\int\limits_0^{ - 2a} {f(x)dx}  = .........

If the system of equations 2x – y + z =0, x- 2y + z = 0 and ax – y + 2z = 0 has infinitely many solutions and f(x) is continuous function satisfying f(x)+f(x+5) = 2, then

\int\limits_0^{ - 2a} {f(x)dx}
is equal to

a) 0
b) 5
c) a
d) –2a

Solution

For system of given equations to have infinitely many solutions, we must have

\left| {\begin{array}{ccccccccccccccc} 2&{ - 1}&1\\ 1&{ - 2}&1\\ a&{ - 1}&2 \end{array}} \right| = 0

or, 2x(-3) + 2-a + (-1+2a) = 0

or, a = 5

Now,

\int\limits_0^{ - 2a} {f(x)d} x

 = \int\limits_0^{ - 10} {f(x)d} x

 = \int\limits_0^{ - 5} {f(x)d} x + \int\limits_{ - 5}^{ - 10} {f(x)d} x

 = \int\limits_0^{ - 5} {f(x)d} x + \int\limits_0^{ - 5} {f(u - 5)du} ,{\rm{ putting x + 5 = u for the second integral}}

 = \int\limits_0^{ - 5} {f(x)d} x + \int\limits_0^{ - 5} {2 - f(u)du} ,                              \left[ \begin{array}{l} {\rm{Given, f(x) + f(x + 5) = 2}}\\ {\rm{Replacing x by u - 5, }}f(u - 5) + f(u) = 2 \end{array} \right]

 = \int\limits_0^{ - 5} {f(x)d} x + \int\limits_0^{ - 5} {2 - f(x)dx} ,{\rm{ Replacing u by x in the second integral}}

= –10 = -2a

Hence, (d).

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