The relation between the height of the plant ('y' in cm) with respect to its exposure to the sunlight is governed by the following equation y=4x-12x², where 'x' is the number of days exposed to the sunlight, for x≤3. (i) Find the rate of growth of the plant with respect to the number of days exposed to the sunlight. (ii) Does the rate of growth of the plant increase or decrease in the first three days? What will be the height of the plant after 2 days?

MathsCBSE Class 12 · Sample paper, session 2023-24 · 4 marks · Case-based

CBSE Class 12 Mathematics sample paper 2023-24: question 38

From the official CBSE sample question paper for 2023-24, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedApplication of Derivatives

The relation between the height of the plant ('y' in cm) with respect to its exposure to the sunlight is governed by the following equation $y = 4 x - \frac{1}{2} x^{2}$ , where 'x' is the number of days exposed to the sunlight, for $x \leq 3$ .

(i)
Find the rate of growth of the plant with respect to the number of days exposed to the sunlight.
2 marks
(ii)
Does the rate of growth of the plant increase or decrease in the first three days? What will be the height of the plant after 2 days?
2 marks

Figure in the paper: Photo of a seedling plant bending towards the bright sun. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

Rate of growth means the derivative of height with respect to days.

Hint 2

To see whether the rate increases or decreases, differentiate the rate once more and check its sign.

Hint 3

Height after 2 days: substitute $x = 2$ in the given equation.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Rate of growth $= \frac{d y}{d x} = \frac{d}{d x} (4 x - \frac{1}{2} x^{2}) = 4 - x$ cm per day.
2 marks
(ii) Let $g (x) = \frac{d y}{d x} = 4 - x$ . Then $g^{'} (x) = - 1 < 0$ , so $g (x)$ is decreasing: the rate of growth decreases in the first three days.
1 mark
(ii) Height after 2 days: $y = 4 (2) - \frac{1}{2} (2)^{2} = 8 - 2 = 6$ cm.
1 mark

Answer: (i) $\frac{d y}{d x} = 4 - x$ ; (ii) rate of growth decreases; height after 2 days = 6 cm

Where the marks are

(i) Differentiates to get $d y / d x = 4 - x$ (2 marks)
(ii) Shows $\frac{d}{d x} (d y / d x) = - 1 < 0$ , so rate decreases (1 mark)
(ii) Height at $x = 2$ is 6 cm (1 mark)

Revise the chapter

Chapter: Application of Derivatives (notes, formulas, where marks are lost and every past question)
Topic: Rate of Change of Quantities

Practise this chapterBoard-style questions from Application of Derivatives, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24: every question, by section.

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CBSE Class 12 Mathematics sample paper 2023-24: question 38

Hints

Step-by-step solution

Revise the chapter

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