Question

The period of real valued function f(x) + f(x + 4) = f(x + 2) + f(x + 6)

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\purple{\rm :\longmapsto\:\bf{ \: f(x) + f(x + 4) = f(x + 2) + f(x + 6) \: }}$

Period means after what value of x, the function starts repeating itself. If for a real valued function f(x), if f(x + T) = f(x) then T is called period of f(x).

So, to find the period of f(x),

Let replace x by x + 2, we get

$\purple{\rm :\longmapsto\:{f(x + 2) + f(x + 2 + 4) = f(x + 2 + 2) + f(x +2 + 6) \: }}$

$\purple{\rm :\longmapsto\:{f(x + 2) + f(x + 6) = f(x +4) + f(x + 8) \: }}$

On adding these two equations, we get

$\purple{\rm{f(x) + f(x + 4) + f(x + 2) + f(x + 6) \: \: = }} \\ \purple{ \rm{f(x + 2) + f(x + 4) + f(x + 6) + f(x + 8)}}$

$\purple{\bf\implies \:f(x) = f(x + 8)}$

Hence, period of real valued function is 8.

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf T - Function & \bf Period \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx & \sf 2\pi \\ \\ \sf cosx & \sf 2\pi \\ \\ \sf sin(ax + b) & \sf \dfrac{2\pi}{ |a| } \\ \\ \sf cos(ax + b) & \sf \dfrac{2\pi}{ |a| } \end{array}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

Correct option is

8

Given f(x)+f(x+4)=f(x+2)+f(x+6) ....(1)

Replacing x by x−2

f(x−2)+f(x+2)=f(x)+f(x+4) ....(2)

Adding (1) and (2), we get

f(x−2)=f(x+6)

Replacing x by x+2, we get

f(x)=f(x+8)

Hence, period is 8