Question

If the system of equation $2x-y+z=0,x-2y+z=0,tx-y+2z=0$ has infinitely many solutions and f(x) be a continuous function, such that $f(5+x)+f\left(x\right)=2$, then ${\int }_0^{-2t}f\left(x\right)dx=$

• A. $0$
• B. $-2t$
• C. $5$
• D. $t$

Answer 1

Answer: (B)

$-2t$

The set of equations have infinitely many solutions. Hence,
$\left|\begin{array}{ccc}2& -1& 1\\ 1& -2& 1\\ t& -1& 2\end{array}\right|=0$
$\Rightarrow -6+2-t-1+2t=0$
$\Rightarrow t=5$
$I={\int }_0^{-2t}f\left(x\right)dx$
$\Rightarrow I={\int }_0^{-10}f\left(x\right)dx$
$\Rightarrow I={\int }_0^{-5}f\left(x\right)dx+{\int }_{-5}^{-10}f\left(x\right)dx$
For the second integral, substitute $x+5=u$
$\Rightarrow I={\int }_0^{-5}f\left(x\right)dx+{\int }_0^{-5}f(u+5)dt$
$\Rightarrow I={\int }_0^{-5}f\left(x\right)dx+{\int }_0^{-5}2-f\left(u\right)dt$
Changing the variable in the second integral,
$\Rightarrow I={\int }_0^{-5}f\left(x\right)dx+{\int }_0^{-5}2-f\left(x\right)dx$
$\Rightarrow I={\int }_0^{-5}2dt$
$\Rightarrow I=-10$
Hence, option B is correct.