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Mathematics Problem & Solution

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).