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GATE | GATE CS 2008 | Question 21

  • Last Updated : 16 Feb, 2018

The minimum number of equal length subintervals needed to approximate 24 to an accuracy of at least 25 using the trapezoidal rule is
(A) 1000 l
(B) 1000
(C) 100 l
(D) 100


Answer: (A)

Explanation: Trapezoidal rule error:
 E_n = -\frac{(b-a)^3}{12N^2}f^{''}(c)

Maximum error = 1/3 * 10^{-6} (given)

Therefore, |En| < 1/3 * 10^{-6}

a = 1 and b = 2 (given)

Therefore,

 N^2 > \frac{(b-a)^3}{12*\frac{1}{3}*10^{-6}}|f^n(c)| = \frac{(2-1)^3}{4*10^{-6}}|f^n(c)| = \frac{10^6}{4}|f^n(c)
 N > \frac{10^3}{2}\sqrt{|f^n(c)|}



f(x) = xe^x + 2e^x

f(x) is maximum at x=2.
Therefore, f(x) = 4e^2

 N > \frac{10^3}{2}\sqrt{4e^2} = \frac{10^3}{2}*2e = 1000e

Thus, option (A) is correct.

Reference: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2008.html

Please comment below if you find anything wrong in the above post.

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