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Applied Maths 2 Marks Questions

Differentiation and it's Applications · CBSE STD 12 Science (CBSE - English Medium). 22 questions.

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Question 12 Marks
Read the following text and answer the following questions on the basis of the same:
In a residential society comprising of 100 houses, there were 60 children between the ages of 10-15 years. They were inspired by their teachers to startcomposting to ensure that biodegradable waste is recycled, For this purpose, instead of each child doing it for only his/her house, children convinced the Residents welfare association to do it as a society initiative. For this they identified a square area in the local park. Local authorities charged amount of ₹ 50 per square metre for space so that there is no misuse of the space and Resident welfare association takes it seriously. Association hired a labourer for digging out $250 m^3$ and he charged ₹ 400 X (depth)². Association will like to have minimum cost.
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Q.1. Let side of square plot is x m and its depth is h metres, then find cost equation C for the pit. Also, find second derivative of cost C w.r.t. h.

Q.2. Find the value of x (in m) for which cost is minimum.
Answer
(1)
$C=\frac{250 \times 50}{h}+400 \times h^2$
$C=\frac{12500}{h}+400 h^2$
$\therefore \quad \frac{d C}{d h}=\frac{-12500}{h^2}+800 h$
and $\frac{d^2 C}{d h^2}=\frac{-(-2) \times 12500}{h^3}+800$
$\Rightarrow$ $\frac{d^2 C}{d h^2}=\frac{25000}{h^3}+800$

(2)
For minimum cost, put $\frac{d C}{d h}=0$,, we get
$\therefore \quad \frac{-12500}{h^2}+800 h=0$
$\Rightarrow \quad 800 h^3=12500$
$\Rightarrow \quad h^3=\frac{125}{8}$
$\Rightarrow \quad h=\frac{5}{2}=2.5 m$
h = 2.5m
At, $h=2.5, \frac{d^2 C}{d h^2}>0$ ( Hence minimum)
Value of x at minimum cost
$x=\frac{400 \times(2.5)^2}{250}$
$=\frac{2500}{250}=10 m$

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Question 22 Marks
Read the following text and answer the following questions on the basis of the same:
The Government declare that farmers can get 500 per quintal for their onions on 1st April and after that, the price will be dropped by ₹ 4 per quintal per extra day.
Roshan's father has 120 quintal of onions in the field on 1 April and estimates that crop is increasing at the rate of ₹ 1 per day.

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Q.1. If x is the number of extra days after 1st April, then find the express for price and quantity of onions. Also, find the excess for reverse, R in terms of $x_1$

Q.2. Find the number of days after $1 ^{ st }$ April, when Roshan's father attain maximum revenue.
Answer
(1)
Given, $x$ is the number of extra days after 1st April
Price, $P(x)=$ ₹ (500-4x)
and Quantity, Q(x) = 120 quintals $\times x(1$ quintal per day $)$
$=(120+x)$ quintals
Now, revenue, $\quad R(x)=$ Quantity $\times$ Price
$=Q(x) \times P(x)$
$\begin{array}{l}=(120+x)(500-4 x) \\ =60000-480 x+500 x-4 x^2 \\ =60,000+20 x-4 x^2\end{array}$
$\therefore$ $R(x)=60,000+20 x-4 x^2$

(2)
We have, $R(x)=60,000+20 x-4 x^2$
$\therefore \quad R^{\prime}(x)=20-8 x$
and $R^{\prime \prime}(x)=-8$
For $R(x)$ to be maximum, $R^{\prime}(x)=0$
$\Rightarrow \quad 20-8 x=0$
$\Rightarrow \quad x=\frac{20}{8}=2.5$
Thus, Roshan's father will attain maximum revenue after 2.5 days.
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Question 32 Marks
Read the following text and answer the following questions on the basis of the same:
Revathi is going to her best friend's birthday party. She has made a beautiful painting for her friend as a birthday gift. Now, she wants to prepare a hand made gift box for that painting. For making lower part of the box, she takes a square piece of cardboard of side 30 cm.

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Q.1. If x can be the length of each side of the square cardboard which is to be cut off from corners of the square piece of side 30 cm, then write the expression for volume of the open box formed by folding up the cutting corners. Also, find $\frac{d V}{d x}$

Q.2. Revathi is inserted in maximising the volume of the box. So, what should be the side of square to be cut off so that volume of the box is maximum?
Answer
(1)
Given, side of square is of length 30 cm.
Also, height of the open box $=x cm$
Length of open box $=30-2 x$
and width of open box $=30-2 x$
$\therefore \quad$ Volume $( V )$ of open box $=x \times(30-2 x) \times(30-2 x)$
$=x(30-2 x)^2$
Now, $\frac{d V}{d x}=x[2(30-2 x)(-2)]+1(30-2 x)^2$
or, $\frac{d V}{d x}=-4 x(30-2 x)+(30-2 x)^2$
or, $\frac{d V}{d x}=(30-2 x)(-4 x+30-2 x)$
or, $\frac{d V}{d x}=(30-2 x)(30-6 x)$

(2)
Since, $\frac{d V}{d x}=(30-2 x)(30-6 x)$ [ from above ]
$\therefore \quad \frac{d^2 V}{d x^2}=-2(30-6 x)+(30-2 x)(-6)$
or, $\frac{d^2 V}{d x^2}=-2(30-6 x+90-6 x)$
or, $\frac{d^2 V}{d x^2}=-2(120-12 x)$
or, $\frac{d^2 V}{d x^2}=-24(10-x)$
Now, put $\frac{d V}{d x}=0$
$\Rightarrow(30-2 x)(30-6 x)=0$
$\Rightarrow \quad x=15$ or $x=5$
At $x=15, \quad \frac{d^2 V}{d x}=-24(10-15)=-24 \times(-5)$
= 120 > 0
At $x=5, \quad \frac{d^2 V}{d x}=-24(10-15)=-24 \times 5$
= 120 < 0
So , volume iwll bw maximum when x = 5 cm.

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Question 42 Marks
The total cost function (in thousands) for manufacturing x manipulators per year is given $C(x)=375+25 x-0.25 x^2 \quad 0 \leq x \leq 50$
(a) Use the marginal cost function to approximate the cost of producing the $31^{s t}$ manipulator.
(b) Use the total cost function to find the exact cost of producing the $31^{s t}$ manipulator.
Answer
(a) The marginal cost is C(x) = 25 - 0.5x
Thus, C(30) = 25 - 0.5(30) = 10
The cost of producing of the $31^{s t}$ manipulator is approximately ₹ 10,000.

(b) The exact cost of producing the $31^{s t}$ manipulator is C(31)-C(30)= 909.75 - 900 9.75, i.e. ₹ 9,750.
The marginal cost of ₹ 10,000 per manipulator is a close approximation to his exact cost.
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Question 52 Marks
Find the interval in which the function f given by $f(x)=2 x^2-3 x$is (a) Increasing (b) Decreasing
Answer
Let, $y=f(x)=2 x^2-3 x$
$\frac{d y}{d x}=4 x-3>0$
For increasing function, $\frac{d y}{d x}>0$ or $4 x-3>0$ or $x>\frac{3}{4}$
Hence, given function is increasing in $\left(\frac{3}{4}, \infty\right)$
For decreasing function, $\frac{d y}{d x}<0$ or $4 x-3<0$ or $x<\frac{3}{4}$
Hence, given function is increasing in $\left(-\infty, \frac{3}{4}\right) \cdot$
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Question 62 Marks
Show that the function $f(x)=4 x^3-18 x^2+27 x-7$ is always increasing on R.
Answer
$f(x)=4 x^3-18 x^2+27 x-7$
$f^{\prime}(x)=12 x^2-36 x+27$
$=3(2 x-3)^2 \geq 0 ; x \in R$
$\therefore f(x)$ is increasing on $R$.
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Question 72 Marks
Show that the function $f(x)=x^3-3 x^2+6 x-100$ is increasing on R.
Answer
$f(x)=x^3-3 x^2+6 x-100$
$f^{\prime}(x)=3 x^2-6 x+6$
$=3\left[x^2-2 x+2\right]=3\left[(x-1)^2+1\right]$
since $f^{\prime}(x)>0 ; x \in R$
$\therefore f(x)$ is increasing on $R$.
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Question 82 Marks
Given $y=e^x(x+1)$. Find the value of second derivative at x = 1
Answer
Given, $y=e^x(x+1)$
Since this function is product of two functions, we will use multiplication rule for derivative.
$\frac{d(f(x) g(x))}{d x}=f(x) \frac{d(g(x))}{d x}+g(x) \frac{d(f(x))}{d x}$
$y^{\prime}=e^x(x+1)+e^x$
Now we can differentiate it again to get the second derivative.
$y^{\prime \prime}=\frac{d\left(e^x(x+1)\right)}{d x}+\frac{d\left(e^x\right)}{d x}$
$y^{\prime \prime}=e^x(x+3)$
at x = 1
$y^{\prime \prime}=e^1(1+3)=4 e$
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Question 92 Marks
Given $y=\frac{x}{x^2+1}$ then find the value of second derivative at x = 1
Answer
Let $y=\frac{x}{x^2+1}$, then
$y=\frac{\left(x^2+1\right) \frac{d}{d x} x-x \frac{d}{d x}\left(x^2+1\right)}{\left(x^2+1\right)^2}$
$\Rightarrow \quad y^{\prime}=\frac{1-x^2}{\left(x^2+1\right)^2}$
$y^{\prime}=\frac{\left(x^2+1\right) \frac{d}{d x} n-x-\frac{d}{d x}\left(x^2+1\right)}{\left(x^2+1\right)^2}$
$y^{\prime \prime}=\frac{\left(x^2+1\right)^2(-2 x)-\left(1-x^2\right)\left[2\left(x^2+1\right) 2 x\right]}{\left(x^2+1\right)^4}$
At x = 1,
$y^{\prime \prime}=\frac{\left(1^2+1\right)^2(-2)-\left(1-1^2\right)\left[2\left(1^2+1\right) 2\right]}{\left(1^2+1\right)^4}$
$\Rightarrow \quad y^{\prime \prime}=\frac{(2)^2(-2)-(1-1)[2(2) 2]}{(2)^4}$
$\Rightarrow \quad y^{\prime \prime}=\frac{-8+0}{(2)^4}$
$\Rightarrow \quad y^{\prime \prime}=\frac{-8+0}{(2)^4}$
$\Rightarrow \quad y^{\prime \prime}=\frac{-1}{2}$
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Question 102 Marks
If $y=x^3 \log x$, then find $\frac{d^4 y}{d x^4}$.
Answer
If
$y=x^3 \log x$
$\therefore \quad \frac{d y}{d x}=\frac{d}{d x}\left(x^3 \log x\right)$
$\frac{d y}{d x}=x^2(1+3 \log x)$
$\frac{d^2 y}{d x^2}=x^2 \frac{d}{d x}(1+3 \log x)+(1+3 \log x) \frac{d}{d x} x^2$
$\frac{d^2 y}{d x^2}=x(5+6 \log x)$
$\frac{d^3 y}{d x^3}=x \frac{d}{d x}(5+6 \log x)+(5+6 \log x) \frac{d}{d x} x$
$\frac{d^3 y}{d x^3}=11+6 \log x$
$\frac{d^4 y}{d x^4}=\frac{6}{x}$
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Question 112 Marks
Check whether the function g(x) = log x is maxima or minima.
Answer
g(x) = log x
$\therefore \quad g^{\prime}(x)=\frac{1}{x}$
Since log(x) is defined for a positive number x,
g'(x) > 0 for any x.
Therefore, there does not exist $c \in R$ such that $g^{\prime}(c)$
Hence, function g does not have maxima or minima.
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Question 122 Marks
Find the least value of a function such that the function f given by $f(x)=x^2+a x+1$ is strictly increasing on (1, 2).
Answer
$f(x)=x^2+a x+1$
$\therefore f^{\prime}(x)=2 x+a$
Now, increasing in (1, 2)
i.e. $f^{\prime}(x)>0$
$\therefore \quad 2 x+a>0$
$a>-2 x$
We have to find least value of a, such that
$x \in(1,2)$
a > - 2
$\Rightarrow \quad a \in(-2, \infty)$
So, least value of a is - 2
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Question 132 Marks
Find the maximum profit that a company can make, if the profit function is given by p(x) =41-$72 x-18 x^2$
Answer
Let profit function
$p(x)=41-72 x-18 x^2$
$\therefore \quad p^{\prime}(x)=-72-36 x$
$\Rightarrow \quad x=-\frac{72}{36}=-2$
$p^{\prime \prime}(x)=-36$
Also, $p \ddot{ u }-2=$         
=-36< 0
By second derivative test, x = - 2 is the point of local maxima of p.
$\therefore \quad$ Maximum profit $=p(-2)$
$=41-72(-2)-18(-2) 2$
$=41+144-72=113$
Hence, the maximum profit that the company can make is 113 units.
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Question 142 Marks
The total revenue in Rupees received from the sale of x units of a product is given by R(x)=$3 x^2$+26 x+15 Find the marginal revenue, when x = 15.
Answer
Here,
$R(x)=3 x^2+26 x+15$
$\frac{d R}{d x}=6 x+26$
$\Rightarrow \quad\left(\frac{d R}{d x}\right)_{\text {st } x=15}=6(15)+26$
= ₹ 116
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Question 152 Marks
If at x = 1 function $x^4-62 x^2+a x+9$ attains its maximum value, in the interval [0, 2], then find the value of a.
Answer
$f(x)=x^4-62 x^2+a x+9$
Therefore, $f^{\prime}(x)=4 x^3-124 x+a$
f attains its maximum value on the interval [0, 2] at x = 1
Therefore, $f^{\prime}(1)=0$
4 - 124 + a = 0
a = 120
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Question 162 Marks
The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its total revenue (marginal revenue). If the total revenue (in) received from the sale of x units of a product is given by $R(x)=3 x^2+36 x+5$find the marginal revenue, when x = 5
Answer
Total revenue is given by $R(x)=3 x^2+36 x+5$
Marginal revenue $=\frac{d R}{d x}=6 x+36$At $x=5$,
$\left(\frac{d R}{d x}\right)_{x=5}=6 \times 5+36=66$
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Question 172 Marks
The total expenditure (in ₹) required for providing the cheap edition of a book for poor and deserving students is given by $C(x)=3 x^2+36 x$, where x is the number of set of books. If the marginal expenditure is defined as $\frac{d C}{d x}$ , write the marginal expenditure required for 1200 such sets.
Answer
Here, $C(x)=3 x^2+36 x$, where $x$ is the number of sets of book$\begin{aligned} \therefore & \frac{d C}{d x}=6 x+36 \\& \left(\frac{d C}{d x}\right)_{x=1200}=6(1200)+36=7,236 \end{aligned}$
So, the marginal cost of books for 1200 such sets = ₹ 7,236
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Question 182 Marks
A firm is selling 100 units at a price of ₹ 250/unit. However, to sell 110 units, they need to cut the price down to 240/unit. What is the level of marginal revenue at this higher level of sales?
Answer
The total revenue function for 100 units as:
$R(100)=100 \times 250=\text {₹} 25000$
Similarly, for 110 units:
$R(110)=110 \times 240= \text {₹} 26400$
Thus, Marginal Revenue $=\frac{26400-25000}{10}$
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Question 192 Marks
Show that the function given by $f(x)=3 x+17$ is increasing on R.
Answer
Let $y=f(x)=3 x+17$
$\frac{d y}{d x}=3>0 \text { for all } x \in R$
Hence, given function is increasing on $R$.
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Question 202 Marks
Show that function y = 4x - 9 is increasing for all $x \in R$.
Answer
Given, $y=4 x-9$
$\frac{d y}{d x}=4>0 \text { for all } x \in R$
Hence, function is increasing for all $x \in R$.
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Question 212 Marks
Given $f(x)=e^x+\log x$. Find the value of $f^{\prime \prime}(x)$
Answer
Given $f(x)=e^x+\log x$
The first derivative will be,
$f^{\prime}(x)=e^x+\frac{1}{x}$
Differentiating it again,
$f^{\prime \prime}(x)=e^x-\frac{1}{x^2}$
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Question 222 Marks
If $x=t^2$ and $y=t^3$ then find $\frac{d^2 y}{d x^2}$
Answer
Given that, $x=t^2$ and $y=t^3$
Then, $\frac{d x}{d t}=2 t$
and $\frac{d y}{d t}=3 t^2$
Thus, $\frac{d y}{d x}=\frac{3 t^2}{2 t}=\frac{3 t}{2}$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{3}{2} \frac{d t}{d x}$
$\frac{3}{2} \cdot \frac{1}{2 t}=\frac{3}{4 t}$
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2 Marks Questions - Applied Maths STD 12 Science Questions - Vidyadip