²||sin(1/x)| + 3|x| ≤ x² + 3|x|

As x → 0, x² + 3|x| → 0.

∴ lim(x→0) f(x) = 0 = f(0).

∴ f(x) is continuous at x = 0. [1 mark]

---

(ii) For differentiability at x = 0, we compute f'(0) by definition:

f'(0) = lim(h→0) [f(h) − f(0)] / h
= lim(h→0) [h² sin(1/h) + 3h − 0] / h
= lim(h→0) [h sin(1/h) + 3]

Now, |h sin(1/h)| ≤ |h| → 0 as h → 0.

∴ lim(h→0) [h sin(1/h) + 3] = 0 + 3 = 3.

∴ f'(0) = 3, which exists.

∴ f(x) is differentiable at x = 0. [1 mark]

---

(iii) Given: x² + y² − 2x − 4y + 3 = 0

Differentiating both sides w.r.t. x,

2x + 2y(dy/dx) − 2 − 4(dy/dx) = 0

⟹ 2y(dy/dx) − 4(dy/dx) = 2 − 2x

⟹ dy/dx (2y − 4) = 2 − 2x

⟹ dy/dx = (2 − 2x) / (2y − 4) = (1 − x) / (y − 2)

At (2, 1):

dy/dx = (1 − 2) / (1 − 2) = (−1) / (−1) = 1

∴ dy/dx at (2, 1) = 1. [1 + 1 = 2 marks]

---

OR

Given: y = (cos x)^(sin x), x ∈ (0, π/2)

Taking log on both sides,

log y = sin x · log(cos x)

Differentiating both sides w.r.t. x,

(1/y)(dy/dx) = cos x · log(cos x) + sin x · (1/cos x) · (−sin x)

(1/y)(dy/dx) = cos x · log(cos x) − sin²x / cos x

∴ dy/dx = y [cos x · log(cos x) − sin²x / cos x]

∴ dy/dx = (cos x)^(sin x) [cos x · log(cos x) − sin²x / cos x] [1 + 1 = 2 marks]
Q3Case-based4 marks

A civil engineer is designing a flyover ramp whose elevation profile is modelled by the function f(x) = { (√(1 + 3x) − 1) / x for x ≠ 0; k for x = 0 }, where x represents horizontal displacement (in metres) from the ramp entry and f(x) represents the normalised height. The ramp must have a smooth, continuous profile with no abrupt changes in elevation.

A civil engineer is designing a flyover ramp whose elevation profile is modelled by the function

f(x) = { (√(1 + 3x) − 1) / x, for x ≠ 0
{ k, for x = 0

where x (in metres) represents the horizontal displacement from the entry point of the ramp and f(x) represents the normalised height.

(i) For the ramp profile to be smooth (no sudden jump), the function f must be continuous at x = 0. Find the value of k that ensures this. [1 mark]

(ii) Once k is set as found in part (i), verify that f is indeed continuous at x = 0 by checking all three conditions. [1 mark]

(iii) The engineer also considers a modified steepness function g(x) = √(log(1 + sin(x²))) for x near the ramp entry. Find dg/dx. [2 marks]

Show answer
(i) For continuity at x = 0, we need k = lim(x→0) f(x).

lim(x→0) [√(1 + 3x) − 1] / x

Rationalising the numerator by multiplying and dividing by [√(1 + 3x) + 1],

= lim(x→0) [(√(1 + 3x) − 1)(√(1 + 3x) + 1)] / [x · (√(1 + 3x) + 1)]

= lim(x→0) [(1 + 3x) − 1] / [x · (√(1 + 3x) + 1)]

= lim(x→0) 3x / [x · (√(1 + 3x) + 1)]

= lim(x→0) 3 / (√(1 + 3x) + 1)

= 3 / (√1 + 1) = 3 / 2

∴ k = 3/2

(ii) For continuity at x = 0, we verify three conditions:

Condition 1 — f(0) exists: f(0) = k = 3/2. ✓

Condition 2 — lim(x→0) f(x) exists:
As computed in part (i), lim(x→0) f(x) = 3/2. ✓

Condition 3 — lim(x→0) f(x) = f(0):
3/2 = 3/2. ✓

∴ All three conditions are satisfied, so f is continuous at x = 0.

(iii) Let g(x) = √(log(1 + sin(x²))).

This is a composition of three functions. Applying the chain rule,

dg/dx = d/dx [log(1 + sin(x²))]^(1/2)

= (1/2) · [log(1 + sin(x²))]^(−1/2) · d/dx [log(1 + sin(x²))]

= 1 / [2√(log(1 + sin(x²)))] · [1 / (1 + sin(x²))] · d/dx [1 + sin(x²)]

= 1 / [2√(log(1 + sin(x²)))] · [1 / (1 + sin(x²))] · cos(x²) · 2x

∴ dg/dx = x · cos(x²) / [(1 + sin(x²)) · √(log(1 + sin(x²)))]
Q4Case-based4 marks

A civil engineer is modelling the vertical cross-section of a road ramp. The height (in metres) at horizontal distance x (in metres) is given by:

f(x) = { 2x² + x, x ≤ 1
{ ax + b, x > 1

For the ramp to be smooth, f must be continuous and differentiable at x = 1.

Definitions provided:
L.H.D.: Lf′(a) = lim[h→0⁻] [f(a−h) − f(a)]/(−h)
R.H.D.: Rf′(a) = lim[h→0⁺] [f(a+h) − f(a)]/h
f is differentiable at x = a if L.H.D. = R.H.D.

A civil engineer is modelling the vertical cross-section of a road ramp that joins a lower road to an elevated highway. The height of the ramp (in metres) at a horizontal distance x (in metres) from the start is modelled by the function

f(x) = { 2x² + x, x ≤ 1
{ ax + b, x > 1

For the ramp profile to be smooth (no abrupt bends or breaks), the height function must be both continuous and differentiable at x = 1. Using the definitions:

Left Hand Derivative (L.H.D.): Lf′(a) = lim[h→0⁻] [f(a − h) − f(a)] / (−h)
Right Hand Derivative (R.H.D.): Rf′(a) = lim[h→0⁺] [f(a + h) − f(a)] / h

Answer the following questions:
(i) Find the value of f(1) using the expression valid at x = 1. [1]
(ii) Find the condition on a and b so that f is continuous at x = 1. [1]
(iii) Find the values of a and b so that f is also differentiable at x = 1, and hence write the complete smooth function. [2]

Show answer
Sub-part (i) [1 mark]

For x ≤ 1, f(x) = 2x² + x.

∴ f(1) = 2(1)² + 1 = 2 + 1 = 3

∴ f(1) = 3

──────────────────────────────────
Sub-part (ii) [1 mark]

For continuity at x = 1, we need:

lim[x→1⁺] f(x) = f(1)

For x > 1, f(x) = ax + b.

∴ lim[x→1⁺] f(x) = a(1) + b = a + b

For continuity at x = 1:

a + b = 3 ...(i)

∴ The condition for continuity is a + b = 3.

──────────────────────────────────
Sub-part (iii) [2 marks]

Computing L.H.D. at x = 1:

Lf′(1) = lim[h→0⁻] [f(1 − h) − f(1)] / (−h)

Since 1 − h < 1 for small h > 0, f(1 − h) = 2(1 − h)² + (1 − h).

= lim[h→0] [2(1 − h)² + (1 − h) − 3] / (−h)

= lim[h→0] [2(1 − 2h + h²) + 1 − h − 3] / (−h)

= lim[h→0] [2 − 4h + 2h² + 1 − h − 3] / (−h)

= lim[h→0] [−5h + 2h²] / (−h)

= lim[h→0] (5 − 2h)

∴ Lf′(1) = 5

Computing R.H.D. at x = 1:

Rf′(1) = lim[h→0⁺] [f(1 + h) − f(1)] / h

Since 1 + h > 1, f(1 + h) = a(1 + h) + b.

= lim[h→0] [a(1 + h) + b − 3] / h

= lim[h→0] [a + ah + b − 3] / h

Using condition (i), a + b = 3, so a + b − 3 = 0.

= lim[h→0] [ah] / h

= lim[h→0] a

∴ Rf′(1) = a

For differentiability at x = 1, L.H.D. = R.H.D.:

a = 5 ...(ii)

Substituting a = 5 in condition (i):

5 + b = 3 ⟹ b = −2 ...(iii)

∴ a = 5 and b = −2.

The complete smooth ramp function is:

f(x) = { 2x² + x, x ≤ 1
{ 5x − 2, x > 1

∴ The values are a = 5 and b = −2, and the smooth height function is f(x) = 2x² + x for x ≤ 1, and f(x) = 5x − 2 for x > 1.
Q5Case-based4 marks

A civil engineer is modelling the elevation profile of a bridge ramp. The height (in metres) of the ramp at a horizontal distance x (in metres) from the start is described by the function f(x) = (e^(2x) − 1)/sin x for x ≠ 0, and f(0) = k.

A civil engineer is modelling the elevation profile of a bridge ramp. The height (in metres) of the ramp at a horizontal distance x (in metres) from the start is described by the function

f(x) = { (e^(2x) − 1) / (sin x), for x ≠ 0
k, for x = 0 }

(i) For the ramp profile to be physically realizable (no sudden jump in height), the function f must be continuous at x = 0. Find the value of k that ensures this continuity.

(ii) Once k is fixed as found in (i), determine whether f is differentiable at x = 0 by computing f′(0) from first principles (using the definition of the derivative).

Show answer
Part (i) — Finding k for continuity at x = 0 [2 marks]

For continuity at x = 0, we require:

lim(x→0) f(x) = f(0) = k

Now,

lim(x→0) f(x) = lim(x→0) [e^(2x) − 1] / sin x

Multiply and divide numerator and denominator by x:

= lim(x→0) { [e^(2x) − 1] / x } / { sin x / x }

= lim(x→0) [e^(2x) − 1] / x ÷ lim(x→0) (sin x / x)

Using standard limits:

lim(x→0) (e^(2x) − 1) / x = lim(x→0) 2 · (e^(2x) − 1) / (2x) = 2 × 1 = 2

lim(x→0) (sin x) / x = 1

∴ lim(x→0) f(x) = 2 / 1 = 2

For continuity, k = lim(x→0) f(x)

∴ k = 2

──────────────────────────────────────
Part (ii) — Differentiability at x = 0 [2 marks]

With k = 2, so f(0) = 2.

By the definition of the derivative:

f′(0) = lim(h→0) [f(h) − f(0)] / h

= lim(h→0) [ (e^(2h) − 1)/sin h − 2 ] / h

Write e^(2h) − 1 = sin h · f(h), and expand using Taylor series (or standard limits):

Numerator = (e^(2h) − 1) / sin h − 2

= (e^(2h) − 1 − 2 sin h) / sin h

So:

f′(0) = lim(h→0) (e^(2h) − 1 − 2 sin h) / (h · sin h)

Since lim(h→0) sin h / h = 1, we have h · sin h ~ h² as h→0.

Expanding using standard limits (e^(2h) = 1 + 2h + 2h² + … and sin h = h − h³/6 + …):

e^(2h) − 1 − 2 sin h = (2h + 2h² + …) − 2(h − h³/6 + …)

= 2h + 2h² + … − 2h + h³/3 + …

= 2h² + higher order terms

∴ f′(0) = lim(h→0) 2h² / h² = 2

∴ f′(0) = 2, which is finite and well-defined.

∴ f is differentiable at x = 0, and f′(0) = 2.
Q6Case-based4 marks

A civil engineer models the deflection of a suspension bridge cable using the function y = (sin x)^(cos x) + (cos x)^(sin x), where x ∈ (0, π/2) represents the horizontal position along the bridge deck. To study the rate of change of deflection, the engineer needs to differentiate y with respect to x.

A civil engineer models the deflection of a suspension bridge cable using the function

y = (sin x)^(cos x) + (cos x)^(sin x),

where x ∈ (0, π/2) represents the horizontal position along the bridge deck.

(i) To analyse the rate of change of deflection, the engineer needs dy/dx. Let u = (sin x)^(cos x) and v = (cos x)^(sin x). Find du/dx. [2]

(ii) Using a similar method, find dv/dx, and hence write down dy/dx in simplified form. [2]

Show answer
Part (i): Finding du/dx where u = (sin x)^(cos x) [2 marks]

Let u = (sin x)^(cos x).

Taking log on both sides,

log u = cos x · log(sin x) ...(i)

Differentiating both sides w.r.t. x,

(1/u) · du/dx = (d/dx)(cos x) · log(sin x) + cos x · (d/dx)(log(sin x))

⟹ (1/u) · du/dx = (−sin x) · log(sin x) + cos x · (1/sin x) · cos x

⟹ (1/u) · du/dx = −sin x · log(sin x) + cos²x/sin x

∴ du/dx = (sin x)^(cos x) · [cos²x/sin x − sin x · log(sin x)]

[Award 1 mark for correct logarithmic differentiation setup of log u = cos x · log(sin x); 1 mark for correct du/dx]

―――――――――――――――――――――――――――――

Part (ii): Finding dv/dx and hence dy/dx [2 marks]

Let v = (cos x)^(sin x).

Taking log on both sides,

log v = sin x · log(cos x) ...(ii)

Differentiating both sides w.r.t. x,

(1/v) · dv/dx = (d/dx)(sin x) · log(cos x) + sin x · (d/dx)(log(cos x))

⟹ (1/v) · dv/dx = cos x · log(cos x) + sin x · (1/cos x) · (−sin x)

⟹ (1/v) · dv/dx = cos x · log(cos x) − sin²x/cos x

∴ dv/dx = (cos x)^(sin x) · [cos x · log(cos x) − sin²x/cos x]

Since y = u + v,

dy/dx = du/dx + dv/dx

∴ dy/dx = (sin x)^(cos x) · [cos²x/sin x − sin x · log(sin x)] + (cos x)^(sin x) · [cos x · log(cos x) − sin²x/cos x]

[Award 1 mark for correct dv/dx; 1 mark for writing dy/dx = du/dx + dv/dx and stating the final combined expression]
Q7Case-based4 marks

A highway engineer is designing a smooth road transition. The elevation profile of the road (in metres) over a stretch is modelled by the function:

f(x) = { ax + b, x ≤ 1
x² + 3, x > 1

where x is the horizontal distance (in km) from a reference point.

A highway engineer is designing a smooth road transition. The elevation profile of the road (in metres) over a stretch is modelled by the function:

f(x) = { ax + b, x ≤ 1
x² + 3, x > 1

where x is the horizontal distance (in km) from a reference point.

(i) For the road profile to have no sudden jumps in elevation, the function f(x) must be continuous at x = 1. Find the condition relating a and b. [1 mark]

(ii) For a smooth gradient transition (no abrupt change in slope), f(x) must also be differentiable at x = 1. Using this condition, find the value of a. [1 mark]

(iii) Using the values found above, determine the value of b, and hence write the complete elevation function f(x). [2 marks]

Show answer
(i) For continuity at x = 1, we check LHL = RHL = f(1).

LHL = lim x→1⁻ f(x) = a(1) + b = a + b

RHL = lim x→1⁺ f(x) = (1)² + 3 = 4

f(1) = a(1) + b = a + b

For continuity at x = 1: LHL = RHL

⟹ a + b = 4 ...(i)

∴ The required condition is a + b = 4.

(ii) For differentiability at x = 1, we compute LHD and RHD.

LHD = lim x→1⁻ [f(x) − f(1)]/(x − 1) = lim x→1⁻ [(ax + b) − (a + b)]/(x − 1)
= lim x→1⁻ a(x − 1)/(x − 1) = a

RHD = lim x→1⁺ [f(x) − f(1)]/(x − 1) = lim x→1⁺ [(x² + 3) − 4]/(x − 1)
= lim x→1⁺ (x² − 1)/(x − 1)
= lim x→1⁺ (x + 1) = 2

For differentiability at x = 1: LHD = RHD

⟹ a = 2

∴ a = 2.

(iii) Substituting a = 2 in equation (i):

2 + b = 4 ⟹ b = 2

∴ b = 2.

The complete elevation function is:

f(x) = { 2x + 2, x ≤ 1
x² + 3, x > 1

∴ The required values are a = 2, b = 2, and the elevation profile is f(x) = 2x + 2 for x ≤ 1 and f(x) = x² + 3 for x > 1.
Q8Case-based4 marks

A civil engineer models the vertical profile of a curved road using the function y = f(x) = √(1 + sin²x), x ∈ ℝ, where y (in metres) represents the height at horizontal position x.

A civil engineer models the vertical profile of a curved road using the function

y = f(x) = √(1 + sin²x), x ∈ ℝ

where y (in metres) represents the height at horizontal position x.

(i) Find f'(x), the rate of change of height with respect to horizontal position. [1]
(ii) Evaluate f'(π/2) and interpret its meaning in context. [1]
(iii) The engineer claims that the road profile satisfies the identity

2y · (dy/dx) = sin 2x

Verify this claim analytically. [2]

Show answer
(i) Finding f'(x):

Given: y = √(1 + sin²x) = (1 + sin²x)^(1/2)

Differentiating both sides w.r.t. x,

dy/dx = (1/2)(1 + sin²x)^(−1/2) · d/dx(sin²x)

⟹ dy/dx = (1/2)(1 + sin²x)^(−1/2) · 2 sin x · cos x

⟹ dy/dx = (sin x cos x) / √(1 + sin²x)

∴ f'(x) = (sin x cos x) / √(1 + sin²x)

(ii) Evaluating f'(π/2):

f'(π/2) = (sin(π/2) · cos(π/2)) / √(1 + sin²(π/2))

⟹ f'(π/2) = (1 · 0) / √(1 + 1)

∴ f'(π/2) = 0

Interpretation: At horizontal position x = π/2, the rate of change of height is zero, meaning the road profile has a local maximum (or minimum) at that point — the road is momentarily level (neither rising nor falling) at x = π/2.

(iii) Verification of the identity 2y · (dy/dx) = sin 2x:

L.H.S. = 2y · (dy/dx)

Substituting y = √(1 + sin²x) and dy/dx = (sin x cos x) / √(1 + sin²x),

L.H.S. = 2 · √(1 + sin²x) · (sin x cos x) / √(1 + sin²x)

⟹ L.H.S. = 2 sin x cos x

⟹ L.H.S. = sin 2x [∵ sin 2x = 2 sin x cos x]

= R.H.S.

∴ L.H.S. = R.H.S. Hence proved.

The engineer's claim is verified.
Q9MCQ1 mark

If y = log(sin x), then dy/dx is equal to:

Show answer
Option (A) is correct.

Explanation: Using the chain rule, d/dx[log(sin x)] = (1/sin x) · cos x = cot x.

∴ dy/dx = cot x
Q10MCQ1 mark

The derivative of log(sin x) with respect to x is:

Show answer
Option (A) is correct.

Explanation: Using the chain rule, d/dx[log(sin x)] = (1/sin x) · cos x = cos x/sin x = cot x.
Q11MCQ1 mark

If y = sin x + cos x, find dy/dx.

Show answer
Option (B) is correct.

Explanation: Differentiating both sides w.r.t. x,

dy/dx = d/dx(sin x) + d/dx(cos x) = cos x + (−sin x) = cos x − sin x.
Q12MCQ1 mark

If f(x) = \begin{cases} \dfrac{\tan 3x}{\sin 2x}, & x \neq 0 \\ k, & x = 0 \end{cases} is continuous at x = 0, then the value of k is:

Show answer
Option (B) is correct.

Explanation: For continuity at x = 0, we need lim_{x→0} f(x) = f(0) = k.

lim_{x→0} \dfrac{\tan 3x}{\sin 2x} = lim_{x→0} \dfrac{\tan 3x}{3x} · \dfrac{2x}{\sin 2x} · \dfrac{3}{2} = 1 · 1 · \dfrac{3}{2} = \dfrac{3}{2}.

∴ k = \dfrac{3}{2}.
Q13Short Answer2 marks

Differentiate x^(sin x) with respect to x.

Show answer
Let y = x<super>sin x</super>.

Taking log on both sides,

log y = sin x · log x

Differentiating both sides w.r.t. x,

(1/y) · dy/dx = cos x · log x + sin x · (1/x)

∴ dy/dx = x<super>sin x</super> (cos x · log x + sin x / x)
Q14Short Answer2 marks

If x³ + y³ = 3xy, find dy/dx.

Show answer
Differentiating both sides w.r.t. x,

3x² + 3y² · dy/dx = 3y + 3x · dy/dx

⟹ 3y² · dy/dx − 3x · dy/dx = 3y − 3x²

⟹ dy/dx (y² − x) = y − x²

∴ dy/dx = (y − x²)/(y² − x)
Q15Short Answer2 marks

Check whether the function

is continuous at or not.

Show answer
For continuity at x = 2, we check LHL = RHL = f(2).

LHL = lim x→2⁻ f(x) = lim x→2⁻ |x−2|/(x−2)

For x < 2, |x−2| = −(x−2), so

LHL = lim x→2⁻ −(x−2)/(x−2) = −1

RHL = lim x→2⁺ f(x) = lim x→2⁺ |x−2|/(x−2)

For x > 2, |x−2| = (x−2), so

RHL = lim x→2⁺ (x−2)/(x−2) = 1

∵ LHL = −1 ≠ RHL = 1, the limit does not exist at x = 2.

∴ f is not continuous at x = 2.
Q16Short Answer2 marks

Differentiate sin(x²) with respect to x.

Show answer
Let y = sin(x²).

Differentiating both sides w.r.t. x, by the Chain Rule,

dy/dx = cos(x²) · d/dx(x²)

⟹ dy/dx = cos(x²) · 2x

∴ dy/dx = 2x cos(x²)
Q17Short Answer3 marks

If f(x) = |cos 3x|, find f'(x) at x = π/6.

Show answer
We have f(x) = |cos 3x|.

Near x = π/6, we check the sign of cos 3x.

At x = π/6: cos 3(π/6) = cos(π/2) = 0.

For x slightly less than π/6, say x = π/6 − h (h > 0):
cos 3(π/6 − h) = cos(π/2 − 3h) = sin 3h > 0.

For x slightly greater than π/6, say x = π/6 + h (h > 0):
cos 3(π/6 + h) = cos(π/2 + 3h) = −sin 3h < 0.

∴ f(x) = |cos 3x| = ⎧ cos 3x, x < π/6
⎨ 0, x = π/6
⎩ −cos 3x, x > π/6

Now we compute the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) at x = π/6.

LHD = lim_{h→0⁺} [f(π/6 − h) − f(π/6)] / (−h)
= lim_{h→0⁺} [cos(π/2 − 3h) − 0] / (−h)
= lim_{h→0⁺} sin 3h / (−h)
= lim_{h→0⁺} (−3) · (sin 3h)/(3h)
= −3 · 1 = −3.

RHD = lim_{h→0⁺} [f(π/6 + h) − f(π/6)] / h
= lim_{h→0⁺} [−cos(π/2 + 3h) − 0] / h
= lim_{h→0⁺} sin 3h / h
= lim_{h→0⁺} 3 · (sin 3h)/(3h)
= 3 · 1 = 3.

∵ LHD ≠ RHD, f(x) = |cos 3x| is not differentiable at x = π/6.

∴ f'(π/6) does not exist.
Q18Short Answer3 marks

If x = a(cos t + t sin t) and y = a(sin t − t cos t), find d²y/dx².

Show answer
Given: x = a(cos t + t sin t) and y = a(sin t − t cos t).

Differentiating x w.r.t. t,
dx/dt = a(−sin t + sin t + t cos t) = at cos t

Differentiating y w.r.t. t,
dy/dt = a(cos t − cos t + t sin t) = at sin t

Using the parametric derivative formula,
dy/dx = (dy/dt)/(dx/dt) = (at sin t)/(at cos t) = tan t ...(i)

Differentiating both sides of (i) w.r.t. x,
d²y/dx² = d/dx(tan t) = (d/dt(tan t)) · (dt/dx)

Now, d/dt(tan t) = sec²t and dt/dx = 1/(dx/dt) = 1/(at cos t)

∴ d²y/dx² = sec²t · 1/(at cos t) = sec³t/(at)

∴ d²y/dx² = sec³t/(at)
Q19Short Answer3 marks

Find the value of the constants a and b so that the function f defined below is continuous at x = 1 and x = 3:

f(x) = { 3, if x ≤ 1
ax + b, if 1 < x < 3
7, if x ≥ 3 }

Show answer
For continuity at x = 1, we require:

lim(x→1⁺) f(x) = f(1)

Now, f(1) = 3.

lim(x→1⁺) f(x) = lim(x→1⁺) (ax + b) = a(1) + b = a + b

∴ a + b = 3 ...(i)

For continuity at x = 3, we require:

lim(x→3⁻) f(x) = f(3)

Now, f(3) = 7.

lim(x→3⁻) f(x) = lim(x→3⁻) (ax + b) = a(3) + b = 3a + b

∴ 3a + b = 7 ...(ii)

Subtracting (i) from (ii),

(3a + b) − (a + b) = 7 − 3

⟹ 2a = 4

⟹ a = 2

Substituting a = 2 in (i),

2 + b = 3 ⟹ b = 1

∴ a = 2 and b = 1.
Q20Short Answer3 marks

Find the derivative of log(sin²x) with respect to √(cos x).

Show answer
Let u = log(sin²x) and v = √(cos x).

We need to find du/dv = (du/dx)/(dv/dx).

Finding du/dx:

Differentiating u = log(sin²x) w.r.t. x,

du/dx = (1/sin²x) · d/dx(sin²x)

⟹ du/dx = (1/sin²x) · 2 sin x · cos x

⟹ du/dx = 2 cos x / sin x = 2 cot x ...(i)

Finding dv/dx:

Differentiating v = √(cos x) = (cos x)<super>1/2</super> w.r.t. x,

dv/dx = (1/2)(cos x)<super>−1/2</super> · (−sin x)

⟹ dv/dx = −sin x / (2√(cos x)) ...(ii)

Finding du/dv:

From (i) and (ii),

du/dv = (du/dx)/(dv/dx) = (2 cot x) / (−sin x / (2√(cos x)))

⟹ du/dv = 2 cot x × (−2√(cos x) / sin x)

⟹ du/dv = −4 cot x · √(cos x) / sin x

⟹ du/dv = −4 (cos x / sin x) · √(cos x) / sin x

⟹ du/dv = −4 cos x · √(cos x) / sin²x

∴ Derivative of log(sin²x) with respect to √(cos x) = −4(cos x)<super>3/2</super> / sin²x
Q21Short Answer3 marks

If x = a(θ − sin θ) and y = a(1 − cos θ), find d²y/dx² at θ = π/2.

Show answer
Given: x = a(θ − sin θ) and y = a(1 − cos θ).

Differentiating both w.r.t. θ,

dx/dθ = a(1 − cos θ) ...(i)

dy/dθ = a sin θ ...(ii)

Using the parametric derivative formula,

dy/dx = (dy/dθ)/(dx/dθ)

⟹ dy/dx = (a sin θ) / (a(1 − cos θ))

⟹ dy/dx = sin θ / (1 − cos θ) ...(iii)

Now, to find d²y/dx², differentiate dy/dx w.r.t. x:

d²y/dx² = (d/dθ)(dy/dx) · (1/(dx/dθ))

Differentiating (iii) w.r.t. θ,

d/dθ(dy/dx) = [(1 − cos θ)(cos θ) − sin θ(sin θ)] / (1 − cos θ)²

= [cos θ − cos²θ − sin²θ] / (1 − cos θ)²

= [cos θ − (cos²θ + sin²θ)] / (1 − cos θ)²

= (cos θ − 1) / (1 − cos θ)²

= −(1 − cos θ) / (1 − cos θ)²

= −1 / (1 − cos θ)

∴ d²y/dx² = [d/dθ(dy/dx)] / (dx/dθ)

= [−1/(1 − cos θ)] / [a(1 − cos θ)]

= −1 / [a(1 − cos θ)²]

Substituting θ = π/2,

1 − cos(π/2) = 1 − 0 = 1

∴ d²y/dx²|_(θ=π/2) = −1 / [a(1)²]

∴ d²y/dx² at θ = π/2 is −1/a.
Q22Long Answer4 marks

A civil engineer models the bending moment of a beam using the function

f(x) = x^(sin x) + (3x² + 2)/(x² + x + 1), x > 0.

The engineer needs to find how rapidly the bending moment changes with position x, and also needs the second component's rate of change separately.

(i) Let u = x^(sin x). By taking logarithms, find du/dx. [2 marks]
(ii) Let v = (3x² + 2)/(x² + x + 1). Find dv/dx and hence find df/dx = d/dx[f(x)]. [2 marks]

Show answer
Part (i): Finding du/dx where u = x^(sin x)

Let u = x^(sin x).

Taking log on both sides,

log u = sin x · log x

Differentiating both sides w.r.t. x,

(1/u) · du/dx = cos x · log x + sin x · (1/x)

⟹ du/dx = u · [cos x · log x + (sin x)/x]

∴ du/dx = x^(sin x) · [cos x · log x + (sin x)/x] ...(i)

Part (ii): Finding dv/dx where v = (3x² + 2)/(x² + x + 1), and hence df/dx

Using the quotient rule,

dv/dx = [(6x)(x² + x + 1) − (3x² + 2)(2x + 1)] / (x² + x + 1)²

Expanding the numerator:

Numerator = 6x³ + 6x² + 6x − (6x³ + 3x² + 4x + 2)

= 6x³ + 6x² + 6x − 6x³ − 3x² − 4x − 2

= 3x² + 2x − 2

∴ dv/dx = (3x² + 2x − 2) / (x² + x + 1)² ...(ii)

Since f(x) = u + v, we have df/dx = du/dx + dv/dx.

From (i) and (ii),

∴ df/dx = x^(sin x) · [cos x · log x + (sin x)/x] + (3x² + 2x − 2)/(x² + x + 1)²
Q23Case-based4 marks

A civil engineer is designing a road whose elevation profile is modelled by f(x) = x² sin(1/x) for x ≠ 0, and f(0) = 0. Continuity ensures no sudden jumps in elevation; differentiability ensures no sharp corners in the gradient.

A civil engineer is designing a road that follows a curved path. The elevation profile of the road is modelled by the function:

f(x) = { x² sin(1/x), x ≠ 0
{ 0, x = 0

(i) Check whether f(x) is continuous at x = 0. [1 mark]
(ii) Find f'(x) for x ≠ 0. [1 mark]
(iii) Check whether f(x) is differentiable at x = 0. Hence comment on the smoothness of the road profile at x = 0. [2 marks]

Show answer
(i) Continuity at x = 0:

For continuity at x = 0, we check whether lim(x→0) f(x) = f(0).

Now, f(0) = 0.

For x ≠ 0:
|f(x) − 0| = |x² sin(1/x)| = x² |sin(1/x)| ≤ x², since |sin(1/x)| ≤ 1.

∵ x² → 0 as x → 0,

∴ lim(x→0) f(x) = 0 = f(0).

∴ f(x) is continuous at x = 0. [1]

(ii) Derivative for x ≠ 0:

Differentiating f(x) = x² sin(1/x) w.r.t. x,

f'(x) = 2x · sin(1/x) + x² · cos(1/x) · (−1/x²)

∴ f'(x) = 2x sin(1/x) − cos(1/x), x ≠ 0. [1]

(iii) Differentiability at x = 0:

By definition, f'(0) = lim(h→0) [f(0 + h) − f(0)] / h

= lim(h→0) [h² sin(1/h) − 0] / h

= lim(h→0) h sin(1/h).

Now, |h sin(1/h)| ≤ |h| → 0 as h → 0. [1]

∴ lim(h→0) h sin(1/h) = 0, i.e., f'(0) = 0.

∵ the limit exists and is finite, f(x) is differentiable at x = 0.

∴ The road profile is smooth (no sharp corner) at x = 0, meaning the gradient changes gradually through that point. [1]
Q24Case-based4 marks

A structural engineer is modelling the deflection of a composite beam using a piecewise function f(x), where x is the horizontal distance (in metres) from one end of the beam. The function must be both continuous and differentiable at the junction point x = 2 to ensure the deflection model is physically realistic and smooth.

A structural engineer is modelling the deflection of a composite beam. The deflection function is given by:

f(x) = { x² + ax + b, if x ≤ 2
{ 3x + 1, if x > 2

where x represents the horizontal distance (in metres) from one end of the beam.

(i) For the deflection to be physically realistic, f must be continuous at x = 2. Find the relationship between a and b. [1 mark]

(ii) The engineer also requires the deflection curve to be smooth (i.e., no sharp bends), which mathematically means f must be differentiable at x = 2. Use this condition to find the value of a. [1 mark]

(iii) Hence find the value of b, and determine the value of f'(1), interpreting what it represents in the context of the problem. [2 marks]

Show answer
Solution:

(i) For continuity at x = 2, we require:

lim x→2⁻ f(x) = lim x→2⁺ f(x) = f(2)

L.H.L. = lim x→2⁻ (x² + ax + b) = 4 + 2a + b

R.H.L. = lim x→2⁺ (3x + 1) = 7

Also, f(2) = 4 + 2a + b

For continuity: 4 + 2a + b = 7

∴ 2a + b = 3 ...(i)

(ii) For differentiability at x = 2, we require L.H.D. = R.H.D.

L.H.D. = lim h→0 [f(2 − h) − f(2)] / (−h)
= lim h→0 [(2 − h)² + a(2 − h) + b − (4 + 2a + b)] / (−h)
= lim h→0 [4 − 4h + h² + 2a − ah + b − 4 − 2a − b] / (−h)
= lim h→0 [−4h + h² − ah] / (−h)
= lim h→0 (4 − h + a)
= 4 + a

R.H.D. = lim h→0 [f(2 + h) − f(2)] / h
= lim h→0 [3(2 + h) + 1 − 7] / h
= lim h→0 [6 + 3h + 1 − 7] / h
= lim h→0 (3h / h)
= 3

For differentiability: L.H.D. = R.H.D.

4 + a = 3

∴ a = −1

(iii) Substituting a = −1 in equation (i):

2(−1) + b = 3
⟹ b = 5

For x ≤ 2, f(x) = x² − x + 5

Differentiating both sides w.r.t. x,

f'(x) = 2x − 1

∴ f'(1) = 2(1) − 1 = 1

Interpretation: f'(1) = 1 means that at a horizontal distance of 1 metre from the end of the beam, the deflection curve has a slope (rate of change of deflection) of 1 metre per metre, indicating the beam is deflecting at a moderate and increasing rate at that point.
Q25Case-based4 marks

A civil engineer models the elevation profile of a road transition using a piecewise function defined as above. Continuity and differentiability at x = 2 ensure a smooth, safe road transition.

A civil engineer models the elevation profile of a road transition between two sections using the function

f(x) = { kx² + 3x ; 0 < x ≤ 2
{ mx + n ; 2 < x < 5

where x is the horizontal distance (in metres) and f(x) is the elevation (in metres). The road profile must be both continuous and differentiable at the joining point x = 2 to ensure a smooth, safe transition (no sudden change in slope or elevation).

(i) Using the continuity condition at x = 2, write the equation relating k, m, and n. [1]
(ii) Using the differentiability condition at x = 2, find a relationship between k and m. [1]
(iii) If it is further given that the elevation at x = 1 is f(1) = 5 metres, find the values of k, m, and n. [2]

Show answer
(i) For continuity at x = 2, we check LHL = RHL.

LHL = lim x→2⁻ f(x) = k(2)² + 3(2) = 4k + 6

RHL = lim x→2⁺ f(x) = m(2) + n = 2m + n

For continuity: LHL = RHL

∴ 4k + 6 = 2m + n ...(i)

(ii) For differentiability at x = 2, we compute LHD and RHD.

LHD = lim x→2⁻ f'(x) = d/dx(kx² + 3x) at x = 2 = 2kx + 3 |_(x=2) = 4k + 3

RHD = lim x→2⁺ f'(x) = d/dx(mx + n) at x = 2 = m

For differentiability: LHD = RHD

∴ 4k + 3 = m ...(ii)

(iii) Given: f(1) = 5.

Since 0 < 1 ≤ 2, we use f(x) = kx² + 3x.

∴ k(1)² + 3(1) = 5

⟹ k + 3 = 5

∴ k = 2

Substituting k = 2 in equation (ii):

4(2) + 3 = m

∴ m = 11

Substituting k = 2 and m = 11 in equation (i):

4(2) + 6 = 2(11) + n

⟹ 14 = 22 + n

∴ n = −8

∴ k = 2, m = 11, and n = −8.
Q26Case-based4 marks

A civil engineer models the vertical displacement y (in metres) of a suspension cable at horizontal distance x (in metres) from a support tower. The relationship is given by:

y = (x + √(x² + 4))³

The engineer needs to analyse the rate of change of this displacement to ensure structural safety standards are met.

A civil engineer is modelling the vertical displacement y (in metres) of a cable at horizontal distance x (in metres) from the support. The displacement is given by:

y = (x + √(x² + 4))³

(i) Find dy/dx. [1 mark]
(ii) Show that dy/dx = 3y · (dy/dx of the inner function) / (x + √(x² + 4)). [1 mark]
(iii) Hence show that (x² + 4)(dy/dx)² = 9y². [2 marks]

Show answer
Sub-part (i): [1 mark]

Let u = x + √(x² + 4), so that y = u³.

Differentiating both sides w.r.t. x,

dy/dx = 3u² · du/dx

Now, du/dx = 1 + (2x)/(2√(x² + 4)) = 1 + x/√(x² + 4) = (√(x² + 4) + x)/√(x² + 4)

∴ dy/dx = 3(x + √(x² + 4))² · (x + √(x² + 4))/√(x² + 4)

∴ dy/dx = 3(x + √(x² + 4))³ / √(x² + 4)

∴ dy/dx = 3y / √(x² + 4) ... (i)

─────────────────────────────────────────
Sub-part (ii): [1 mark]

From the working in (i), the inner function is u = x + √(x² + 4) and

du/dx = (x + √(x² + 4)) / √(x² + 4)

Since y = u³, we have

dy/dx = 3u² · du/dx = 3(x + √(x² + 4))² · (x + √(x² + 4))/√(x² + 4)

= 3y · [du/dx] / (x + √(x² + 4))

∴ dy/dx = 3y · (du/dx) / (x + √(x² + 4)) Hence shown.

─────────────────────────────────────────
Sub-part (iii): [2 marks]

From result (i):

dy/dx = 3y / √(x² + 4)

Squaring both sides,

(dy/dx)² = 9y² / (x² + 4)

⟹ (x² + 4)(dy/dx)² = 9y²

∴ L.H.S. = R.H.S. Hence proved.
Q27Case-based4 marks

A civil engineer is designing a road profile that must be smooth (no sharp bends) to ensure vehicle safety. The height (in metres) of the road above ground level is modelled by the function

f(x) = { x² + 2x + 3, x ≤ 1
{ ax + b, x > 1

where x is the horizontal distance (in km) from the starting point.

A civil engineer is designing a road profile that must be smooth (no sharp bends) to ensure vehicle safety. The height (in metres) of the road above ground level is modelled by the function

f(x) = { x² + 2x + 3, x ≤ 1
{ ax + b, x > 1

where x is the horizontal distance (in km) from the starting point.

(i) Find the value of f(1) and the left-hand limit of f at x = 1. [1]
(ii) For the road to be continuous at x = 1 (no height gap), determine the condition relating a and b. [1]
(iii) For the road to also be smooth (differentiable at x = 1, i.e., no sharp bend), find the values of a and b, and hence write the complete model f(x). [2]

Show answer
(i)

For x ≤ 1, f(x) = x² + 2x + 3.

∴ f(1) = (1)² + 2(1) + 3 = 1 + 2 + 3 = 6.

Left-hand limit: lim x→1⁻ f(x) = lim x→1 (x² + 2x + 3) = 1 + 2 + 3 = 6.

∴ f(1) = 6 and LHL = 6.

(ii)

For continuity at x = 1, we need LHL = RHL = f(1).

LHL = 6 (from part (i)).

RHL: lim x→1⁺ f(x) = lim x→1 (ax + b) = a(1) + b = a + b.

For continuity: a + b = 6 ...(i)

∴ The required condition is a + b = 6.

(iii)

For the road to be differentiable at x = 1, we need LHD = RHD at x = 1.

Computing LHD:

For x ≤ 1, f(x) = x² + 2x + 3.

Differentiating, f'(x) = 2x + 2.

∴ LHD = lim x→1⁻ f'(x) = 2(1) + 2 = 4.

Computing RHD:

For x > 1, f(x) = ax + b.

Differentiating, f'(x) = a.

∴ RHD = lim x→1⁺ f'(x) = a.

For differentiability: LHD = RHD

⟹ a = 4 ...(ii)

Substituting a = 4 in equation (i):

4 + b = 6 ⟹ b = 2.

∴ a = 4 and b = 2.

The complete road-profile model is:

f(x) = { x² + 2x + 3, x ≤ 1
{ 4x + 2, x > 1

∴ With a = 4 and b = 2, the road is both continuous (no height gap) and smooth (no sharp bend) at x = 1.
Q28Case-based4 marks

A civil engineer models the slope profile of a pedestrian ramp using a piecewise function f(x). For a smooth, jerk-free transition at the entry point x = 0, continuity of f at x = 0 is essential. The engineer then analyses the logarithm of the slope function to study the rate of change of slope.

A civil engineer is designing a pedestrian ramp for a public building. The slope profile of the ramp is modelled by the function

f(x) = { (√(9 + x) − 3) / x, x ≠ 0
k, x = 0 }

For the ramp to have a smooth, jerk-free transition at the flat entry point x = 0, the slope function must be continuous there.

(i) Find the value of k that makes f continuous at x = 0. [1 mark]
(ii) Once k is fixed, verify that f is continuous at x = 0 by checking the three conditions of continuity. [1 mark]
(iii) The engineer also studies the rate of change of slope. If g(x) = log(f(x)) for x ≠ 0 (using the value of k found in part (i) to extend f smoothly), find the derivative of g(x) = log(√(9 + x) − 3) − log x with respect to x at x = 7. [2 marks]

Show answer
(i) Finding k for continuity at x = 0:

For continuity at x = 0, we need lim(x→0) f(x) = f(0) = k.

lim(x→0) f(x) = lim(x→0) [√(9 + x) − 3] / x

Rationalising the numerator:

= lim(x→0) [√(9 + x) − 3] / x × [√(9 + x) + 3] / [√(9 + x) + 3]

= lim(x→0) [(9 + x) − 9] / [x(√(9 + x) + 3)]

= lim(x→0) x / [x(√(9 + x) + 3)]

= lim(x→0) 1 / [√(9 + x) + 3]

= 1 / [√9 + 3] = 1 / (3 + 3) = 1/6

∴ k = 1/6

(ii) Verification of continuity at x = 0 (with k = 1/6):

For continuity at x = 0, we check three conditions:

Condition 1 — f(0) is defined: f(0) = k = 1/6 ✓

Condition 2 — lim(x→0) f(x) exists: From part (i), lim(x→0) f(x) = 1/6 ✓

Condition 3 — lim(x→0) f(x) = f(0): 1/6 = 1/6 ✓

Since all three conditions are satisfied,
∴ f is continuous at x = 0.

(iii) Finding g'(x) at x = 7, where g(x) = log[√(9 + x) − 3] − log x:

Differentiating both sides w.r.t. x,

g'(x) = d/dx [log(√(9 + x) − 3)] − d/dx [log x]

For the first term, applying the chain rule:

d/dx [log(√(9 + x) − 3)] = 1 / [√(9 + x) − 3] × d/dx [√(9 + x) − 3]

= 1 / [√(9 + x) − 3] × 1 / [2√(9 + x)]

For the second term:

d/dx [log x] = 1/x

∴ g'(x) = 1 / [2√(9 + x)(√(9 + x) − 3)] − 1/x

At x = 7:

√(9 + 7) = √16 = 4

g'(7) = 1 / [2 × 4 × (4 − 3)] − 1/7

= 1 / [2 × 4 × 1] − 1/7

= 1/8 − 1/7

= 7/56 − 8/56

= −1/56

∴ g'(7) = −1/56
Q29Case-based4 marks

A civil engineer models the elevation profile of a road using piecewise and exponential functions. Continuity at a join point ensures no abrupt height change, while differentiability of the modified profile determines the smoothness of the road slope.

A civil engineer is designing a road that follows a piecewise-smooth elevation profile. The height (in metres) of the road at a horizontal distance x (in metres) from a reference point is modelled by:

f(x) = { (x² − 4x + 3)/(x − 1), x ≠ 1
{ k, x = 1

For the road to have no abrupt change in elevation (i.e., no sudden jump), the function f(x) must be continuous at x = 1.

(i) Find the value of k so that f(x) is continuous at x = 1. [1 mark]
(ii) Using the value of k found in part (i), state the value of f(1) and verify that lim(x→1) f(x) = f(1). [1 mark]
(iii) After repair work, the engineer modifies the elevation model to:

g(x) = x^(sin x) for x > 0.

Find dg/dx (i.e., g'(x)). [2 marks]

Show answer
(i)

For continuity at x = 1, we require:

lim(x→1) f(x) = f(1) = k

For x ≠ 1:

f(x) = (x² − 4x + 3)/(x − 1)

Factorising the numerator:

x² − 4x + 3 = (x − 1)(x − 3)

∴ f(x) = (x − 1)(x − 3)/(x − 1) = x − 3, for x ≠ 1

∴ lim(x→1) f(x) = 1 − 3 = −2

For continuity, k = −2.

∴ k = −2

(ii)

With k = −2:

f(1) = k = −2

Also, lim(x→1) f(x) = −2 [as shown in part (i)]

∴ lim(x→1) f(x) = f(1) = −2

Hence f(x) is continuous at x = 1. ✓

(iii)

Let g(x) = x<super>sin x</super>, x > 0.

Taking log on both sides,

log g(x) = sin x · log x

Differentiating both sides w.r.t. x,

(1/g(x)) · g'(x) = cos x · log x + sin x · (1/x)

⟹ g'(x) = g(x) · [cos x · log x + (sin x)/x]

∴ g'(x) = x<super>sin x</super> · [cos x · log x + (sin x)/x]
Q30Case-based4 marks

A highway engineer models a road profile using a piecewise function f(x) = ax² + bx + 1 for x ≤ 1, and f(x) = 2x³ − 5 for x > 1. Smoothness of the road requires f to be continuous and differentiable at the join point x = 1.

A highway engineer is designing a smooth transition curve between two road segments. The road profile over a stretch is modelled by the function

f(x) = { ax² + bx + 1, x ≤ 1
{ 2x³ − 5, x > 1

For the road to be 'smooth' (no sudden jerks or breaks), the engineer requires the function f to be both continuous and differentiable at x = 1.

(i) Find the value of 'a' and 'b' so that f(x) is continuous at x = 1. [1 mark]
(ii) Using the values found in (i), verify whether f is differentiable at x = 1. [1 mark]
(iii) The engineer realises that the condition from (i) gives infinitely many pairs (a, b). To make the design unique, she imposes the additional constraint that the slope of the tangent to f at x = 1 equals 6. Find the unique values of a and b, and hence write the complete piecewise function. [2 marks]

Show answer
Solution:

(i) For continuity at x = 1, we check LHL = RHL = f(1).

f(1) = a(1)² + b(1) + 1 = a + b + 1

lim x→1⁺ f(x) = 2(1)³ − 5 = 2 − 5 = −3

For continuity at x = 1:

a + b + 1 = −3

∴ a + b = −4 ...(i)

(ii) For differentiability at x = 1, we compute LHD and RHD.

For x ≤ 1: f(x) = ax² + bx + 1 ⟹ f'(x) = 2ax + b
∴ LHD = f'(1⁻) = 2a(1) + b = 2a + b

For x > 1: f(x) = 2x³ − 5 ⟹ f'(x) = 6x²
∴ RHD = f'(1⁺) = 6(1)² = 6

For differentiability at x = 1, we need LHD = RHD:

2a + b = 6 ...(ii)

Using a + b = −4 from (i), this gives one constraint on (a, b).
Since equation (i) alone does not fix a and b uniquely, f is differentiable at x = 1 only if the pair (a, b) also satisfies 2a + b = 6.
∵ from (i) alone infinitely many pairs satisfy continuity, differentiability requires the additional condition 2a + b = 6.
∴ f is differentiable at x = 1 if and only if condition (ii) is also satisfied.

(iii) The engineer's additional constraint: the slope of the tangent at x = 1 equals 6.

Slope of tangent from the right-hand piece = f'(1⁺) = 6(1)² = 6. ✓ (already 6)

For a unique smooth design, both continuity and the slope condition must hold simultaneously:

From the differentiability condition: 2a + b = 6 ...(ii)
From the continuity condition: a + b = −4 ...(i)

Subtracting (i) from (ii):

(2a + b) − (a + b) = 6 − (−4)

⟹ a = 10

Substituting a = 10 in (i):

10 + b = −4 ⟹ b = −14

∴ a = 10, b = −14

The complete piecewise function is:

f(x) = { 10x² − 14x + 1, x ≤ 1
{ 2x³ − 5, x > 1

∴ The unique values are a = 10 and b = −14.

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Continuity and Differentiability Class 12 Maths Questions