KVPY SX SB Exam 2015 Question Paper

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Question : 14 of 120

Marks: +1, -0

Let x > 0 be a fixed real number. Then the integral

∫↖{∞}↙{0}e^{–1} {| x - t |} dt

is equal to -

$x + 2e^{-x} - 1$
$x-2e^{-x }+ 1$
$x + 2e^{-x} + 1$
$- x - 2e^{-x} + 1$

Solution:

\f(x)=∫↖{∞}↙{0}e^{–t} {| x - t |} dt

\f(x)=∫↖{x}↙{0}e^{–t} ( x - t ) dt+ ∫↖{∞}↙{x}e^{–t} (t-x) dt

f'(x) =e^{–x} (x - x) - e ^{- 0} (x- 0). 0

+∫↖{x}↙{0}e^{–t}(1)dt+0- e ^{-x} (x- x).1 +∫↖{∞}↙{x}-e^{–t}dt

=

[-e^{-t}]^x_0+[-e^{-t}]^∞_x

= – e ^{–x} + 1 + 0- e ^{–x}

\f'(x) = 1- 2e^{–x}

dy = 1- 2e^{–x}dx

y = x + 2e^{–x} + c

\f(0) =∫↖{∞}↙{0}e^{-t} t\ dt

=[(-e^{-t})t]^∞_0+∫↖{∞}↙{0}e^{-t} dt

=[0-e^{-t }]^{∞} _{0}

= 0 + 1
f(0) = 1
f(0) = 1 = 0 + 2e

^{–0}

+ c
c = –1
f(x) = x + 2e

^{–x}

– 1

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