Question

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Answer 1

Answer:

$\mathrm{x \in (-\infty,-1] \cup (0,1]\cup(2,3]}$

Step-by-step explanation:

Given :

$\mathrm{\big(x^2-2x\big)(2x-2) - 9\dfrac{2x-2}{x^2-2x} \leq 0}$

To find :

Solution of the following inequality

Solution :

$\longmapsto \mathrm{\big(x^2-2x\big)(2x-2) - 9\dfrac{2x-2}{x^2-2x} \leq 0}$

$\implies \mathrm{\dfrac{\big(x^2-2x\big)^2(2x-2) - 9(2x-2)}{x(x-2)}\leq 0}$

Multiplying and dividing x(x - 2)

$\implies \mathrm{\dfrac{x(x-2)\Big(\big(x^2-2x\big)^2(2x-2) - 9(2x-2)\Big)}{x^2(x-2)^2}\leq 0}$

As x²(x - 2)² is always positive, we can send denominator to RHS

$\implies \mathrm{x(x-2)\Big(\big(x^2-2x\big)^2(2x-2) - 9(2x-2)\Big)\leq 0}$

$\implies \mathrm{x(x-2)(2x-2)\big((x^2-2x)^2-9\big)\leq 0}$

$\implies \mathrm{2x(x-2)(x-1)\big((x^2-2x)^2-3^2\big)\leq 0}$

We know that,

$\boxed{\mathrm{a^2-b^2 = (a + b)(a - b)}}$

Here, a = x² - 2x ; b = 3. So,

$\implies \mathrm{x(x-2)(x-1)\big((x^2-2x-3)(x^2-2x+3)\big)\leq 0}$

$\implies \mathrm{x(x-2)(x-1)(x^2-3x+x-3)(x^2-2x+3)\leq 0}$

$\implies \mathrm{x(x-2)(x-1)(x(x-3)+1(x-3))(x^2-2x+3)\leq 0}$

$\implies \mathrm{x(x-2)(x-1)(x-3)(x+1)(x^2-2x+3)\leq 0}$

x² - 2x + 3 has imaginary roots as Δ < 0

=> Δ = 4 - 4(3) = 4 - 12 = -8 < 0

So, we can ignore that in find the solutions as it doesn't have real roots

$\implies \mathrm{x(x-2)(x-1)(x-3)(x+1)\leq 0}$

Now using Wavey - Curve method, [Refer Attachment]

$\therefore \underline{\boxed{\mathrm{x \in (-\infty,-1] \cup (0,1]\cup(2,3]}}}$

[Note : We have open brackets for 0 and 2 because, in the original question they are in the denominator. If we substitute 0 or 2 we get undefined number. ]

Hope it helps!!