Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 5 Numerical Methods Miscellaneous Problems Text Book Back Questions and Answers, Notes.

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

Question 1.

If f (x) = e^{ax} then show that f(0), Δf(0), Δ²f(0) are in G.P

Solution:

Given f(x) = e^{ax}

f(0) = e° = 1 ……… (1)

Δf(x) = e^{a(x+h)} – e^{ax}

= e ^{ax+ah} – e^{ax}

= e^{ax}. e^{ah} – e^{ax}

= e^{ax} (e^{ah} – 1)

Δf(0) = e° (e^{ah} – 1)

= (e^{ah} – 1) …….. (2)

Δ²f(x)= Δ [Δf(x)]

= Δ [e^{a(x+h)} – e^{ax}]

[e^{a(x+h+h)} – e^{a(x+h)}] – [e^{a(x+h)} – e^{ax}]

= e^{a(x+2h)} – e^{a(x+h)} – e^{a(x+h)} + e^{ax}

Δ²f(0) = Δ [Δf(x)]

= e^{a(2h)} – e^{a(h)} – e^{a(h)} + e^{0}

= e^{2ah} – e^{ah} – e^{ah} + 1

= (e^{ah})² – 2e^{ah} + 1

= [e^{ah} – 1]² ………… (3)

from (1), (2) & (3)

[t_{2}]² =[Δf(0)]² = (e^{ah} – 1)²

t_{1} × t_{3} = f(0) × Δ²f(0)

= (1)(e^{ah} – 1)² = (e^{ah} – 1)²

⇒ [Δf(0)]² = f(0) × Δ²f(0)

∴ f(0), Δf(0), Δ²f(0) an is G.P.

Question 2.

Prove that

(i) (1 + Δ) (1 – ∇) = 1

(ii) Δ∇ = Δ – ∇

(iii) EV = Δ = ∇E

Solution:

(i) LHS = (1 + Δ) (1 – ∇)

= (E) (E^{-1}) = E^{1-1}

= E° = 1

= RHS

Hence proved.

(ii) LHS = Δ∇

= (E – 1)(1 – E^{-1})

= E – EE^{-1} + E^{-1}

= E – 1 – 1 – E^{-1}

= E – 2 – E^{-1} ………… (1)

RHS = Δ – ∇

= (E – 1) -(1 – E^{-1})

= E – 1 – 1 + E^{-1}

= E – 2 + E^{-1} ………. (2)

from (1) & (2) LHS = RHS

Hence proved.

(iii) E∇ = EE^{-1}Δ [∵ ∇ = E^{-1}Δ]

= Δ ……… (1)

∇E = E^{-1} ΔE

= E^{-1} EΔ

= Δ ………. (2)

from (1) (2)

E∇ = Δ = ∇E

Question 3.

A second degree polynomial passes though the point (1, -1) (2, -1) (3, 1) (4, 5). Find the polynomial.

Solution:

Points are (1, -1), (2, -1), (3, 1) and (4, 5)

we will use Newton’s backward interpolation formula to find the polynomial.

= 5 + (x – 4) (4) + (x – 4) (x – 3) + 0

= 5 + 4x – 16 + x² – 7x + 12

y(x) = x² – 3x + 1

Question 4.

Find the missing figures in the following table

Solution:

Here y_{0} = 7; y_{1} = 11; y_{2} = ?; y_{3} = 18; y_{4} = ?; y_{5} = 32

Since only four values of f(x) are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.

Δ^{4}y_{k} = 0

(ie) (E – 1)^{4} y_{k} = 0

(i.e) (E^{4} – 4E³ + 6E² – 4E + 1)y_{k} = 0 ……….. (1)

Put k = 0 in (1)

(E^{4} – 4E³ + 6E² – 4E + 1)y_{0} = 0

E^{4} y_{0} – 4E^{3} y_{0} + 6E² y_{0} – 4E y_{0} + y_{0} = 0

y_{4} – 4y_{3} + 6y_{2} – 4y_{1} + y_{0} = 0

y_{4} – 4(18) + 6y_{2} – 4(11) + 7 = 0

y_{4} – 72 + 6y_{2} – 44 + 7 = 0

y_{4} + 6y_{2} = 109

(2)

Put k = 1 in (1)

(E^{4} – 4E^{3} + 6E² – 4E + 1)y_{1} = 0

[E^{4} y_{1} – 4E y_{1} + 6E² y_{1} – 4Ey_{1} + y] = 0

y_{5} – 4y_{4} + 6y_{3} – 4y_{2} + y_{1} = 0

32 – 4 (y_{4}) + 6(18) — 4(y_{2}) + 11 = 0

32 – 4y_{4} + 108 – 4y_{2} + 11 = 0

-4y_{4} – 4y_{2} + 151 = 0

4y_{4} + 4y_{2} = 151 ,……. (3)

Solving equation (1) & (2)

Substitute y_{2} = 14.25 in eqn (1)

y_{4} + 6(14.25) = 109

y_{4} + 25.50 = 109

y_{4} = 109 – 85.5

∴ y_{4} = 23.5

∴ Required two missing values are 14.25 and 23.5.

Question 5.

Find f (0.5) if f(-1) = 202, f(0) = 175, f(1) = 82 and f(2) = 55

Solution:

From the given data

Here we have to apply Newton’s forward interpolation formula, since the value of f(x) is required near the beginning of the table.

y_{(x= x0+nh)} =f(x_{0}) + \(\frac { n }{1!}\) Δf(x_{0}) + \(\frac { n(n-1) }{2!}\) Δ²f(x_{0}) + \(\frac { n(n-1)(n-2) }{3!}\) Δ³f(x_{0}) + ………

Given:

x = 0.5 and h = 1

x_{0} + nh = x

-1 + n(1) = 0.5

n = 1 + 0.5

∴ n = 1.5

= 202 – 40.5 – 24.75 – 8.25

= 202 – 73.5

f(0.5) = 128.5

Question 6.

From the following data find y at x = 43 and x = 84

Solution:

To find y at x = 43

Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.

= 184 + (0.3) (20) + (0.3) (-0.7)

= 184 + 6.0 – 0.21

= 190 + 0.21

y_{(x=43)} = 189.79

To find y at x = 84

Since the value of y is required at the end of the table, we apply backward interpolation formula.

x_{n} + nh = x

90 + n(10) = 84

10n = 84 – 90

10n = -6

∴ n = -0.6

y_{(x=84)} = 304 + \(\frac { (0.6) }{1!}\) (28) + \(\frac {(0.6)(-0.6 + 1) }{2!}\)(2) +

= 304 + (0.6) (28) + \(\frac { (-0.6)(0.4) }{2}\) + 2

= 304 – 16.8 – 0.24

= 304 – 17.04

= 286.96

Question 7.

The area A of circle of diameter ‘d’ is given for the following values

Find the approximate values for the areas of circles of diameter 82 and 91 respectively.

Solution:

To find A at D = 82

Since the value of A is required near the beginning of the table. We use the Newton’s forward interpolation formula.

= 5026 + 259.2 – 4.8 – 0.128 – 0.1664

= 5285.2 – 5.0944

= 5280.1056

A = 5280.11

To find Δ at D = 91

Since the value of A is required near the beginning of the table. We use the Newton’s forward interpolation formula.

= 7854 – 1378.8 + 28.8 + 0.096 + 0.0576

= 7882.9536 – 1378.8

= 6504.1536

= 6504.15

Question 8.

If u_{0} = 560, u_{1} = 556, u_{2} = 520, u_{4} = 385, show that u_{3} = 465

Solution:

U_{0} = 560; U_{1} = 556; U_{2} = 520; U_{4} = 385

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

Δ^{4}U_{0}

(E – 1)^{4} U_{0} = 0

⇒ (E^{4} – 4E³ + 6E² – 4E + 1) U_{0} = 0

⇒ E^{4}U_{0} – 4E³U_{0} + 6E²U_{0} – 4EU_{0} + U_{0} = 0

U_{4} – 4U_{3} + 6U_{2} – 4U_{1} + U_{0} = 0

385 – 4(U_{3}) + 6 (520) – 4 (556) + 560 = 0

385 – 4(U_{3}) + 3120 – 2224 + 560 = 0

1841 – 4U_{3} = 0

4U_{3} = 1841 ⇒ U_{3} = \(\frac { 1841 }{4}\)

U_{3} = 460.25

Question 9.

From the following table obtain a polynomial of degree y in x

Solution:

We will use Newton’s backward interpolation formula to find the polynomial.

To find y in terms of x

x_{n} + nh = x

5 + n(1) = x

∴ n = x – 5

= 1 + 2x – 10 + 2 (x² – 9x + 20) + \(\frac { 4 }{3}\) (x – 5) (x² – 7x + 12) + \(\frac { 2 }{3}\)(x² – 9x + 20)(x² – 5x + 6)

= 1 + 2x – 10 + 2x² – 18x + 40 + \(\frac { 4 }{3}\)

[x³ – 7x² + 12x – 5x² + 35x – 60] + \(\frac { 2 }{3}\) [x^{4} – 5x³ + 6x² – 9x³ + 45x² – 54x + 20x² – 100x + 120]

Question 10.

Using Lagrange’s interpolation formula find a polynominal which passes through the points (0, -12), (1, 0), (3, 6) and (4, 12).

Solution:

We can construct a table using the given points.

Here x_{0} = 0; x_{1} = 1; x_{2} = 3; x_{3} = 4,

y_{0} = -12; y_{1} = 0; y_{2} = 6; y_{3} = 12

= (x³ – 7x² + 12x – x² + 7x – 12) – (x³ – 5x² + 4x) + (x³ – 4x² + 3x)

= (x³ – 8x² + 19x – 12) – (x³ – 5x² + 4x) + (x³ – 4x² + 3x)

= x³ – 8x² + 19x – 12 – x³ + 5x² – 4x + x³ – 4x² + 3x

∴ y = x³ – 7x² + 18x – 12