Question

If f(x)= x^2-166=0 then the iterative formula for Newton raphson method is

Answer 1

Answer:

Answer:

Solution:

Let the number of Rs.50 notes be A and the number of Rs.100 notes be B

According to the given information,

A + B = 25 ……………………………………………………………………….. (i)

50A + 100B = 2000 ………………………………………………………………(ii)

When equation (i) is multiplied with (ii) we get,

50A + 50B = 1250 …………………………………………………………………..(iii)

Subtracting the equation (iii) from the equation (ii) we get,

50B = 750

B = 15

Substituting in the equation (i) we get,

A = 10

Hence, Manna has 10 notes of Rs.50 and 15 notes of Rs.100.Answer:

Solution:

Let the fixed charge be Rs x and per km charge be Rs y.

According to the question,

x + 10y = 105 …………….. (1)

x + 15y = 155 …………….. (2)

From (1), we get x = 105 – 10y ………………. (3)

Substituting the value of x in (2), we get

105 – 10y + 15y = 155

5y = 50

y = 10 …………….. (4)

Putting the value of y in (3), we get

x = 105 – 10 × 10 = 5

Hence, fixed charge is Rs 5 and per km charge = Rs 10

Charge for 25 km = x + 25y = 5 + 250 = Rs 255

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∣hopethishelpsu∣

Answer:

Solution :-

↦Let, the first number be x

↦ Second number be x + 1

↦ Third number be x + 2

↦ Fourth number be x + 3

↦ Fifth number be x + 4

↦ And, the addition of five consecutive numbers is 270.

According to the question,

⇒ x + x + 1 + x + 2 + x + 3 + x + 4 = 270

⇒ x + x + x + x + x + 1 + 2 + 3 + 4 = 270

⇒ 5x + 10 = 270

⇒ 5x = 270 - 10

⇒ 5x = 260

⇒ < /p > < p > x = < /p > < p > \dfrac{260}{5} \: ⇒</p><p>x=</p><p>

5

260

➠x=52

Hence, the required numbers are,

✦ First number = x = 52

✦ Second number = x + 1 = 52 + 1 = 53

✦ Third number = x + 2 = 52 + 2 = 54

✦ Fourth number = x + 3 = 52 + 3 = 55

✦ Fifth number = x + 4 = 52 + 4 = 56

But, in question we have to find the addition of second and fifth number,

➳ Second number + Fifth number

➳ 53 + 56

\implies < /p > < p > 109 \: ⟹</p><p>109

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hopethishelpsu

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Hope this helps u

Answer 2

Iterative formula for newton raphson method is

$x_{n + 1} \: = x_{n} - \frac{ f(x_{n}) }{ {f}^{n} (x_{n} )}$

where, fⁿ(xₙ) is the derivative of f(xₙ).

Here f(x) = x² - 166 = 0

Its derivative is 2x

at x = 0 , 2x = 0.

Thus its derivative doesnt exist at x = 0 .

In this case , newton raphson method fails since the function is not analytic everywhere.

Hence iterative formula cannot be found for the function f(x) = x² - 166

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