Question

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

• A. $\frac{17}{6}$
• B. $\frac{53}{6}$
• C. $\frac{31}{3}$
• D. $\frac{35}{3}$

Answer 1

The correct option is D $\frac{35}{3}$

Let $\alpha =[a{]}^2-5[a]+4,\beta =6\{a{\}}^2-5\left\{a\right\}+1,$ then
$f\left(x\right)=\alpha x^3+\beta x-\left(\tan x\right)\times \text{sgn}x$

$\because f\left(x\right)$ is an even function
$\therefore f(-x)=f\left(x\right)$
$\Rightarrow -\alpha x^3-\beta x-(-\tan x)\times (-\text{sgn}x)=\alpha x^3+\beta x-\left(\tan x\right)\times \text{sgn}x$
$\Rightarrow -2(\alpha x^2+\beta )x=0\forall x\in R$
$\Rightarrow \alpha =0,\beta =0$
$\Rightarrow [a{]}^2-5[a]+4=0\text{and}6\{a{\}}^2-5\left\{a\right\}+1=0$
$\Rightarrow \left(\right[a]-1)\left(\right[a]-4)=0\text{and}\left(3\right\{a\}-1)\left(2\right\{a\}-1)=0$
$\Rightarrow a=1+\frac{1}{3},1+\frac{1}{2},4+\frac{1}{3},4+\frac{1}{2}$
$\Rightarrow a=\frac{4}{3},\frac{3}{2},\frac{13}{3},\frac{9}{2}$

$\therefore$ Sum of values of $a$
$=\frac{35}{3}$