RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Maths · 11 questions · 26 marks
Derivatives measure instantaneous rates of change, and this chapter turns that single idea into a toolkit: finding how fast one quantity changes as another does, deciding where a function climbs or falls, drawing tangents and normals, and pinning down maximum or minimum values. The same first-derivative-equals-zero condition that locates a stationary point also solves practical optimisation problems, from minimising cost to maximising the volume of a box cut from a sheet of metal.
The rate of change of the area of a circle with respect to its radius r, at r = 5 cm, is:
Answer
The rate of change is 10π cm²/cm. Area A = πr², so dA/dr = 2πr. At r = 5, dA/dr = 2π(5) = 10π.
The function f(x) = x² - 4x + 6 is increasing on the interval:
Answer
f is increasing on (2,∞). f'(x) = 2x - 4. f is increasing where f'(x) > 0, i.e. 2x - 4 > 0, giving x > 2. So f is increasing on the interval (2,∞).
The slope of the tangent to the curve y = x³ - 3x + 2 at x = 2 is:
Answer
The slope is 9. dy/dx = 3x² - 3. At x = 2: dy/dx = 3(4) - 3 = 12 - 3 = 9.
The maximum value of f(x) = -x² + 4x + 1 is:
Answer
The maximum value is 5. f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. f''(x) = -2 < 0, confirming x=2 gives a maximum. f(2) = -4 + 8 + 1 = 5.
Assertion (A): The function f(x) = x³ has a point of inflection at x = 0, not a local extremum. Reason (R): f'(0) = 0 and f''(0) = 0, so the second derivative test fails, and f'(x) = 3x² ≥ 0 does not change sign around x = 0.
Answer
Both A and R are true and R is the correct explanation of A. f'(x) = 3x² gives f'(0) = 0, and f''(x) = 6x gives f''(0) = 0, so R correctly identifies that the second derivative test is inconclusive here. Since f'(x) = 3x² is non-negative on both sides of x=0 (it does not switch from positive to negative or vice versa), the first derivative test shows x=0 is neither a maximum nor a minimum — it is a point of inflection, confirming A and matching R's reasoning.
Find the equation of the tangent to the curve y = x² + 3x - 2 at the point (1,2).
Answer
First check the point lies on the curve: at x=1, y = 1 + 3 - 2 = 2. Confirmed. Find the slope: dy/dx = 2x + 3, so at x=1, slope = 2(1) + 3 = 5. Using point-slope form y - y0 = m(x - x0): y - 2 = 5(x - 1), which simplifies to y = 5x - 3.
Using differentials, find the approximate value of √25.3.
Answer
Let f(x) = √x, and take x = 25 (a nearby perfect square) with Δx = 0.3. f'(x) = 1/(2√x), so f'(25) = 1/(2×5) = 1/10 = 0.1. Δy ≈ f'(x)Δx = 0.1 × 0.3 = 0.03. So √25.3 ≈ f(25) + Δy = 5 + 0.03 = 5.03.
Find the intervals in which f(x) = 2x³ - 15x² + 36x + 1 is increasing and the intervals in which it is decreasing.
Answer
f'(x) = 6x² - 30x + 36 = 6(x² - 5x + 6) = 6(x-2)(x-3). The critical points are x=2 and x=3, dividing the real line into three intervals. For x < 2: both (x-2) and (x-3) are negative, so their product is positive — f'(x) > 0, increasing. For 2 < x < 3: (x-2) is positive and (x-3) is negative, so f'(x) < 0, decreasing. For x > 3: both factors are positive, so f'(x) > 0, increasing. So f is increasing on (-∞,2)∪(3,∞) and decreasing on (2,3).
A ladder 13 m long is leaning against a vertical wall. The bottom of the ladder is being pulled away from the wall at 2 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 5 m from the wall?
Answer
Let x be the distance of the bottom of the ladder from the wall, and y be the height of the top on the wall, so x² + y² = 13² = 169 at all times. When x = 5, y = √(169-25) = √144 = 12. Differentiate x² + y² = 169 with respect to time t: 2x(dx/dt) + 2y(dy/dt) = 0. Solving for dy/dt: dy/dt = -(x/y)(dx/dt). Substitute x=5, y=12, dx/dt=2: dy/dt = -(5/12)(2) = -10/12 = -5/6 m/s. The negative sign shows y is decreasing, so the top of the ladder is sliding down the wall at 5/6 m/s.
Find two positive numbers whose sum is 24 and whose product is maximum. Also find the maximum product.
Answer
Let the two numbers be x and 24 - x, with 0 < x < 24. Their product is P(x) = x(24 - x) = 24x - x². P'(x) = 24 - 2x. Setting P'(x) = 0 gives x = 12. P''(x) = -2 < 0, confirming x=12 gives a maximum. The two numbers are x = 12 and 24 - x = 12. Maximum product = 12 × 12 = 144.
Read the following and answer the questions that follow: An open rectangular box is made from a square metal sheet of side 24 cm, by cutting equal squares of side x cm from each of the four corners and folding up the sides. (a) Write the volume V(x) of the resulting box in terms of x. (b) Find dV/dx. (c) Find the value of x that gives the maximum volume. (d) Find the maximum volume.
Answer
(a) The base becomes a square of side (24-2x) and the height is x, so V(x) = x(24-2x)² = 4x³ - 96x² + 576x. (b) dV/dx = 12x² - 192x + 576 = 12(x² - 16x + 48) = 12(x-4)(x-12). (c) Setting dV/dx = 0 gives x = 4 or x = 12. Since x=12 would make the base side 24-24=0 (no box), the feasible critical point is x=4. Checking: d²V/dx² = 24x - 192, which at x=4 gives 96-192 = -96 < 0, confirming a maximum. (d) V(4) = 4(24-8)² = 4(16)² = 4(256) = 1024 cm³.
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