Continuity and Differentiability — Chapter Test | Class 12 IMO

Q1

The revenue received by E-learning Hub Yudgam from selling x digital mock test series is R(x) = 3x² + 36x + 5 (in rupees). The marginal revenue when x = 5 is:

Marginal Revenue R'(x) = 6x + 36. Substituting x = 5, we get 6(5) + 36 = 30 + 36 = 66 rupees.

Q2

Consider f(x) = |x − 1| + |x − 2|. The number of points where f(x) is not differentiable is:

The sum of absolute value functions has sharp turns at the roots of the expressions inside the modulus. Here, the roots are x = 1 and x = 2. So, it is not differentiable at 2 points.

Q3

Let f(x) = x³ − x² + x + 1 and g(x) = max{f(t): 0 ≤ t ≤ x} for 0 ≤ x ≤ 1, then g(x) is not differentiable at x = 1/3. True or False: g is discontinuous at x = 1/3.

g(x) is always continuous as the maximum function of a continuous function. It may not be differentiable where f stops being the maximum, but it remains continuous.

Q4

If y = cos⁻¹(√(1 − x²)/√(1 + x²)), then dy/dx is positive for:

Let x = tan θ, cos⁻¹(cos θ/√(1+tan²θ)) = cos⁻¹(cos θ/sec θ) = cos⁻¹(cos² θ) = 2θ? Actually cos⁻¹(cos² θ) = 2θ for θ in [0,π/2] i.e. x≥0. dy/dx = 2/(1+x²) positive. For x<0, sign changes. Derivative positive for x<0? Derive: y = cos⁻¹(√(1−x²)/√(1+x²)) = tan⁻¹|x|? Actually derivative = 1/(1+x²) * sign(x). We do properly: y = cos⁻¹(√((1−x²)/(1+x²))). Let x = tan θ, θ∈(−π/2, π/2). So inside sqrt is cos 2θ. Thus y = cos⁻¹(√(cos 2θ)). For cos 2θ ≥ 0, √(cos 2θ) = √(cos 2θ). Then y = sin⁻¹(√(1−cos 2θ)) = sin⁻¹(√(2 sin² θ)) = sin⁻¹(√2|sin θ|). Complicated. Simpler: derivative of given y is 1/(1+x²) for x>0 and −1/(1+x²) for x<0. So positive for x<0.

Q5

Find the derivative of cos⁻¹(sin x) with respect to x, for x in (0, π/2).

We can write sin x as cos(π/2 − x). So, cos⁻¹(cos(π/2 − x)) = π/2 − x. The derivative of π/2 − x is −1.

Q6

A shopkeeper in Karol Bagh models his daily profit P(x) = 100x − x² for selling x silk sarees. To find the rate of change of profit when 20 sarees are sold, he calculates P'(20). What is the value?

Marginal profit P'(x) = 100 − 2x. For x = 20, P'(20) = 100 − 40 = 60.

Q7

If y = sin x + cos x, which of the following differential equations is satisfied?

dy/dx = cos x − sin x. d²y/dx² = −sin x − cos x = −(sin x + cos x) = −y. Therefore, d²y/dx² + y = 0.

Q8

Derivative of tan⁻¹((√x + √a) / (1 − √(ax))) with respect to x is:

Using the formula tan⁻¹((A + B)/(1 − AB)) = tan⁻¹A + tan⁻¹B. Let A = √x, B = √a. y = tan⁻¹(√x) + tan⁻¹(√a). dy/dx = (1 / (1 + (√x)²)) × (1 / 2√x) = 1 / (2√x(1 + x)).

Q9

If y = (log x)ˣ + x^(log x), then at x = e, dy/dx is:

From earlier calculation, u′ = (log x)ˣ [log(log x) + 1/log x], v′ = x^(log x) [2 log x / x]. At x=e, log x=1, u=1, u′ = 1[0+1]=1. v=e, v′ = e[2/e]=2. Total = 3.

Q10

If y = e^(a sin⁻¹ x), then (1 − x²)y₂ − xy₁ equals:

y₁ = e^(a sin⁻¹ x) × a/√(1−x²) = a y/√(1−x²). So √(1−x²) y₁ = a y. Squaring: (1−x²)y₁² = a² y². Differentiating: −2x y₁² + 2(1−x²)y₁ y₂ = 2a² y y₁. Divide by 2y₁: −x y₁ + (1−x²)y₂ = a² y.

Q11

A factory in New Delhi produces x units of smartwatches. The cost function is C(x) = 0.005x³ − 0.02x² + 30x + 5000 in rupees. The marginal cost when 3 units are produced is:

Marginal Cost is C'(x) = 0.015x² − 0.04x + 30. At x = 3, C'(3) = 0.015(9) − 0.04(3) + 30 = 0.135 − 0.120 + 30 = 30.015.

Q12

Find dy/dx for the parametric equations x = a cos θ, y = a sin θ.

dx/dθ = −a sin θ. dy/dθ = a cos θ. dy/dx = (dy/dθ) / (dx/dθ) = (a cos θ) / (−a sin θ) = −cot θ.

Q13

How many real solutions does the inverse trigonometric equation sin⁻¹x = πx/2 have?

Let f(x) = sin⁻¹x and g(x) = πx/2. Plotting both on their valid domain [−1, 1], they intersect at the origin (x = 0) and at the boundaries x = 1 (where both equal π/2) and x = −1 (where both equal −π/2). Thus, there are exactly three real solutions.

Q14

If y = logₑ((x² + 1)/√(x² − 1)), then dy/dx is:

y = log(x²+1) − ½ log(x²−1). y′ = 2x/(x²+1) − (1/2)·2x/(x²−1) = 2x/(x²+1) − x/(x²−1) = (2x(x²−1) − x(x²+1))/((x²+1)(x²−1)) = (x³ − 3x)/(x⁴−1).

Q15

If y = (tan⁻¹ x)², then (1 + x²)² y₂ + 2x(1 + x²)y₁ is equal to:

y₁ = 2 tan⁻¹ x / (1+x²). So (1+x²)y₁ = 2 tan⁻¹ x. Differentiating: 2x y₁ + (1+x²)y₂ = 2/(1+x²). Multiply by (1+x²): 2x(1+x²)y₁ + (1+x²)² y₂ = 2.