If x2-1x-2x-12-x-2-0- then x lies in the interval

(x^2 1)(x+2)(x+1)^2(x 2) 0 ⇒(x+1)(x 1)(x+2)(x+1)^2(x 2) 0 ⇒(x 1)(x+2)(x+1)^3(x 2) 0 ⇒Pivot points= 2, 1,1,2 Now, draw wavy curve for the given condition. From ...

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Standard XII

Mathematics

Graph of Quadratic Expression

If x2-1x+2x+1...

Question

$If \frac{(x^{2} - 1) (x + 2) (x + 1)^{2}}{(x - 2)} < 0,$
then $x$ lies in the interval

(-2, -1) ∪ (1, 2)

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(-∞, -2) ∪ (2, ∞)

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(-2, -1) ∪ (2, ∞)

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(-2, -1) ∪ (1, ∞)

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Solution

The correct option is A (-2, -1) ∪ (1, 2)
$\frac{(x^{2} - 1) (x + 2) (x + 1)^{2}}{(x - 2)} < 0$
$\Rightarrow \frac{(x + 1) (x - 1) (x + 2) (x + 1)^{2}}{(x - 2)} < 0$
$\Rightarrow \frac{(x - 1) (x + 2) (x + 1)^{3}}{(x - 2)} < 0$
$\Rightarrow Pivot points = - 2, - 1, 1, 2$
Now, draw wavy curve for the given condition.

From Wavy Curve, we get
$- 2 < x < - 1 or 1 < x < 2 \Rightarrow x \in (- 2, - 1) \cup (1, 2)$
Hence, The correct answer is option (a)

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If (x2−1)(x+2)(x+1)2(x−2)<0, then x lies in the interval

$If \frac{(x^{2} - 1) (x + 2) (x + 1)^{2}}{(x - 2)} < 0,$
then $x$ lies in the interval