CBSE Class 12 Mathematics Applications of Derivatives Q #2099
COMPETENCY BASED
APPLY
4 Marks 2026-27 CBSE Sample QP SUBJECTIVE
Water is essential for life, and rivers are vital water sources across the world. The Yamuna Action Plan (YAP), one of India's largest river restoration projects, was launched to clean the Yamuna, but its effectiveness (WQI) the Water Quality Index must be measured through cost-benefit, economic and environmental analysis. Therefore, comprehensive pollutant treatment, prevention of further contamination, and community awareness are crucial for sustainable river management.
Implementing the same concept to the river Yamuna, if an initial base line WQI is 35, a quadratic expression is formed to measure the impact of investment of ₹$x$ crore on the WQI i.e., $W(x)$ as $W(x) = 35 + 1.6x - 0.02x^2$, where $x$ is in Rupees in crores Use the given $W(x)$ to answer the following questions:
(i) Find the value of $x$ at the stationary point for the function $W(x)$. [1 mark]
(ii) Find the maximum value of the WQI using the second derivative test? [1 mark]
(iii) (a) If the investment is ₹ 35 crores, find by how much the WQI is short of its maximum achievable value.[2 marks]
Or
(b) Evaluate the WQI when the investment is ₹60 crore. Determine whether this indicates an enhancement or a diminishing return in our objective when compared with the optimum WQI. [2 marks]
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Detailed Solution

Step 1: Find the stationary point

To find the stationary point, we calculate the first derivative of $W(x) = 35 + 1.6x - 0.02x^2$ with respect to $x$ and set it to zero: $$W'(x) = 1.6 - 0.04x$$ Setting $W'(x) = 0$: $$1.6 - 0.04x = 0 \implies x = \frac{1.6}{0.04} = 40$$ The stationary point is at $x = 40$.

Step 2: Second derivative test for maximum

We find the second derivative: $$W''(x) = -0.04$$ Since $W''(40) = -0.04 < 0$, the function has a local maximum at $x = 40$. The maximum WQI is: $$W(40) = 35 + 1.6(40) - 0.02(40)^2 = 35 + 64 - 32 = 67$$

Step 3: Evaluate investment scenarios

For (a), if $x = 35$: $$W(35) = 35 + 1.6(35) - 0.02(35)^2 = 35 + 56 - 24.5 = 66.5$$ Shortfall = $67 - 66.5 = 0.5$. For (b), if $x = 60$: $$W(60) = 35 + 1.6(60) - 0.02(60)^2 = 35 + 96 - 72 = 59$$ Since $59 < 67$, this indicates a diminishing return compared to the optimum.

Final Answer: (i) 40, (ii) 67, (iii)(a) 0.5, (iii)(b) 59 (Diminishing return)

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to utilize the concept of derivatives to solve a real-world optimization problem.
Knowledge Dimension: PROCEDURAL
Justification: The student must follow a specific sequence of mathematical steps (differentiation, setting to zero, second derivative test) to reach the solution.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It integrates environmental awareness (YAP) with the Application of Derivatives (Maxima/Minima), aligning with NEP 2020's focus on interdisciplinary learning.

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