Question

find a positive real root of x-cosx=0 by bisection method correct up to four decimal places between 0 and 1​

Answer 1

0.75 is the positive real root of x-cosx=0 by bisection method.

Given,

x - cos x = 0.

To Find,

Positive real roots of x - cos x = 0 by bisection method.

Solution,

Assume the function f(x) = x - cos x

∴ f(0) = 0 - cos 0 = -1

∴ f(1)  = 1 -  cos 1  = 0.46

∴ Root lies between [0,1]

∴ c = $\frac{a +b}{2}$

    =  $\frac{0 + 1}{2}$

    = 0.5

∴ x = 0.75 is the required root.

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Answer 2

0.75 is the positive real root of  x - cos x  by bisection method corrected up to four decimal places between 0 and 1​.

Let, f(x) = x - cos x

⇒ f(0) = 0 - cos 0 = -1

⇒ f(1) = 1 - cos 1 = 0.46

∴ the root of  f(x) = x - cos x lies between [0,1].

$c = \frac{a+b}{2}$

   =  $\frac{0 + 1}{2}$

   = 0.5

∴ 0.75 is the positive real root of  x - cos x  by bisection method corrected up to four decimal places between 0 and 1​.

#SPJ2