Question

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

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

Answer 1

Answer: (D)

$\frac{35}{3}$

$x^3$and are odd functions. Also $\tan x$ and $\text{sgn}x$ are odd functions; therefore $\left(\tan x\right)\text{sgn}x$ is an even function. So for $f\left(x\right)$ to be an even function, coefficients of $x^3$and $x$ must be equal to zero.

$\Rightarrow [a{]}^2-5[a]+4=0$.........$\left(1\right)$

$6{\left\{a\right\}}^2-5\left\{a\right\}+1=0$.......$\left(2\right)$

And $a=\left[a\right]+\left\{a\right\}$..................$\left(3\right)$

From $\left(1\right)$ and $\left(2\right)$, $\left[a\right]=1,4$ and $\left\{a\right\}=\frac{1}{2},\frac{1}{3}$

Hence, $a$ can take values $\frac{3}{2}$,$\frac{9}{2}$, $\frac{4}{3}$ and $\frac{13}{3}$.

Their sum is $\frac{35}{3}$.