Question

The common solution set for real x, of inequation x4– x2 – 12 ≤ 0 and x2 – 5x + 4 ≥ 0 is


[–2, 2]


[–2, 4]


[–2, 1]


[4, ∞)

Answer 1

Answer:

sorry I didn't understand your question

Answer 2

The common solution set for the inequality $x^4-x^2-12\le0$ and $x^2-5x+4\ge0$ is [-2,1]

Therefore, option (3) is correct.

Step-by-step explanation:

The given inequalities

$x^4-x^2-12\le0$

And $x^2-5x+4\ge0$

From second inequality

$x^2-5x+4\ge0$

$\implies x^2-4x-x+4\ge0$

$\implies x(x-4)-1(x-4)\ge0$

$\implies (x-4)(x-1)\ge0$

$\implies x\in (-\infty,1]\cup[4,\infty)$  ............ (1)

Again from first inequality

$x^4-x^2-12\le0$

$\implies x^4-4x^2+3x^2-12\le0$

$\implies x^4-4x^2+3x^2-12\le0$

$\implies x^2(x^2-4)+3(x^2-4)\le0$

$\implies (x^2+3)(x^2-4)\le0$

Now for all the values of x $x^2+3$ will always be positive

Therefore, for $x^4-x^2-12\le0$ to be satisfied,

$x^2-4\le0$

$\implies (x-2)(x+2)\le0$

$\implies x\in [-2,2]$  ............ (2)

The final answer will be the intersection of the results in (1) and (2)

Which will be

$x\in [-2,1]$

Hope this answer is helpful.

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