Question

Plz solve question no. 14......fast..

Answer 1

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Answer 2

Answer:

x ∈ (4,6)∪(-3,-2)

Step-by-step explanation:

Given :

To find all possible values of x if,

⇒ $\frac{(x^2+1)(x^2 + 3x + 2)(4-x)^4}{(x^2+4x+3)(x - 4)}\geq 0$

Solution :

⇒ $\frac{(x^2+1)(x^2 + 3x + 2)(4-x)^4}{(x^2+4x+3)(x - 4)}\geq 0$

⇒ $\frac{(x^2+1)(x^2 + x + 2x + 2)[-1(x-4)]^4}{(x^2+x + 3x+3)(x - 4)}\geq 0$

⇒ $\frac{(x^2+1)(x+2)(x+1)[-1^4](x-4)^4}{(x + 1)(x+3)(x - 4)}\geq 0$

⇒ $\frac{(x^2+1)(x+2)(x-4)^3}{(x+3)}\geq 0$

⇒ x² + 1≠0 (if x is real),

⇒ x + 2 ≥ 0 (or) x + 2 ≤ 0⇒ x → -2

⇒  x + 4 ≥ 0 (or) x + 4 ≤ 0⇒ x → -4

⇒ x + 3 ≠ 0 ⇔ x ≠ -3  x→-3

⇒ ∴ x ∈ (4,6)∪(-3,-2)