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 :
Solution
<p>$- 1 \le {{{x^2} - 4x + 2} \over {{x^2} + 3}} \le 1$</p>
<p>$\Rightarrow - {x^2} - 3 \le {x^2} - 4x + 2 \le {x^2} + 3$</p>
<p>$\Rightarrow 2{x^2} - 4x + 5 \ge 0$ & $- 4x \le 1$</p>
<p>$x \in R$ & $x \ge - {1 \over 4}$</p>
<p>So domain is $\left[ { - {1 \over 4},\infty } \right)$</p>
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
This question is part of PrepWiser's free JEE Main question bank. 65 more solved questions on Inverse Trigonometric Functions are available — start with the harder ones if your accuracy is >70%.