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}

$\int\limits_0^{ – 5} {f(x)d} x + \int\limits_0^{ – 5} {2 – f(u)du}$

$\left[ \begin{array}{l} {\rm{Given, f(x) + f(x + 5) = 2}}\\ {\rm{Replacing x by u – 5, }}f(u – 5) + f(u) = 2 \end{array} \right]$

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

= –10 = -2a

Hence, (d).