The domain of the function f(x)={^{-1}≤ft(x²-5x+6/x²-9)}{_e(x²-3x+2)}\;{ is : }\; [24-Jun-2022-Shift-1]

Correct Answer is : [−12,1)∪(2,∞)−{3,3+52,3−52}≤ft[-1/2, 1) (2, ∞) - ≤ft\{3, {3+{5}}{2}, {3-{5}}{2}\}[−21,1)∪(2,∞)−{3,23+5,23−5}

≤ft[-1/2, 1) (2, ∞) - ≤ft\{3, {3+{5}}{2}, {3-{5}}{2}\}

Correct Answer is : [−12,1)∪(2,∞)−{3,3+52,3−52}≤ft[-1/2, 1) (2, ∞) - ≤ft\{3, {3+{5}}{2}, {3-{5}}{2}\}[−21,1)∪(2,∞)−{3,23+5,23−5}

Test Index

Inverse Trigonometric Functions

Section: Mathematics

Question : 60 of 99

Marks: +1, -0

f(x)=\frac{\cos^{-1}\left(\frac{x^2-5x+6}{x^2-9}\right)}{\log_e(x^2-3x+2)}\;\text{ is : }\;

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

Solution: 👈: Video Solution

-1 \leq \frac{x^2-5x+6}{x^2-9} \leq 1

and

x^2-3x+2>0,\ \neq 1

\frac{(x-3)(2x+1)}{x^2-9} \geq 0 \mid \frac{5(x-3)}{x^2-9} \geq 0

The solution to this inequality is

x \in \left[-\frac{1}{2}, \infty\right) - \{3\}

for

x^2-3x+2>0

and

\neq 1

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

Combining the two solution sets (taking intersection)

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

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