વિઘેય f(x)=frac{cos ^{-1}left(frac{x^{2}-5 x+ | vedclass

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Question

$-1 \leq \frac{x^{2}-5 x+6}{x^{2}-9} \leq 1$

$\frac{x^{2}-5 x+6}{x^{2}-9}-1 \leq 0$

$\frac{1}{x+3} \geq 0$

$x \in(-3, \infty) \ldots \ldots(1

Vedclass · Accepted Answer

$\left[-\frac{1}{2}, 1\right) \cup(2, \infty)-\left\{\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}\right\}$

Vedclass · Answer

$-1 \leq \frac{x^{2}-5 x+6}{x^{2}-9} \leq 1$

$\frac{x^{2}-5 x+6}{x^{2}-9}-1 \leq 0$

$\frac{1}{x+3} \geq 0$

$x \in(-3, \infty) \ldots \ldots(1)$

$\frac{x^{2}-5 x+6}{x^{2}-9}+1 \geq 0$

$\frac{2 x+1}{x+3} \geq 0$

$x \in(-\infty,-3) \cup\left[-\frac{1}{2}, \inftyight) \ldots \ldots(2)$

after taking intersection

$x \in\left[-\frac{1}{2}, \inftyight)$

$x^{2}-3 x+2>0$

$x \in(-\infty, 1) \cup(2, \infty)$

$x^{2}-3 x+2 
eq 1$

$x 
eq \frac{3 \pm \sqrt{5}}{2}$

after taking intersection of each solution

$\left[-\frac{1}{2}, 1ight) \cup(2, \infty)-\left\{\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}ight\}$

વિઘેય $f(x)=\frac{\cos ^{-1}\left(\frac{x^{2}-5 x+6}{x^{2}-9}\right)}{\log _{e}\left(x^{2}-3 x+2\right)} $ નો પ્રદેશ ........ છે.

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