Exercise 5.2 - Chapter 5 Numerical Methods 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 5.2
Question 1.
Using graphic method, find the value of $\mathrm{y}$ when $\mathrm{x}=48$ from the following data:

Solution:
The given points are $(40,6.2),(50,7.2)(60,9.1)$ and $(70,12)$.
We plot the points on a graph with suitable scale

The value of $y$ when $x=48$ is $6.8$
Question 2.
The following data relates to indirect labour expenses and the level of output

Estimate the expenses at a level of output of 350 units, by using the graphic method.
Solution:
Take the units of output along the $\mathrm{x}$-axis, labour expenses along the $y$-axis.
The points to be plotted are $(200,2500),(300,2800)(400,3100),(640,3820),(540,3220),(580$, $3640)$

From the graph, the expenses at a level of output of 350 units are $₹ 2940$.
Question 3.
Using Newton's forward interpolation formula find the cubic polynomial.

Solution:
Newton's forward interpolation formula is

$
y_{\left(x=x_0+n h\right)}=y_0+\frac{n}{1 !} \Delta y_0+\frac{n(n-1)}{2 !} \Delta^2 y_0+\frac{n(n-1)(n-2)}{3 !} \Delta^3 y_0+\ldots
$

To find $y$ in terms of $x, x_0+n h=x . \quad x_0=0, h=1 \Rightarrow n=x$
So $y(x)=1+\frac{x}{1 !}(1)+\frac{x(x-1)}{2 !}(-2)+\frac{x(x-1)(x-2)}{3 !}(12)$
$
\begin{aligned}
& y(x)=1+x-\left(x^2-x\right)+2\left[x^3-3 x^2+2 x\right] \\
& y(x)=1+x-x^2+x+2 x^3-6 x^2+4 x \\
& f(x)=y=2 x^3-7 x^2+6 x+1 \text { is the required cubic polynomial }
\end{aligned}
$
Question 4.
The population of a city in a census taken once in 10 years is given below.

Estimate the population in the year 1955.
Solution:

Let ' $x$ ' denote the year of the census. Let ' $y$ ' represent the population in lakhs. We have to find the population in the year 1955 (i.e) the value of $y$ when $x=1955$. Since the value of $y$ is required near the beginning of the table, we use Newton's forward interpolation formula.
$
y_{\left(x=x_0+n h\right)}=y_0+\frac{n}{1 !} \Delta y_0+\frac{n(n-1)}{2 !} \Delta^2 y_0+\frac{n(n-1)(n-2)}{3 !} \Delta^3 y_0+\ldots
$

Now $x=1955 . \therefore x_0+n h=1955, x_0=1951 h=10$
$
\Rightarrow 1951+10 n=1955
$
$
\begin{aligned}
10 n & =4 \\
n & =\frac{4}{10}=0.4
\end{aligned}
$

$
\begin{aligned}
y_{(x=1955)} & =35+\frac{0.4}{1 !}(7)+\frac{(0.4)(0.4-1)}{2 !}(9)+\frac{(0.4)(0.4-1)(0.4-2)}{3 !}(1) \\
& =35+2.8+\frac{(0.4)(-0.6)(9)}{2}+\frac{(0.4)(-0.6)(-1.6)}{6} \\
& =37.8-1.08+0.064 \\
& =36.784
\end{aligned}
$
Thus the estimated population in the year 1955 is $36.784$ lakhs

Question 5.
In an examination the number of candidates who secured marks between certain intervals was as follows:

Estimate the number of candidates whose marks are less than 70.
Solution:
Let $\mathrm{x}$ be the marks and $\mathrm{y}$ be the number of candidates. The given class intervals are not continuous. So we make them continuous and find the cumulative frequency.

The difference table is as follows

Since the required value of $y$ is near the end of the table, we use Newton's backward interpolation formula

$
y_{\left(x=x_n+n h\right)}=y_n+\frac{n}{1 !} \nabla y_n+\frac{n(n+1)}{2 !} \nabla^2 y_n+\frac{n(n+1)(n+2)}{3 !} \nabla^3 y_n+\ldots
$
Now
$
\begin{aligned}
& x=70, h=20, x_n=99.5 \\
& x_n+n h=70 \\
& 99.5+20 n=70 \\
& n=\frac{70-99.5}{20}=-1.475 \\
& y=235+\frac{(-1.475)}{1 !}(17)+\frac{(-1.475)(-1.475+1)}{2 !}(-33) \\
& +\frac{(-1.475)(-1.475+1)(-1.475+2)}{3 !}(-18) \\
& y=235-25.08-\frac{(1.475)(0.475)(33)}{2}-\frac{(1.475)(0.475)(0.525)(18)}{6} \\
& y=235-25.08-11.56-1.10 \\
& y=197.26 \\
&
\end{aligned}
$
Hence the estimated value of the number of candidates whose marks are less than 70 is 197
Question 6.
Find the value of $f(x)$ when $x=32$ from the following table

Solution:
To find $y=f(x)$ when $x=32$ from the given table. Since the required value of $y$ is near the beginning of the table, we use Newton's forward interpolation formula.
$
\begin{aligned}
& 32=30+n(5) \\
& n=\frac{32-30}{5}=0.4 \\
&
\end{aligned}
$
The difference table is as follows

$
\begin{aligned}
y_{(x=32)}= & 15.9+\frac{0.4}{1 !}(-1)+\frac{(0.4)(0.4-1)}{2 !}(0.2)+ \\
& \frac{(0.4)(0.4-1)(0.4-2)}{3 !}(-0.2)+\frac{(0.4)(0.4-1)(0.4-2)(0.4-3)}{4 !}(0.2) \\
y= & 15.9-0.4-0.024-0.0128-0.00832 \\
y= & 15.45488
\end{aligned}
$
Hence the value of $f(x)$ when $x=32$ is $15.45$
Question 7.
The following data gives the melting point of an alloy of lead and zinc where ' $t$ ' is the temperature in degree $\mathrm{c}$ and $\mathrm{P}$ is the percentage of lead in the alloy

Find the melting point of the alloy containing 84 per cent lead.
Solution:
To find $\mathrm{T}$ when $\mathrm{P}=84$. The required value of $\mathrm{T}$ is near the end of the table. So we use Newton's backward interpolation formula.

$\begin{aligned}
& t_{\left(p+p_{\mathrm{n}}+n h\right)=t_n}+\frac{n}{1 !} \nabla t_n+\frac{n(n+1)}{2 !} \nabla^2 t_n+\frac{n(n+1)(n+2)}{3 !} \nabla^3 t_n+\ldots . \\
& \quad \text { Now } p_n+n h \quad=84, h=10 \\
& \Rightarrow 90+\mathrm{n}(10)=84 \\
& \Rightarrow \mathrm{n}=\frac{84-90}{10}=\frac{-6}{10}=-0.6
\end{aligned}$

$
\begin{aligned}
t_{(p=84)}= & 304+\frac{(-0.6)}{1 !}(28)+\frac{(-0.6)(-0.6+1)}{2 !}(2)+\frac{(-0.6)(-0.6+1)(-0.6+2)}{3 !}(0) \\
& +\frac{(-0.6)(-0.6+1)(-0.6+2)(-0.6+3)(0)}{4 !}(0) \\
& +\frac{(-0.6)(-0.6+1)(-0.6+2)(-0.6+3)(-0.6+4)}{5 !}(4) \\
t= & 304-16.8-0.24-0.0914 \\
t= & 286.8686
\end{aligned}
$
Hence the melting point of the alloy containing 84 per cent lead is $286.9^{\circ} \mathrm{C}$
Question 8.
Find $f(2.8)$ from the following table.

Solution:
To find $\mathrm{y}=\mathrm{f}(\mathrm{x})$ at $\mathrm{x}=2.8$ from the given table. We use Newton's backward interpolation formula since the required value is near the end of the table.
$
y_{\left(x=x_n+n h\right)}=y_n+\frac{n}{1 !} \nabla y_n+\frac{n(n+1)}{2 !} \nabla^2 y_n+\frac{n(n+1)(n+2)}{3 !} \nabla^3 y_n+\ldots
$
Now $x_n=3, h=1, x=2.8$
$
\begin{aligned}
& 2.8=3+\mathrm{n}(1) \\
& \mathrm{n}=2.8-3=-0.2
\end{aligned}
$
The difference table given below

$
\begin{aligned}
& y=34+\frac{(0.2)}{1 !}(23)+\frac{(-0.2)(-0.2+1)}{2 !}(14)+\frac{(-0.2)(-0.2+1)(-0.2+2)}{3 !}(6) \\
& y=34-4.6-1.12-0.288 \\
& y=27.992
\end{aligned}
$
Hence the value of $f(x)$ at $x=2.8$ is $27.992$
Question 9.
Using interpolation estimate the output of a factory in 1986 from the following data

Solution:
Let $\mathrm{x}$ denote the year and $\mathrm{y}$ represent the output. The $\mathrm{x}$ values are not equidistant. So we use

Lagrange's formula
$
\begin{aligned}
& \mathrm{x}_0=1974, \mathrm{y}_0=25 \\
& \mathrm{x}_1=1978, \mathrm{y}_1=60 \\
& \mathrm{x}_2=1982, \mathrm{y}_2=80 \\
& \mathrm{x}_3=1990, \mathrm{y}_3=170
\end{aligned}
$
For $\mathrm{x}=1986$ we have to find $\mathrm{y}$ value
$
\begin{aligned}
y= & \frac{\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_0-x_1\right)\left(x_0-x_2\right)\left(x_0-x_3\right)} \times y_0+\frac{\left(x-x_0\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_1-x_0\right)\left(x_1-x_2\right)\left(x_1-x_3\right)} y_1 \\
& +\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_3\right)}{\left(x_2-x_0\right)\left(x_2-x_1\right)\left(x_2-x_3\right)} \times y_2+\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right)}{\left(x_3-x_0\right)\left(x_3-x_1\right)\left(x_3-x_2\right)} y_3
\end{aligned}
$
We find the different values separately and substitute in the formula.

$
\begin{aligned}
& y=\frac{(8)(4)(-4)}{(-4)(-8)(-16)}(25)+\frac{(12)(4)(-4)}{(4)(-4)(-12)}(60)+\frac{(12)(8)(-4)}{(8)(4)(-8)}(80)+\frac{(12)(8)(4)}{(16)(12)(8)}(170) \\
& y=6.25-60+120+42.5 \\
& y=108.75
\end{aligned}
$
The output of the factory in 1986 is 109 (thousand tonnes)
Question 10.
Use Lagrange's formula and estimate from the following data the number of workers getting income not exceeding Rs. 26 per month.

Solution:
Let $\mathrm{x}$ represent the income per month and $y$ denote the number of workers.
From the given data,
$
\begin{aligned}
& x_0=15, y_0=36 \\
& x_1=25, y_1=40 \\
& x_2=30, y_2=45 \\
& x_3=35, y_3=48
\end{aligned}
$
We have to find the value of $y$ at $x=26$
By Lagrange's interpolation formula,
$
\begin{aligned}
y= & \frac{\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_0-x_1\right)\left(x_0-x_2\right)\left(x_0-x_3\right)} \times y_0+\frac{\left(x-x_0\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_1-x_0\right)\left(x_1-x_2\right)\left(x_1-x_3\right)} y_1 \\
& +\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_3\right)}{\left(x_2-x_0\right)\left(x_2-x_1\right)\left(x_2-x_3\right)} \times y_2+\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right)}{\left(x_3-x_0\right)\left(x_3-x_1\right)\left(x_3-x_2\right)} y_3
\end{aligned}
$
The different values are given in the table below

$
\begin{aligned}
& y=\frac{(1)(-4)(-9)}{(-10)(-15)(-20)}(36)+\frac{(11)(-4)(-9)}{(10)(-5)(-10)}(40)+\frac{(11)(1)(-9)}{(15)(5)(-5)}(45)+\frac{(11)(1)(-4)}{(20)(10)(5)}(48) \\
& y=-0.432+31.68+11.88-2.112 \\
& y=41.016
\end{aligned}
$
Thus the number of workers getting income not exceeding Rs. 26 per month is 42
Question 11.
Using interpolation estimate the business done in 1985 from the following data.

Solution:
Let $\mathrm{x}$ denote the year of business and $\mathrm{y}$ (in lakhs) denote the amount of business.
From the given data,
$
\begin{aligned}
& \mathrm{x}_0=1982, \mathrm{y}_0=150 \\
& \mathrm{x}_1=1983, \mathrm{y}_1=235 \\
& \mathrm{x}_2=1984, \mathrm{y}_2=365 \\
& \mathrm{x}_3=1986, \mathrm{y}_3=525
\end{aligned}
$
We have to find the value of $y$ when $x=1985$.
By Lagrange's interpolation formula,
$
\begin{aligned}
& y=\frac{\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_0-x_1\right)\left(x_0-x_2\right)\left(x_0-x_3\right)} \times y_0+\frac{\left(x-x_0\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_1-x_0\right)\left(x_1-x_2\right)\left(x_1-x_3\right)} \\
&+\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_3\right)}{\left(x_2-x_0\right)\left(x_2-x_1\right)\left(x_2-x_3\right)} \times y_2+\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right)}{\left(x_3-x_0\right)\left(x_3-x_1\right)\left(x_3-x_2\right)} y_3
\end{aligned}
$
We form a table for the different values

$
\begin{aligned}
& y=\frac{(2)(1)(-1)}{(-1)(-2)(-4)}(150)+\frac{(3)(1)(-1)}{(1)(-1)(-3)}(235)+\frac{(3)(2)(-1)}{(2)(1)(-2)}(365)+\frac{(3)(2)(1)}{(4)(3)(2)}(525) \\
& y=37.5-235+547.5+131.25 \\
& y=481.25
\end{aligned}
$
Thus the business done in the year 1985 is estimated as $481.25$ lakhs
Question 12.
Using interpolation, find the value of $\mathrm{f}(\mathrm{x})$ when $\mathrm{x}=15$

Solution:
We have to find the value of $y$ when $x=15$.
From the given data,
$
\begin{aligned}
& \mathrm{x}_0=3, \mathrm{y}_0=42 \\
& \mathrm{x}_1=7, \mathrm{y}_1=43 \\
& \mathrm{x}_2=11, \mathrm{y}_2=47 \\
& \mathrm{x}_3=19, \mathrm{y}_3=60
\end{aligned}
$
Since the intervals are unequal, we use the Lagrange's interpolation formula,
$
\begin{aligned}
& y=\frac{\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_0-x_1\right)\left(x_0-x_2\right)\left(x_0-x_3\right)} \times y_0+\frac{\left(x-x_0\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_1-x_0\right)\left(x_1-x_2\right)\left(x_1-x_3\right)} y_1 \\
&+\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_3\right)}{\left(x_2-x_0\right)\left(x_2-x_1\right)\left(x_2-x_3\right)} \times y_2+\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right)}{\left(x_3-x_0\right)\left(x_3-x_1\right)\left(x_3-x_2\right)} y_3
\end{aligned}
$
The different values are given in the table below.

Using the above values,
$
\begin{aligned}
& y=\frac{(8)(4)(-4)}{(-4)(-8)(-16)}(42)+\frac{(12)(4)(-4)}{(4)(-4)(-12)}(43)+\frac{(12)(8)(-4)}{(8)(4)(-8)}(47)+\frac{(12)(8)(4)}{(16)(12)(8)}(60) \\
& y=10.5-43+70.5+15 \\
& y=53
\end{aligned}
$
Hence the value of $f(x)$ when $x=15$ is 53