Domain and range

The real-valued arcsin function \(\sin^{-1}x\) is defined only for \(x\in[-1,1]\). Since \(2\notin[-1,1]\), \(\sin^{-1}2\) is not defined as a real number. (Note: arcsin can be extended to complex values, but the question concerns real definition.) Exam tip: first check the domain of the inverse trigonometric function before evaluating.

The function \(\cos^{-1}x\) (arccos) has real values only for \(x\in[-1,1]\); its principal range for real inputs is \([0,\pi]\). Since \(\tfrac{3}{2}=1.5>1\), \(\cos^{-1}\left(\tfrac{3}{2}\right)\) has no real value. Option \(\dfrac{\pi}{3}\) is a tempting distractor but incorrect because \(\cos\dfrac{\pi}{3}=\tfrac{1}{2}\), not \(\tfrac{3}{2}\). Exam tip: always check the domain of inverse trigonometric functions before attempting to find a value.

By definition \(\sec^{-1}u\) is defined only when \(|u|\ge1\). With \(u=2x+1\) this gives \(|2x+1|\ge1\). Solving yields \(2x+1\ge1\Rightarrow x\ge0\) or \(2x+1\le-1\Rightarrow x\le-1\). Option B is the reverse inequality and is incorrect. Option C (requiring positivity) is not sufficient because negative values with magnitude ≥1 are also allowed. Option D (nonzero) is necessary but not sufficient — values like \(2x+1=0.5\) are nonzero yet outside the domain. Exam tip: always convert \(|u|\ge1\) into the two linear inequalities to get the domain explicitly (here \(x\le-1\) or \(x\ge0\)).

For \(cosec^{-1}(u)\) the argument must satisfy \(|u|\ge1\) because cosecant values lie in \((-
\infty,-1]\cup[1,\infty)\). Here \(u=x-2\), so require \(|x-2|\ge1\), giving \(x\le1\) or \(x\ge3\). Thus the domain is \((-
\infty,1]\cup[3,\infty)\). Option C is a close distractor but wrong because it excludes the endpoints \(x=1,3\), which are allowed (\(|x-2|=1\) is permitted). Exam tip: convert inverse trigonometric domains into inequalities on the argument (e.g. \(|\cdot|\ge1\) for \(cosec^{-1}\)) and check inclusivity of boundary points.