Question

Let $f\left(x\right)=\left(\right[a{]}^2-5\left[a\right]+4)x^3-(6\{a{\}}^2-5\{a\}+1)x-\left(\tan x\right)\times sgnx$ be an even function for all $x\in R$, then the sum of all possible values of ${}^{\prime }3a^{\prime }$
is
(where, [.] and {.} denote the greatest integer function and fractional part functions, respectively)

Answer 1

Let $\alpha =[a{]}^2-5[a]+4and\beta =6\{a{\}}^2-5\left\{a\right\}+1$
$\therefore f\left(x\right)=\alpha x^3-\beta x-\left(\tan x\right)\times sgnx$
Also, as $f\left(x\right)$ is an even function.$\therefore f(-x)=f\left(x\right)$
$\Rightarrow -\alpha x^3+\beta x-\left(\tan x\right)\times sgnx=\alpha x^3-\beta x-\left(\tan x\right)\times sgnx$
$\Rightarrow 2(-\alpha x^2-\beta )x=0forallx\in R$
$\Rightarrow \alpha =0and\beta =0$
$\therefore [a{]}^2-5[a]+4=0and6\{a{\}}^2-5\left\{a\right\}+1=0$
$\Rightarrow \left(\right[a]-1)\left(\right[a]-4)=0and\left(3\right\{a\}-1)\left(2\right\{a\}-1)=0$
$\Rightarrow \left[a\right]=1or4and\left\{a\right\}=\frac{1}{3}or\frac{1}{2}$
$\Rightarrow a=1+\frac{1}{3},1+\frac{1}{2},4+\frac{1}{3},4+\frac{1}{2}$
Therefore, sum of all possible values of a is $\frac{35}{3}\therefore 3a=35$