CBSE Class 12 Mathematics Application of Derivatives Case Studies

Answer: (b) 4 − x

Explanation:
1. Rate of growth = \( \frac{dy}{dx} \)
2. \( \frac{dy}{dx} = 4 - \frac{1}{2}(2x) \) = 4 − x

Teacher's Note:
1. Rate of change of a quantity means its derivative, not the quantity itself, so option (a) is just y.
2. x − 4 has the wrong sign: it would make the plant shrink in the first few days.

Answer: (a) 4

Explanation:
1. For maximum height, \( \frac{dy}{dx} \) = 4 − x = 0, so x = 4.
2. \( \frac{d^2y}{dx^2} \) = −1 < 0, so y is maximum at x = 4.
3. The plant reaches its maximum height after 4 days.

Teacher's Note:
1. Always confirm a maximum with the second derivative (or a sign change of dy/dx).
2. For y = ax² + bx + c with a < 0, the maximum is at x = −b/(2a) = −4/(2 × (−½)) = 4.

Answer: (c) 8 cm

Explanation:
1. Maximum occurs at x = 4.
2. y(4) = 4(4) − \( \frac{1}{2} \)(4)² = 16 − 8
3. Maximum height = 8 cm

Teacher's Note:
1. Find where the maximum occurs first, then substitute that x into y itself, not into dy/dx.
2. Substituting into dy/dx would give 0, which is the slope at the top, not the height.

Answer: (b) 6 cm

Explanation:
1. Put x = 2 in y = 4x − \( \frac{1}{2} \)x².
2. y = 8 − \( \frac{1}{2} \)(4) = 8 − 2
3. Height after 2 days = 6 cm

Teacher's Note:
1. This is a plain substitution; no derivative is needed.
2. 6 cm is less than the maximum 8 cm, which fits because 2 days is before the peak at day 4.

Answer: (d) 1

Explanation:
1. 4x − \( \frac{1}{2} \)x² = \( \frac{7}{2} \); multiply by 2: 8x − x² = 7.
2. x² − 8x + 7 = 0, i.e. (x − 1)(x − 7) = 0, so x = 1 or x = 7.
3. Among the options only x = 1 appears, so the answer is 1 day.

Teacher's Note:
1. A downward parabola reaches each height below its maximum twice, once while rising and once while falling.
2. Quick check: y(1) = 4 − 0.5 = 3.5 = 7/2 cm ✓

Answer: (b) 12.5

Explanation:
1. P′(x) = −10x + 125
2. P′(x) = 0 gives x = 12.5; P″(x) = −10 < 0, so this is a maximum.
3. Production for maximum profit = 12.5

Teacher's Note:
1. Production cannot be negative, so options (c) and (d) are ruled out at sight.
2. 37500 is P(0), the profit with zero production, not a production level.

Answer: (b) Rs 38281.25

Explanation:
1. Maximum occurs at x = 12.5.
2. P(12.5) = −5(156.25) + 125(12.5) + 37500 = −781.25 + 1562.5 + 37500
3. Maximum profit = Rs 38281.25

Teacher's Note:
1. Square first: 12.5² = 156.25, then multiply by −5; doing it in the wrong order is a common slip.
2. The maximum is only a little above P(0) = 37500, which rules out the much larger Rs 38,28,125.

Answer: (d) (0, 12.5)

Explanation:
1. P is strictly increasing where P′(x) = −10x + 125 > 0, i.e. x < 12.5.
2. Production x cannot be negative, so we take x > 0.
3. Profit is strictly increasing on (0, 12.5).

Teacher's Note:
1. On (12.5, ∞) the derivative is negative, so profit decreases after the maximum point.
2. In real-life problems combine the calculus condition with the natural restriction on the variable.

Answer: (b) 37,730

Explanation:
1. P(2) = −5(2)² + 125(2) + 37500
2. P(2) = −20 + 250 + 37500
3. P(2) = 37730

Teacher's Note:
1. Keep the sign of −5x²: using +20 instead of −20 gives the trap answer 37,770.
2. 37500 is the profit with no production, so it cannot be P(2).

Answer: (a) 15

Explanation:
1. −5x² + 125x + 37500 = 38250 gives −5x² + 125x − 750 = 0.
2. Divide by −5: x² − 25x + 150 = 0, i.e. (x − 10)(x − 15) = 0.
3. x = 10 or x = 15; of these only 15 is among the options.

Teacher's Note:
1. Profit values below the maximum are reached at two production levels placed symmetrically about 12.5.
2. Check: P(15) = −1125 + 1875 + 37500 = 38250 ✓

Answer: (b) \( \pm\frac{1}{2} \)

Explanation:
1. f(x) = 12x⁴ − 6x², so f′(x) = 48x³ − 12x = 12x(4x² − 1).
2. f′(x) = 0 gives x = 0 or 4x² = 1, i.e. x = \( \pm\frac{1}{2} \).
3. Among the options, the abscissa of a critical point is \( \pm\frac{1}{2} \).

Teacher's Note:
1. Factor out the common 12x before solving; it gives x = 0 as a critical point too.
2. Critical points are where f′(x) = 0 (or does not exist), not where f(x) = 0.

Answer: (c) 3588

Explanation:
1. f′(x) = 48x³ − 12x
2. f″(x) = 144x² − 12
3. f″(5) = 144(25) − 12 = 3600 − 12 = 3588

Teacher's Note:
1. Multiply out f(x) = 12x⁴ − 6x² first; it makes repeated differentiation easier.
2. Forgetting the outer factor 6 gives 598, which is one sixth of the correct value.

Answer: (b) (−1/2, 0) ∪ (1/2, ∞)

Explanation:
1. f′(x) = 12x(2x − 1)(2x + 1); critical points −1/2, 0, 1/2 split the line into four intervals.
2. Signs of f′: (−∞, −1/2): −; (−1/2, 0): +; (0, 1/2): −; (1/2, ∞): +
3. f is increasing on (−1/2, 0) ∪ (1/2, ∞).

Teacher's Note:
1. For a product of linear factors, signs alternate across the critical points starting with + on the far right.
2. Test one value per interval, e.g. f′(1) = 36 > 0 and f′(−1/4) = 2.25 > 0.