CBSE 2026 · Region 1 · Set 1 · Q36 · 4 marks
An online delivery company in a city has $\displaystyle 5000$ subscribers and collects annual subscription fees of ₹ $\displaystyle 300$ per subscriber for unlimited free deliveries.
The company wishes to increase the annual subscription fee. It is predicted that, for every increase of ₹$\displaystyle 1$, ten subscribers will discontinue. Assume that the company increased the annual fee by $\displaystyle ₹ x$. Based on the given information, answer the following questions :(i)How many subscribers will discontinue after an increase of $\displaystyle ₹ x$ in annual fee?(ii)If $\displaystyle \mathrm{R}(x)$ denotes the total revenue collected after the increase of $\displaystyle ₹ x$ in subscription fee, express $\displaystyle \mathrm{R}(x)$ as a function of $\displaystyle x$.(iii)Find the value of $\displaystyle x$ for which $\displaystyle \mathrm{R}(x)$ is maximum.Find the sub-intervals of $\displaystyle (0,5000)$ in which $\displaystyle \mathrm{R}(x)$ is increasing and decreasing.
An online delivery company in a city has $\displaystyle 5000$ subscribers and collects annual subscription fees of ₹ $\displaystyle 300$ per subscriber for unlimited free deliveries.
The company wishes to increase the annual subscription fee. It is predicted that, for every increase of ₹$\displaystyle 1$, ten subscribers will discontinue. Assume that the company increased the annual fee by $\displaystyle ₹ x$.
Based on the given information, answer the following questions :
(i)
How many subscribers will discontinue after an increase of $\displaystyle ₹ x$ in annual fee?
(ii)
If $\displaystyle \mathrm{R}(x)$ denotes the total revenue collected after the increase of $\displaystyle ₹ x$ in subscription fee, express $\displaystyle \mathrm{R}(x)$ as a function of $\displaystyle x$.
(iii)
Find the value of $\displaystyle x$ for which $\displaystyle \mathrm{R}(x)$ is maximum.
Find the sub-intervals of $\displaystyle (0,5000)$ in which $\displaystyle \mathrm{R}(x)$ is increasing and decreasing.
Official answer
From CBSE’s own marking scheme for this paper.
x = $\displaystyle 100$
Marking-scheme solution
(i)
$\displaystyle 10 x$
(ii)
$\displaystyle R(x)=(5000-10 x)(300+x)$
(iii)
$\displaystyle R^{\prime}(x)=2000-20 x$
$\displaystyle R^{\prime}(x)=0 \Rightarrow x=100$
$\displaystyle R^{\prime \prime}(x)=-20<0$ at $\displaystyle x=100$
At $\displaystyle x=100, R(x)$ is maximum
$\displaystyle R^{\prime}(x)=2000-20 x$
$\displaystyle R^{\prime}(x)=0 \Rightarrow x=100$
In $\displaystyle (0,100), R^{\prime}(x)>0$. Therefore, $\displaystyle R(x)$ is increasing in $\displaystyle (0,100)$ or $\displaystyle (0,100]$
In $\displaystyle (100,5000), R^{\prime}(x)<0$. Therefore, $\displaystyle R(x)$ is decreasing in $\displaystyle (100,5000)$ or $\displaystyle [100,5000)$
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.