Question

QUESTION 1 (a) Show that the equation (x - 2) = has a root between x = 2 and x = 3. Using the x+2 first approximation as 2.7 and the Newton-Raphson method, calculate this root correct to two decimal places. (8 marks) (b) Show that e' +x-2 = 0 has a root in interval [0, 1]. Using basic iteration method, calculate this root correct to four decimal places. (12 marks) 1 (C) Find an approximate value for the integral S dx using trapezoidal rule with 5 1+ x equal strips. (5 marks)

Answer 1

(x-2)² = x x+2 (x-2) ² - x = 0 = f(x) let. x+2 f(2) (2-2² - 2 -0.5 1 2+2 f(3) = (3-2}- 3 = 1-3 ترال 3+2 0.4 a 2 f(3) = 2 f (2) is negative & f(3) is positive, Therefore there must be value between 2 & 3 at which f(x) is zero. Newton's Raphson Method 26=2.7 f(x) = (xe- 2²-a x+2 Il f(x) = 2(x-2) - 2 (x+2)2 iteration » f(xo) = f(2.7) = (2.7-2) ²_ 2.7 = -0.08 2.7 +2 f'(x) = f'(2-7)= ax (2.7-2) - 2 2 = 1:31 (2.7+2) x= floo) to- = 2.7- -0.08 1:31 x1= 2.76

and iteration flo,) = f(2-76)= (2.76-2) ² - 2.76 2.76+2 flol)=0 :. the root is "=2.76 for (x-2)²= a+2 e +2-2 let fow f(o)= e to-2=-1 f(1) = e +1-2 = 1.72 fo) co l+1) > 0 there must be in interval to, 1] . a root Uxing Newton Raphson Method : flow= e²+2-2 f(x)=e" + 26=0 - Ist Iteration f(x) = flo)= e'+0-2=-1 f'(x) = f'(o)=8+1=2 21 = No - f(ao) 아을 f'(o) 2,= 0.5 and iteration - f(x) = f(o.s) = 0.1487 f'(X) = f'(0.5) = 2.6487 = 0.5- 0.1487 f(x) 2.6487 X₂= 0.4439 x2 = x, f(x)

Ird stelation 0.0026 f(x) = f(0.4439)= fG) = f(0.4439) = 2.5587 X₃ = H2- f(2) = 0.4439 - 0.0026 2.5587 F'(x₂) &z= 0.4429 Tth iteration . 0.4429 f(xg) = f(0.4429) = ? +0.4429 -2=0 flo. 4429)=0 root is x=0.4429 an dal It su 0 no. of equal strips ,n=5 step size ,h= b-a 4-o 11 0.8 Table for a dys 0 0.8 2. 4 3.2 4 0.4415 0.5279 0.3923 0.3586 - 0.3333 ac Using trapezoidal Rule I- h [ Got Ys+ 2(y,t yz + 43+ YA)] 0.3333 + 210.5279+0.4415 +0.3923 +0.3586)] 1 - 02 [1+0.3333 +2x17202] 0:9 [ + I= 1.9095