# GATE2012-28

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The bisection method is applied to compute a zero of the function $f(x) =x ^{4} – x ^{3} – x ^{2} – 4$ in the interval [1,9]. The method converges to a solution after ––––– iterations.

(A) 1
(B) 3
(C) 5
(D) 7
0
actually if they would have given 2 as option then as per me if i am applying Binary search then the answer should then have be 2.

I illustrate it below:

1,2,3,4,5,6,7,8,9

First Binary Search Algo:

Binary Search(a,x)

{

do{

mid=floor(beg+end/2);

if(a[mid]==x)

return TRUE

else if(x>a[mid])

lo=mid+1

else

hi=mid-1

}while(beg<=end)

}

Considering array indices startted at 1 and went to 9

lo=1

hi=9

mid=floor(1+9/2)=5 @ f(5) !=0 Fail

lo=1

hi=4 (its pritty intuitive why i chose lower half!)

mid=floor(1+4/2)=2 @ f(2)=0 <----tada!!! we have got it

SO just 2 is the answer

THE QUESTION MUST BE DONE THIS WAY IF WE ARE USING BINARY SEARCH I FEEL IF ANY BODY HAS A DFFERENT OPINION THEN PLZ LET ME KNOW

Bisection method is exactly like binary search on a list.

In bisection method, in each iteration, we pick the mid point of the interval as approxiamation of the root, and see where are we, i.e. should we choose left sub-interval, or right-subinterval, and we continue until we find the root, or we reach some error tolerance.

So in first iteration, our guess for root is mid point of [1,9] i.e. 5. Now f(5) > 0, so we choose left sub-interval [1,5] (as any value in right sub-interval [5,9] would give more positive value of $f$).

In second iteration, we choose mid point of [1,5] i.e. 3, but again f(3) > 0, so we again choose left sub-interval [1,3].

In third iteration, we choose mid point of [1,3] i.e. 2, now f(2) = 0

So we found root in 3 iterations. So answer is option (B).

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0
is it still in the syllabus?
0
Not in syllabus
0
okk..thanks buddy!
0
newton rapson method, simpson or kutta methods, etc What else are not in syllabus. Thanks..

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