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If $\mathrm{x}$ is real and $\mathrm{x}^{2}-3 \mathrm{x}+2>0, \mathrm{x}^{2}-3 \mathrm{x}-4 \leq 0$, then which on
of the following is correct?
Options:
of the following is correct?
Solution:
2891 Upvotes
Verified Answer
The correct answer is:
$-1 \leq x < 1$ or $2 < x \leq 4$
Consider first: $\mathrm{x}^{2}-3 \mathrm{x}+2>0$ $\Rightarrow \quad(x-1)(x-2)>0$...(1)
$\Rightarrow \quad x < 1$ or $x>2$
and $x^{2}-3 x-4 \leq 0$
$\Rightarrow \quad(x-4)(x+1) \leq 0$
$\Rightarrow-1 \leq x \leq 4$...(2)
Combining $(1)$ and $(2)$ $-1 \leq x < 1$ or $2 < x \leq 4$
Drawing on number line:
$\Rightarrow \quad x < 1$ or $x>2$
and $x^{2}-3 x-4 \leq 0$
$\Rightarrow \quad(x-4)(x+1) \leq 0$
$\Rightarrow-1 \leq x \leq 4$...(2)
Combining $(1)$ and $(2)$ $-1 \leq x < 1$ or $2 < x \leq 4$
Drawing on number line:
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