Application of Derivatives — Chapter Test | Class 12 IMO

Q1

The absolute maximum value of f(x) = x³ - 3x on [-2, 2] is:

f'(x) = 3x² - 3 = 0 → x = ±1. Evaluate f(x) at boundaries and critical points: f(-2) = -2, f(-1) = 2, f(1) = -2, f(2) = 2. Absolute maximum is 2.

Q2

The maximum slope of the curve y = -x³ + 3x² + 9x - 27 is:

Slope m = dy/dx = -3x² + 6x + 9. To maximize m, dm/dx = -6x + 6 = 0 → x = 1. Max slope = -3(1)² + 6(1) + 9 = 12.

Q3

The function f(x) = 2x³ - 3x² - 36x + 7 is strictly decreasing in the interval:

f'(x) = 6x² - 6x - 36 = 6(x² - x - 6) = 6(x - 3)(x + 2). For strictly decreasing, f'(x) < 0, which happens in the interval (-2, 3).

Q4

Find the interval in which the function f(x) = x³ - 12x² + 36x + 17 is strictly increasing.

f'(x) = 3x² - 24x + 36 = 3(x - 2)(x - 6). For f'(x) > 0, x < 2 or x > 6. Interval is (-∞, 2) ∪ (6, ∞).

Q5

A manufacturer sells a product at ₹x per unit. The demand function is x = 100 − 2p. The marginal revenue when p = 20 is:

p = (100 − x)/2. Revenue R = p x = x(100 − x)/2 = 50x − x²/2. MR = dR/dx = 50 − x. When p = 20, x = 100 − 40 = 60. MR = 50 − 60 = −10? That's negative. We check: if x = 100 − 2p, then p = 50 − x/2. R = 50x − x²/2. MR = 50 − x. At p = 20, x = 100 − 40 = 60, MR = −10. Not in options. Maybe marginal revenue with respect to p? No. Perhaps the question means p = 20 is the selling price, and demand is x = 100 − 2p. Then x = 60. R = 20 × 60 = 1200. Marginal revenue usually means derivative of R with respect to quantity. dR/dx = 50 − x = −10. To get a positive option, we change to x = 200 − 2p. Then p = 100 − x/2. R = 100x − x²/2. MR = 100 − x. At p = 20, x = 200 − 40 = 160, MR = 100 − 160 = −60. Still negative. Maybe it's a linear demand and marginal revenue is always half? Actually, if p = a − bx, R = ax − bx², MR = a − 2bx. With demand x = 100 − 2p → p = 50 − 0.5x. R = 50x − 0.5x², MR = 50 − x. At p = 20, x = 60, MR = −10. Not matching. We'll adjust options to include −10 or change numbers. We set x = 100 − p. Then p = 100 − x. R = 100x − x². MR = 100 − 2x. At p = 20, x = 80, MR = 100 − 160 = −60. No. We use p = 50, then x = 0, MR = 100. We'll just make the correct answer something else and ensure it's in options. We do: demand x = 200 − 10p, then p = 20 − x/10. R = 20x − x²/10. MR = 20 − x/5. At p = 10, x = 100, MR = 20 − 20 = 0. Not good. We'll scrap and write a simpler one: R(x) = 5x − 0.01x², find MR at x = 100. MR = 5 − 0.02x = 5 − 2 = 3. Options: ₹3, etc.

Q6

At what point does the tangent to the curve y = eˣ intersect the y-axis if the tangent is drawn at the point (0, 1)?

y = eˣ implies dy/dx = eˣ. At (0, 1), m = e⁰ = 1. Equation of tangent is y - 1 = 1(x - 0), which is y = x + 1. The y-intercept is at x = 0, giving y = 1. Point is (0, 1).

Q7

The interval on which f(x) = x/log x is increasing is:

f'(x) = (log x − 1)/(log x)². f'(x) > 0 when log x > 1 → x > e. Domain is x > 0, x ≠ 1. Increasing on (e, ∞).

Q8

For the curve y = x² - 5x + 6, the tangent at point (2, 0) is parallel to the line:

dy/dx = 2x - 5. At x = 2, slope m = 4 - 5 = -1. The line x + y = 0 can be written as y = -x, which has slope -1. Thus, they are parallel.

Q9

The point on the curve y² = 4x at which the tangent is perpendicular to the line 2x + y = 3 is:

Line slope = −2. Tangent slope = 1/2 (perpendicular). dy/dx for y² = 4x is 2y dy/dx = 4 → dy/dx = 2/y. Set 2/y = 1/2 → y = 4. Then 4² = 4x → 16 = 4x → x = 4. Point (4, 4) has tangent slope 2/4 = 1/2. Correct. But option (4, 4) is there. So point (4, 4). Tangent slope = 2/4 = 1/2, which is perpendicular to slope −2. That's correct. But option (1/4, −1) has y = −1, slope = 2/(−1) = −2, not 1/2. So (4, 4) is correct. We'll set correct index to 1.

Q10

A particle moves along a straight line such that its displacement s at any time t is given by s = t³ - 6t² + 3t + 4. Find the velocity when the acceleration is zero.

Velocity v = ds/dt = 3t² - 12t + 3. Acceleration a = dv/dt = 6t - 12. If a = 0, t = 2. At t = 2, v = 3(2)² - 12(2) + 3 = 12 - 24 + 3 = -9.

Q11

The normal to the curve y = x² − 4x + 3 at x = 2 passes through the point:

dy/dx = 2x − 4 = 0 at x = 2, so the tangent is horizontal and the normal is the vertical line x = 2, which passes through (2, 4).

Q12

Find two positive numbers whose sum is 15 and the sum of whose squares is minimum. The numbers are:

Let numbers be x and 15-x. Sum of squares S = x² + (15-x)². S' = 2x - 2(15-x) = 4x - 30. S' = 0 gives x = 7.5. The numbers are 7.5 and 7.5.

Q13

The function f(x) = xˣ is strictly increasing for:

Let y = xˣ. Taking log, log y = x log x. Differentiating, (1/y)(dy/dx) = 1 + log x. So f'(x) = xˣ(1 + log x). f'(x) > 0 implies log x > -1, so x > 1/e.

Q14

The function f(x) = x³ − 6x² + 12x − 8 is increasing on the interval:

f'(x) = 3x² − 12x + 12 = 3(x² − 4x + 4) = 3(x − 2)² ≥ 0 for all x. So f is increasing on R.

Q15

The cost function of a firm is C(x) = 3x² + 5x + 27 (in ₹). The marginal cost when x = 3 is:

Marginal cost = dC/dx = 6x + 5. At x = 3, MC = 18 + 5 = 23.