Mathematics - Functions Question with Solution | TestHub

Since period of f(x) is $2(10-2)=16$ $\therefore \mathrm{f}(0)=\mathrm{f}(16)=\mathrm{f}(32)=\dots =\mathrm{f}(160=5$ $\therefore$ there are atleast 11 values of x for which $\mathrm{f}(\mathrm{x})=5$ $\mathrm{F}(0)=\mathrm{f}(4)=\mathrm{f}(16)$ Due to symmetry in one period length $\mathrm{f}(\mathrm{x})=5$ one solution other then 0, 16, 32, $\_\_\_\_$ ∴ at least minimum possible number of values of x is $10+11=21$