Let `f(x)=([a]^2-5[a]+4)x^3-(6{a}^2-5{a}+1)x-(tanx)xsgnx` be an even function for all `x in Rdot` Then the sum of all possible values of `a` is (where `[dot]a n d{dot}` denote greatest integer function and fractional part function, respectively). `(17)/6` (b) `(53)/6` (c) `(31)/3` (d) `(35)/3`

Options

A. `(17)/(6)`
B. `(53)/(6)`
C. `(31)/(3)`
D. `(35)/(3)`

Explanation

To determine the sum of all possible values of \( a \) such that the function

\[ f(x) = (a^2 - 5a + 4)x^3 - (6a^2 - 5a + 1)x - (\tan x)x \sgn x \]

is an even function for all \( x \in \mathbb{R} \), we need to analyze the conditions for \( f(x) \) to be even.

### Step 1: Identify the conditions for \( f(x) \) to be even

A function \( f(x) \) is even if \( f(-x) = f(x) \) for all \( x \). Therefore, we need to compute \( f(-x) \):

\[ f(-x) = (a^2 - 5a + 4)(-x)^3 - (6a^2 - 5a + 1)(-x) - (\tan(-x))(-x) \sgn(-x) \]

Using the properties of odd functions, we know \( \tan(-x) = -\tan(x) \) and \( \sgn(-x) = -\sgn(x) \). Thus, we have:

\[ f(-x) = -(a^2 - 5a + 4)x^3 + (6a^2 - 5a + 1)x + (\tan x)x \sgn x \]

### Step 2: Set \( f(-x) = f(x) \)

Equating \( f(-x) \) and \( f(x) \):

\[ -(a^2 - 5a + 4)x^3 + (6a^2 - 5a + 1)x + (\tan x)x \sgn x = (a^2 - 5a + 4)x^3 - (6a^2 - 5a + 1)x - (\tan x)x \sgn x \]

### Step 3: Simplify the equation

This gives us two separate equations by equating coefficients:

1. Coefficient of \( x^3 \):
\[ -(a^2 - 5a + 4) = a^2 - 5a + 4 \]
This simplifies to:
\[ 2(a^2 - 5a + 4) = 0 \]
Thus, we have:
\[ a^2 - 5a + 4 = 0 \]

2. Coefficient of \( x \):
\[ (6a^2 - 5a + 1) = -(6a^2 - 5a + 1) \]
This simplifies to:
\[ 2(6a^2 - 5a + 1) = 0 \]
Thus, we have:
\[ 6a^2 - 5a + 1 = 0 \]

### Step 4: Solve the quadratic equations

**For the first equation**:
\[ a^2 - 5a + 4 = 0 \]
Using the quadratic formula:
\[ a = \frac{5 \pm \sqrt{(5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm 3}{2} \]
This gives:
\[ a = 4 \quad \text{or} \quad a = 1 \]

**For the second equation**:
\[ 6a^2 - 5a + 1 = 0 \]
Using the quadratic formula:
\[ a = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot 1}}{2 \cdot 6} = \frac{5 \pm \sqrt{25 - 24}}{12} = \frac{5 \pm 1}{12} \]
This gives:
\[ a = \frac{6}{12} = \frac{1}{2} \quad \text{or} \quad a = \frac{4}{12} = \frac{1}{3} \]

### Step 5: Sum all possible values of \( a \)

The possible values of \( a \) are \( 4, 1, \frac{1}{2}, \frac{1}{3} \). Now, we sum these values:

\[ 4 + 1 + \frac{1}{2} + \frac{1}{3} = 5 + \frac{3}{6} + \frac{2}{6} = 5 + \frac{5}{6} = \frac{30}{6} + \frac{5}{6} = \frac{35}{6} \]

### Final Answer

The sum of all possible values of \( a \) is \( \frac{35}{6} \).