Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems

Students can download 12th Business Maths Chapter 5 Numerical Methods Miscellaneous Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems

Question 1.
If f(x) = eax then show that f(0), ∆f(0), ∆2 f(0) are in G.P
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q1

Question 2.
Prove that
(i) (1 + ∆) (1 – ∇) = 1
(ii) ∆∇ = ∆ – ∇
(iii) E∇ = ∆ – ∇E
Solution:
(i) To show (1 + ∆) (1 – ∇) = 1
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q2
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q2.1
Hence proved.

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems

Question 3.
A second degree polynomial passes through the point (1, -1) (2, -1) (3, 1) (4, 5). Find the polynomial.
Solution:
Given values can be tabulated as follows
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q3
We have to find a second-degree polynomial.
We use Newton’s forward interpolation formula
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q3.1
The difference table is given below
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q3.2
(or) y = x2 – 3x + 1 is the required second-degree polynomial which passes through the given points

Question 4.
Find the missing figures in the following table
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q4
Solution:
Since only four values of y are given, the polynomial which fits the data is of degree three.
Hence fourth differences are zero.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q4.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q4.2
Thus the missing figures are 14.25 and 23.5

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems

Question 5.
Find f(0.5) if f(-1) = 202, f(0) = 175, f(1) = 82 and f(2) = 55
Solution:
The given data can be written as
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q5
We have to find y when x = 0.5.
We use Newton’s forward interpolation formula
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q5.1
The difference table is
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q5.2
Using these values we get,
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q5.3
Thus the value of f(0.5) = 128.5

Question 6.
From the following data find y at x = 43 and x = 84
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q6
Solution:
We have to find the value of y at (a) x = 43 and (b) x = 84
(a) x = 43.
The value of y is required at the beginning of the table.
So we use Newton’s forward interpolation formula
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q6.1
Here x = 43 , x0 = 40, h = 10
So 43 = 40 + 10n
n = 0.3
The difference table is given below
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q6.2
y = 184 + 6 – 0.21
y = 189.79

(b) x = 84.
The value of y is required at the end of the table.
So we use Newton’s backward interpolation formula
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q6.3
We use the back difference values from the table
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q6.4
Hence the value of y at x = 43 is 189.75 and the value of y at x = 84 is 286.96

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems

Question 7.
The area A of a circle of diameter ‘d’ is given for the following values
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q7
Find the approximate values for the areas of circles of diameter 82 and 91 respectively.
Solution:
Let diameter be x and area be y
We have to find value of y when (a) x = 82 and (b) x = 91
We first find the difference as given below
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q7.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q7.2
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q7.3
Hence the area of a circle when the diameter is 82 is 5281
area of a circle when the diameter is 91 is 6504.

Question 8.
If u0 = 560, u1 = 556, u2 = 520, u4 = 385, show that u3 = 465
Solution:
Given u0 = 560, u1 = 556, u2 = 520, u4 = 385
Since only four values are given,
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q8

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems

Question 9.
From the following table obtain a polynomial of degree y in x
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q9
Solution:
To find a polynomial y = f(x)
Here x0 = 1, h = 1
x = x0 + nh
x = 1 + n(1)
n = x – 1
We find the forward differences as below
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q9.1
Using Newton’s forward interpolation formula,
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q9.2
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q9.3

Question 10.
Using Lagrange’s interpolation formula find a polynomial which passes through the points (0, -12), (1, 0), (3, 6) and (4, 12).
Solution:
The given values are
x0 = 0, y0 = -12
x1 = 1, y1 = 0
x2 = 3, y2 = 6
x3 = 4, y3 = 12
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q10
By Lagrange’s interpolaiton formula,
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Miscellaneous Problems Q10.1
⇒ y = (x – 1) (x – 3) (x – 4) – x (x – 1) (x – 4) + x (x – 1) (x – 3)
⇒ y = (x – 1) (x – 4)[(x – 3) – x] + x (x – 1) (x – 3)
⇒ y = (x – 1) (x – 4) (-3) + x (x – 1) (x – 3)
⇒ y = (x – 1) [-3x + 12 + x2 – 3x]
⇒ y = (x – 1) (x2 – 6x + 12)
⇒ y = x3 – 6x2 + 12x – x2 + 6x – 12
y = x3 – 7x2 + 18x – 12 is the required polynomial which passes through the given points