KVPY SX SB Exam 2015 Question Paper

Question : 12 of 120

Marks: +1, -0

Define a function f : R → R by
f(x) = max

\{|x|, |x - 1|, ... , |x - 2n|\}

where n is a fixed natural number. Then

∫↖{2n} ↙{0} f(x)dx

is -

Solution:

∫↖{2n}↙{0}\f (x)dx =∫↖{ n}↙{0}\f(x)dx +∫↖{2n}↙{n}\f (x)dx

∫↖{ n}↙{0}{|x- 2n|}dx+∫↖{2n}↙{n} {|x |}dx

∫↖{ n}↙{0}{| 2n-x|}dx+∫↖{2n}↙{n} x . dx

=[2nx-x^2/2] ^ n _0+[x^2/2]^{2n}_n

=(2n^2-n^2/2)+({4n^2}/2-n^2/2)

={3n^2}/2+{3n^2}/2=3n^2

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