Question

Let $f(x)$ be a linear function such that $f(0)=-5$ and $f(f(0))=-15$. The number of values of $m$ for which the solutions of the inequality $f(x)f(m-x)\ge 0$ form an interval of length 2 , is.

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

Let $f(x)=ax+b$
$f(0)=-5\Rightarrow b=-5$
$f(f(0))=-15\Rightarrow a=2$
$\therefore \mathrm{f}(\mathrm{x})=2\mathrm{x}-5$
Now $f(x)\cdot f(m-x)\ge 0$
$\therefore \left(\mathrm{x}-\frac{5}{2}\right)\left(\mathrm{x}-\left(\mathrm{m}-\frac{5}{2}\right)\right)⩽0$
Case I $\frac{5}{2}>m-\frac{5}{2}$, here length of interval
$=\frac{5}{2}-\mathrm{m}+\frac{5}{2}=2\Rightarrow \mathrm{m}=3$
Case II $\frac{5}{2}<\mathrm{m}-\frac{5}{2}$, here length of interval
$=m-\frac{5}{2}-\frac{5}{2}=2$
$\Rightarrow m=7$
$\therefore$ two values of $\mathrm{m}$