KVPY SX SB Exam 20141 Question Paper

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Question : 14 of 120

Marks: +1, -0

For a real number x let [x] denote the largest integer less than or equal to x and {x} = x - [x] . Let n be a positive integer. Then

∫↖{n}↙{0}\cos(2π[x]\{x\}) \dx

is equal to

0
1
$n$
$2n - 1$

Solution:

∫↖{n}↙{0}\cos(2π[x]\{x\}) \dx

=

∫↖{1}↙{0}\cos(0) \dx+∫↖{2}↙{1}\cos( 2π (x- 1)) \dx

+∫↖{3}↙{2}\cos(4π (x- 2)) ) \dx+....

+∫↖{n}↙{n-1}\cos(2π (n -1)(x- (n -1)))\dx

= (1 - 0) +∫↖{2}↙{1}\cos2π x dx+∫↖{3}↙{2}\cos4 πx dx

+...+∫↖{n}↙{n-1}\cos(2π (n -1)x) dx

1+{{\sin2πx}/{2π}|}_1^2+{{\sin4πx}/{4π}|}_2^3 +....

+{{{\sin2π(n-1)x}/{2π(n-1)}|}^n}_{n-1}

= 1 + 0 = 1

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