Students can download 12th Business Maths Chapter 5 Numerical Methods Ex 5.1 Questions 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 Ex 5.1

Question 1.
Evaluate (log ax)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q1

Question 2.
If y = x3 – x2 + x – 1 calculate the values of y for x = 0, 1, 2, 3, 4, 5 and form the forward differences table.
Solution:
Given y = x3 – x2 + x – 1
When
x = 0, y = 0 – 0 + 0 – 1 = -1
x = 1, y = 1 – 1 + 1 – 1 = 0
x = 2, y = 8 – 4 + 2 – 1 = 5
x = 3, y = 27 – 9 + 3 – 1 = 20
x = 4, y = 64 – 16 + 4 – 1 = 51
x = 5, y = 125 – 25 + 5 – 1 = 104
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q2

Question 3.
If h = 1 then prove that (E-1 ∆) x3 = 3x2 – 3x + 1.
Solution:
h = 1 To prove (E-1 ∆) x3 = 3×2 – 3x + 1
L.H.S = (E-1 ∆) x3 = E-1 (∆x3)
= E-1[(x + h)3 – x3]
= E-1( x + h)3 – E-1(x3)
= (x – h + h)3 – (x – h)3
= x3 – (x – h)3
But given h = 1
So(E-1 ∆) x3 = x3 – (x – 1)3
= x3 – [x3 – 3x2 + 3x – 1]
= 3x2 – 3x + 1
= RHS

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1

Question 4.
If f(x) = x2 + 3x then show that ∆f(x) = 2x + 4
Solution:
f(x) = x2 + 3 x
∆f(x) = f(x + h) – f(x)
= (x + h)2 + 3(x + h) – x2 – 3x
= x2 + 2xh + h2 + 3x + 3h – x2 – 3x
= 2xh + 3h + h2
Put h = 1, ∆f(x) = 2x + 4

Question 5.
Evaluate \(\Delta\left[\frac{1}{(x+1)(x+2)}\right]\) by taking ‘1’ as the interval of differencing.
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q5
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q5.1

Question 6.
Find the missing entry in the following table
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q6
Solution:
Since only four values of y are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q6.1

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1

Question 7.
Following are the population of a district
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q7
Find the population of the year 1911?
Solution:
Since five values are given, the polynomial which fits the data is of degree four.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q7.1
From the given table
y0 = 363, y1 = 391, y2 = 421, y4 = 467 and y5 = 501
501 – 5(467) + 10y3 – 10(421) + 5(391) – 363 = 0
501 – 2335 + 10y3 – 4210 + 1955 – 363 = 0
-501 + 2335 + 4210 – 1955 + 363 = 10y3
10y3 = 4452
y3 = 445.2
Hence the population of the year 1911 is 445 thousand

Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1

Question 8.
Find the missing entries from the following.
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q8
Solution:
Since only four values of f(x) are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q8.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1 Q8.2

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