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,
$\int\limits_0^{ - 2a} {f(x)d} x$
$= \int\limits_0^{ - 10} {f(x)d} x$
$= \int\limits_0^{ - 5} {f(x)d} x + \int\limits_{ - 5}^{ - 10} {f(x)d} x$
$= \int\limits_0^{ - 5} {f(x)d} x + \int\limits_0^{ - 5} {f(u - 5)du}$ , {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{x \to u - 5, }}f(u - 5) + f(u) = 2\end{array} \right] $
$= \int\limits_0^{ - 5} {f(x)d} x + \int\limits_0^{ - 5} {2 - f(x)dx} ,$ {Replacing u by x in the second integral}
= –10 = -2a
Hence, (d).