First slide

Functions (XII)

Question

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

Moderate

A

$\frac{17}{6}$
B

$\frac{56}{6}$
C

$\frac{31}{3}$
D

$\frac{35}{3}$

Solution

$f (x) = α x^{3} - β x - \tan x \cdot S g n (x), w h e r e α = {[a^{}]}^{2} - 5 [a] + 4, β = 6 {\{a\}}^{2} - 5 \{a\} + 1$
$\sin c e f i s e v e n f u n c t i o n \Rightarrow f (- x) = f (x) \Rightarrow - α x^{3} + β x - \tan x \cdot S g n (x) = α x^{3} - β x - \tan x \cdot S g n (x)$
$\Rightarrow 2 (α x^{3} - β x) = 0 f o r a l l x \in R$
$\Rightarrow α = 0 a n d β = 0 (I d e n t i t y i n x)$
$t h e r e f o r e {[a]}^{2} - 5 [a] + 4 = 0 a n d 6 {\{a\}}^{2} - 5 \{a\} + 1 = 0 \Rightarrow [a] = 1 o r [a] = 4 a n d \{a\} = \frac{1}{3} o r \frac{1}{2}$
$\Rightarrow a = 1 + \frac{1}{3}, 1 + \frac{1}{2}, 4 + \frac{1}{3}, 4 + \frac{1}{2}$
$\Rightarrow s u m o f t h e v a l u e s o f a = \frac{35}{3}$

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