JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 2)
Considering only the principal values of the inverse trigonometric functions, the domain of the function $$f(x)=\cos ^{-1}\left(\frac{x^{2}-4 x+2}{x^{2}+3}\right)$$ is :
$$\left(-\infty, \frac{1}{4}\right]$$
$$\left[-\frac{1}{4}, \infty\right)$$
$$(-1 / 3, \infty)$$
$$\left(-\infty, \frac{1}{3}\right]$$
Explanation
$$ - 1 \le {{{x^2} - 4x + 2} \over {{x^2} + 3}} \le 1$$
$$ \Rightarrow - {x^2} - 3 \le {x^2} - 4x + 2 \le {x^2} + 3$$
$$ \Rightarrow 2{x^2} - 4x + 5 \ge 0$$ & $$ - 4x \le 1$$
$$x \in R$$ & $$x \ge - {1 \over 4}$$
So domain is $$\left[ { - {1 \over 4},\infty } \right)$$
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