Question

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

• A. $-5$
• B. $0$
• C. $\lambda$
• D. $-2\lambda$

Answer 1

The correct option is D $-2\lambda$

Given equations,

$2x-y+z=0\to \left(1\right)$

$x-2y+z=0\to \left(2\right)$

$\lambda x-y+2z=0\to \left(3\right)$

Given system of equations has infinitely many solutions.

$\therefore \left(1\right)\times a+\left(2\right)\times b=\left(3\right)$

$2a+b=-1$

$-a-2b=-1$

$a+b=2$

$(-a-b)+(a+b)=-1+2\Rightarrow b=-1,a=3$

$\lambda =2\left(3\right)-1=5$

Given $f(x+5)+f\left(x\right)=2$

${\int }_0^{-10}f\left(x\right)dx={\int }_0^{-10}f\left(x\right)dx$

$=-{\int }_{-10}^{0}f\left(x\right)dx$

$=-[{\int }_{-10}^{-5}f\left(x\right)dx+{\int }_{-5}^{0}f\left(x\right)dx]$

$=-[{\int }_{-10}^{-5}f\left(x\right)dx+{\int }_{-10}^{-5}f(x+5)dx]$

$=-\left[{\int }_{-10}^{-5}f\left(x\right)+f(5+x)dx\right][\because d(x+5)=dx]$

$=-{\int }_{-10}^{-5}2dx=-2(-5-(-10\left)\right)$

$\therefore {\int }_0^{-2\lambda }f\left(x\right)dx=-10$