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Strictly Increasing Functions

lf x 2 +6 x...

Question

lf $x^{2} + 6 x - 27 > 0$ and $- x^{2} + 3 x + 4 > 0$ , then $x$ lies in the interval

A

(3, 4)

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B

[

3, 4

]

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C

(- \infty, 3) \cup (4, \infty)

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D

(- 9, 4)

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Solution

The correct option is B $(3, 4)$
$x^{2} + 6 x - 27 > 0 \Rightarrow (x + 9) (x - 3) > 0 \Rightarrow x < - 9 o r x > 3$ ----(1)
Also, $- x^{2} + 3 x + 4 > 0$
$\Rightarrow (x - 4) (x + 1) < 0$
$\Rightarrow - 1 < x < 4$ ----(2)
From $(1)$ and $(2)$ , $x \in (3, 4)$ .

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Q. Match the elements of the list I, which contains inequalities, with the elements of list II, which has solution sets.

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D) $x^{2} - 3 x - 4 < 0, x^{2} - 3 x + 2 > 0$	4) $[3, 4]$
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