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

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

Question 1.
Evaluate (log ax)
Solution:

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

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

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:

Question 6.
Find the missing entry in the following table

Solution:
Since only four values of y are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.

Question 7.
Following are the population of a district

Find the population of the year 1911?
Solution:
Since five values are given, the polynomial which fits the data is of degree four.

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

Question 8.
Find the missing entries from the following.

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: