Question

Let $f:R\to R$ be a differentiable function satisfying $f\left(\frac{x+y}{3}\right)=\frac{2+f\left(x\right)+f\left(y\right)}{3}\forall x,y\in R$ and $f^{\prime }\left(2\right)=2$, then answer the following questions:

If $g\left(x\right)=x^4-{f\left(\left|x\right|\right)}^2-6;$ then the value of $g(k-1)$, where $k$ denotes the solution as well as number of solutions of equation $g\left(x\right)=0$ is

• A. $-9$
• B. $8$
• C. $55$
• D. None of these

Answer 1

Answer: (A)

$-9$

Substitute $x=3x,y=0$ in $f\left(\frac{x+y}{2}\right)=\frac{2+f\left(x\right)+f\left(y\right)}{3}$, we get

$f\left(\frac{3x+0}{3}\right)=\frac{2+f\left(3x\right)+f\left(0\right)}{3}\Rightarrow f\left(x\right)=\frac{2+f\left(3x\right)+f\left(0\right)}{3}$ ...(1)

Substitute $x=y=0$

$f\left(0\right)=\frac{2+f\left(0\right)+f\left(0\right)}{3}\Rightarrow 3f\left(0\right)=2+2f\left(0\right)\Rightarrow f\left(0\right)=2$ ....(2)

From (1)

$f\left(x\right)=\frac{2+f\left(3x\right)+2}{3}\Rightarrow 3f\left(x\right)=f\left(3x\right)+4$ ...(3)

Now $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(\frac{3x+3h}{3}\right)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\left\{\frac{\frac{2+f\left(3x\right)+f\left(3h\right)}{3}-f\left(x\right)}{h}\right\}=\underset{h\to 0}{lim}\left[\frac{2+f\left(3x\right)+f\left(3h\right)-3f\left(x\right)}{3h}\right]$

$=\underset{h\to 0}{lim}\left[\frac{2+3f\left(x\right)-4+f\left(3h\right)-3f\left(x\right)}{3h}\right]$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\left[\frac{f\left(3h\right)-2}{3h}\right]$

Since $f\left(x\right)$ is differentiable $\forall x;{lim}_{h\to 0}f\left(3h\right)=2$

$\Rightarrow f\left(0\right)=2\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{3f^{\prime }\left(3h\right)}{3}=f^{\prime }\left(0\right)=k$ (say)

$\Rightarrow f^{\prime }\left(x\right)=k\Rightarrow f\left(x\right)=kx+c$

$f^{\prime }\left(x\right)=k\Rightarrow f^{\prime }\left(2\right)=k$

But $f^{\prime }\left(2\right)=2\Rightarrow k=2$

$\therefore f\left(x\right)=2x+c$

Also $f\left(0\right)=2\Rightarrow c=2$

$\therefore f\left(x\right)=2x+2$

$g\left(x\right)=x^4-f{\left(\left|x\right|\right)}^2-6$$=x^4-(2{\left|x\right|}^2+2)-6$$=x^4-2{\left|x\right|}^2-8$

$={\left|x\right|}^4-2{\left|x\right|}^2-8$$=({\left|x\right|}^2-4)({\left|x\right|}^2+2)$

$\therefore g\left(x\right)=0$

$\Rightarrow {\left|x\right|}^2=4$ or ${\left|x\right|}^2=-2$

$\Rightarrow {\left|x\right|}^2=4(\because \left|x\right|\ge 0)$

$\Rightarrow \left|x\right|=\pm 2(\because \left|x\right|\ne -2)$

$\Rightarrow \left|x\right|=2$$\Rightarrow x=\pm 2$

$\therefore$ Solution of equation $g\left(x\right)=0$ are $2$ and $-2$

$\Rightarrow$ there are two solutions of equation $g\left(x\right)=0$

$\because k$ is the value of solutions as well as number of solutions

$\Rightarrow k=2$

$\therefore g(k-1)=g(2-1)=g\left(1\right)$

$\therefore g(k-1)=g\left(1\right)={\left(1\right)}^4-2{\left|1\right|}^2-8=1-2-8=-9$