CGPET 2016 Solved Paper

By Newton Raphson Formula,
Xn+1=Xn−

f(xn)

f′(xn)

∴ Xn+1=Xn−

xn3−2xn−5

3xn2−2

n=0,1,2, .....
Let x0=2 .......(i)
Then, f(x0)=f(2)=23−2(2)−5=−1
and f′(x0)=f′(2)=3(2)2−2=10
By putting n=0 in Eq. (i),
x1=2−(

−1

10

)=2.1
⇒ f(x1)=f(2.1)=(2.1)3−2(2.1)5=0.061
f′(x1)=f′(2.1)=3(2.1)2−2=1.23
x2=2.1−

0.061

11.23

=2.094
So, the positive roots of the equation lies in interval (2, 3).