Mathematics - Functions Question with Solution | TestHub

$f(x-1)+f(x+1)=\sqrt{3}\cdot f(x)$, ......(1)

Putting $x=x+1$

$$f(x)+f(x+2)=\sqrt{3}\cdot f(x+1)\dots \dots \dots \left(2\right)$$

Again, putting $x=x+1$:

$$f(x+1)+f(x+3)=\sqrt{3}\cdot f(x+2)\dots \dots \dots \left(3\right)$$

      - (4)

Continning like this we get

f (x) is a periodic function with period ‘12’

Then:

$$\begin{array}{rl}g(x)=\sum _{r=0}^{99}f(5+12r)& =100\times f(5)\\ & =1000\end{array}$$

This implies that $g(x)$ is a constant function. Hence, $g^{\prime }(x)=0$.

$f(x-1)+f(x+1)=\sqrt{3}\cdot f(x)$ ......(1)

Putting $x=x+1$:

$$f(x)+f(x+2)=\sqrt{3}\cdot f(x+1)\dots \dots \dots \left(2\right)$$

Again, putting $x=x+1$:

$$f(x+1)+f(x+3)=\sqrt{3}\cdot f(x+2)\dots \dots \dots \left(3\right)$$

$$f(x+2)-f(x-1)=\sqrt{3}[f(x+1)-f(x)]..\dots \left(4\right)$$

$$f(x+3)-f(x)=\sqrt{3}[f(x+2)-f(x+1)]$$

Continning like this we get

f (x) is a periodic function with period ‘12’

Then:

$$\begin{array}{rl}g(x)=\sum _{r=0}^{99}f(5+12r)& =100\times f(5)\\ & =1000\end{array}$$

This implies that $g(x)$ is a constant function. Hence, $g^{\prime }(x)=0$.