Free Differentiation and Optimisation MCQs with Answers

• A. 12x^3 - 6x^2 + 5
• B. 12x^3 - 2x^2 + 5
• C. 3x^3 - 6x^2 + 5
• D. 12x^4 - 6x^3 + 5

Explanation: Apply the power rule to each term: the derivative of 3x^4 is 12x^3, and the derivative of -2x^3 is -6x^2.

• A. x = 0
• B. x = 1
• C. x = 3
• D. x = 2

Explanation: The first derivative is 3x^2 - 12x + 9, which gives critical points x = 1 and x = 3.

• A. 5
• B. 6
• C. 7
• D. 8

Explanation: Differentiate the function to obtain dy/dx = 2x + 3. Substituting x = 2 gives a tangent slope of 7.

• A. 80 cm²
• B. 90 cm²
• C. 100 cm²
• D. 120 cm²

Explanation: If the sides are x and 20 - x, the area is A = x(20 - x). This quadratic is maximised at x = 10, giving a square with area 10 × 10 = 100…

• A. 8
• B. 10
• C. 12
• D. 16

Explanation: Differentiate to get f'(x) = 2x - 16/x². Setting this equal to zero gives x³ = 8, so x = 2.

• A. 1 cm
• B. 2 cm
• C. 3 cm
• D. 4 cm

Explanation: The volume is V(x) = x(12 - 2x)². Differentiating gives critical values x = 2 and x = 6, but x = 6 produces zero volume, so the maximum…

• A. Zero
• B. One
• C. Three
• D. Two

Explanation: The second derivative is f''(x) = 12x(x - 2), which changes sign at both x = 0 and x = 2. Therefore, the curve has two points of inflection.

• A. 10 units
• B. 20 units
• C. 30 units
• D. 40 units

Explanation: The derivative is P'(q) = -2q + 40. Setting marginal profit equal to zero gives q = 20, and the negative coefficient of q² confirms that…

• A. -4/3
• B. -3/4
• C. 3/4
• D. 4/3

Explanation: Implicit differentiation gives 2x + 2y(dy/dx) = 0, so dy/dx = -x/y. At (3, 4), the slope is -3/4.

• A. -4
• B. -2
• C. 2
• D. 4

Explanation: The critical points in the interval are x = -1 and x = 1, and the endpoints are -2 and 2.

• A. 0
• B. e
• C. 2
• D. 1

Explanation: Using the product rule, f'(x) = e^x(x^2 + 2x + 1). Substituting x = 0 gives f'(0) = 1.

• A. 15(3x - 2)^4
• B. 5(3x - 2)^4
• C. 15(3x - 2)^5
• D. (3x - 2)^4

Explanation: The chain rule multiplies the outer derivative by the derivative of 3x - 2. Therefore, f'(x) = 5(3x - 2)^4 × 3 = 15(3x - 2)^4.

• A. A local maximum is certain
• B. The test gives no definite classification
• C. A local minimum is certain
• D. An inflection point is certain

Explanation: When the second derivative is zero at a stationary point, the second derivative test is inconclusive.

• A. x < 1
• B. x < 0
• C. x > 1
• D. x > 0

Explanation: The second derivative is f''(x) = 6x - 6. It is positive when x > 1, so the graph is concave upward on that interval.

• A. h = r
• B. h = 4r
• C. h = 3r
• D. h = 2r

Explanation: For a closed cylinder, minimising surface area subject to fixed volume gives the condition h = 2r.

• A. 50 units
• B. 100 units
• C. 150 units
• D. 200 units

Explanation: Revenue is R(q) = pq = 100q - 0.5q^2. Since R'(q) = 100 - q, the stationary point occurs at q = 100, and R''(q) = -1 confirms a maximum.

• A. 4.83
• B. 4.90
• C. 4.93
• D. 5.07

Explanation: For f(x) = sqrt(x), use f(25) = 5 and f'(25) = 1/10. Thus sqrt(24.3) is approximately 5 + (1/10)(-0.7) = 4.93.

• A. 3/13
• B. -3/13
• C. 4/13
• D. -4/13

Explanation: Differentiation gives 2xy + x^2(dy/dx) + 3y^2(dy/dx) = 0. At (1, 2), this becomes 4 + 13(dy/dx) = 0, so dy/dx = -4/13.

• A. 8 units
• B. 10 units
• C. 12 units
• D. 14 units

Explanation: Average cost is C(q)/q = q - 12 + 100/q. Its derivative is 1 - 100/q^2, which is zero at q = 10, giving the minimum average cost.

• A. 5/2
• B. 5
• C. 25/2
• D. 25

Explanation: Since (x - y)^2 is nonnegative, x^2 + y^2 is at least 2xy. Therefore, 25 is at least 2xy, with equality when x = y, so the maximum is 25/2.