Question

Solve
$\frac{x}{ {x}^{4} - 6 {x}^{3} + 11 {x}^{2} - 6x } \leqslant 0$
​

Answer 1

Step-by-step explanation:

$Given:\frac{x}{x^4-6x^3 + 11x^2-6x} \leq0$

$=\frac{x}{x(x^3 - 6x^2 + 11x - 6)} \leq0$

$=\frac{x}{x(x^3-3x^2-3x^2+11x-6)}\leq0$

$=\frac{x}{x[x^2(x - 3)-(3x^2-11x+6)]}\leq0$

$=\frac{x}{x[x^2(x-3)-(3x^2-9x-2x+6)]}\leq0$

$=\frac{x}{x[x^2(x-3)-(3x(x-3)-2(x-3))]}\leq0$

$=\frac{x}{x[x^2(x-3)-(3x-2)(x-3)]}\leq0$

$=\frac{x}{x[(x-3)(x^2-3x+2)]}\leq0$

$=\frac{x}{x[(x-3)(x^2-2x-x+2)]}\leq0$

$=\frac{x}{x[(x-3)[x(x-2)-(x-2)]]}\leq 0$

$=\frac{x}{x(x-1)(x-2)(x-3)}\leq0$

The zeroes of the denominator are: x = 0, x = 1, x = 2, x = 3.

Solution: x < 0 (or) 0 < x < 1 (or) 2 < x < 3

Interval Notation: (-∞,0) ∪ (0,1) ∪ (2,3).

Hope it helps!

Answer 2

Answer:

x ∈ (-∝ , 0) ∪ (0 , 1) ∪ (2 , 3)

Step-by-step explanation:

Given :

To solve for x,

if $\frac{x}{x^4 - 6x^3 + 11x^2 - 6x} \leq 0$

Solution :

$\frac{x}{x^4 - 6x^3 + 11x^2 - 6x} \leq 0$

⇒ $\frac{x}{x(x^3 - 6x^2 + 11x - 6)} \leq 0$

⇒ $\frac{1}{x^3 - 6x^2 + 11x - 6} \leq 0$

By factorizing the denominator,

we get,

$x^3 - 6x^2 + 11x - 6$

⇒ $x^3 - 5x^2 + 6x - x^2 + 5x - 6$

⇒ $x(x^2 - 5x + 6) - (x^2 - 5x + 6)$

⇒ $(x - 1)(x^2 - 5x + 6)$

⇒ $(x - 1)(x^2 - 3x - 2x + 6)$

⇒ $(x - 1)(x(x - 3) - 2(x - 3))$

⇒ $(x - 1)((x - 2)(x - 3))$

⇒ $(x - 1)(x - 2)(x - 3)$

∴ $\frac{1}{x^3 - 6x^2 + 11x - 6} \leq 0$ ⇒ $\frac{1}{(x -1)(x - 2)(x - 3)} \leq 0$

⇒ $\frac{1}{(x -1)(x - 2)(x - 3)} \leq 0$

As numerator contains only positive values (no variables),

The denominator must be less than 0,.

Hence,

⇒ $(x -1)(x - 2)(x - 3) < 0$

It is possible only when  1 of them less than 0 (or) all 3 less than 0,.

i.e.,

x > 2 and x < 3 (or) x < 1 and x ≠ 0

i.e., x ∈ (-∝ , 0) ∪ (0 , 1) ∪ (2 , 3)