JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 18

This question was dropped in NTA's final key. The working shows why no option is right.

Logarithm. $x^2-3x+2>0$ needs $x<1$ or $x>2$, and the denominator is non-zero only if $x^2-3x+2\ne1$, i.e. $x\ne\dfrac{3\pm\sqrt5}2$.

Inverse cosine. For $x\ne3$ the argument is $\dfrac{x-2}{x+3}$ (and $x=3$ is excluded since the denominator $x^2-9$ vanishes). $-1\le\dfrac{x-2}{x+3}\le1$ gives $x\ge-\frac12$.

Domain. $\left[-\frac12,1\right)\cup(2,\infty)-\left\{3,\dfrac{3\pm\sqrt5}2\right\}$. Option (D) omits the exclusion of $x=3$, and the others are wider still, so none matches.