Domain and range

Values of (\sec y) lie in \(\left(-\infty,-1\right]\cup\left[1,\infty\right)\). Hence this is the domain of \(\sec^{-1}x\).

The value of (\cosec y) never lies in \(\left(-1,1\right)\). So the domain of \(\cosec^{-1}x\) is \(\left(-\infty,-1\right]\cup\left[1,\infty\right)\).

Values of (\sin y) lie only in \(\left[-1,1\right]\). So \(\sin^{-1}x\) is defined when \(x\in\left[-1,1\right]\).

The value of (\cos y) does not go outside \(\left[-1,1\right]\). Hence the domain of \(\cos^{-1}x\) is \(\left[-1,1\right]\).

The inverse tangent function tan⁻¹x is defined as the inverse of tangent after restricting tangent to its principal interval (−π/2, π/2). On that interval, tan y is continuous, strictly increasing, and takes every real value from −∞ to +∞. Therefore every real number x has exactly one principal angle y such that tan y = x. The input set, or domain, of tan⁻¹x is consequently all real numbers, ℝ. Option C is correct. The interval [−1, 1] is the domain of inverse sine and inverse cosine, not inverse tangent. The interval [0, π] is associated with the principal range of inverse cosine, while option D describes neither the domain nor the range of tan⁻¹x.

The function \(\cos^{-1}x\) is defined only for \(x\in\left[-1,1\right]\). The number (-2) is not in this domain.

Since \(\frac{7}{6}\notin\left[-1,1\right]\), \(\sin^{-1}\left(\frac{7}{6}\right)\) is not defined. Check the domain first.

The domain of \(\cos^{-1}x\) is \(\left[-1,1\right]\). Since \(\frac{11}{10}>1\), it is not defined.

For \(\sec^{-1}x\), we need \(\left|x\right|\ge1\). Here \(\left|\frac{3}{2}\right|\ge1\), so it is defined.

The function \(cosec^{-1}x\) is defined when \(\left|x\right|\ge1\). Here \(\left|-\frac{5}{4}\right|\ge1\), so it is defined.

The domain of arccos, \(\cos^{-1}\), is \([-1,1]\). Thus \(\cos^{-1}(-1)\) is defined and equals \(\pi\), so the expression equals \(\pi/\pi=1\). \(\cos^{-1}(3)\) and \(\sin^{-1}(5/4)\) are undefined because 3 and 5/4 lie outside the domain \([-1,1]\). Arcsec has domain \(|x|\ge 1\), so \(\sec^{-1}(1/2)\) is also undefined. Exam tip: always check the allowed domain of the inverse trig function before simplifying; values outside the domain make the expression invalid regardless of algebraic form.

\(\csc^{-1}x\) is defined only for \(|x|\ge 1\), i.e. the domain is \(( -\infty,-1 ]\cup[1,\infty )\). Hence \(x=\tfrac{1}{2}\) is not allowed because \(|\tfrac{1}{2}|<1\). The other choices satisfy the domain: \(x=2\) and \(x=-3\) are valid, and \(x=-1\) is valid as an endpoint. Exam tip: memorize domains of inverse trig functions (especially arccsc/arccot) to avoid common mistakes on boundaries.

For \(\sin^{-1}(2x)\), we need \(-1\le 2x\le 1\). Hence \(-\frac{1}{2}\le x\le \frac{1}{2}\).

\(\sin^{-1}x\) (arcsin) is defined only when \(x\) lies within the range of the sine function. For real \(y\), \(\sin y\) takes values only in \([-1,1]\); moreover on the principal branch \(y\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\) the values \(\pm1\) also occur. Thus \(\sin^{-1}x\) is defined exactly for \(x\in[-1,1]\), so option A is correct. Option B is incorrect because it excludes the endpoints \(\pm1\), but \(\sin^{-1}(\pm1)=\pm\tfrac{\pi}{2}\) are valid. Options C and D are clearly wrong: C allows all real numbers (not possible), and D gives an angle interval, not valid values of \(x\). Exam tip: to find the domain of an inverse trig function, first recall the range of the original trig function (e.g., range(sin) = [-1,1]).

The domain of \(\sin^{-1}x\) is \([-1,1]\), so \(\sin(\sin^{-1}x)=x\) there. In composition, check the inner function domain.

For \(\sin^{-1}(2x)\), we need \(-1\le 2x\le1\). This gives \(x\in[-\frac{1}{2},\frac{1}{2}]\).

The governing concept is the real domain of arccos: its input must lie in the closed interval [−1, 1]. For cos⁻¹(3x − 1), impose −1 ≤ 3x − 1 ≤ 1. Adding 1 throughout gives 0 ≤ 3x ≤ 2, and division by the positive number 3 gives 0 ≤ x ≤ 2/3. The endpoints are included because arccos is defined at both −1 and 1, so the domain is [0, 2/3]. Therefore option A is correct. Option B is the domain of the inner arccos input, not of x; option C results from incomplete algebra, and option D incorrectly excludes valid endpoint values.

From \(-1\le x^2-1\le1\), we get \(0\le x^2\le2\). Thus \(x\in[-\sqrt{2},\sqrt{2}]\), so none of the first three is exact.

The governing domain fact is that tan⁻¹u, or arctan u, is defined for every real number u. The tangent function itself has restrictions, but its inverse has real domain all of ℝ. Since x² + 1 is real for every real x, substituting u = x² + 1 creates no additional restriction on x. Although x² + 1 is always at least 1, that describes the range of the inner expression, not the domain of the complete composite function. Therefore the domain is ℝ and option A is correct. Options B, C, and D impose restrictions that are not required for arctan; in particular, [1, ∞) is only the set of possible inner values.

\(cosec y=\frac{1}{sin y}\), so values of \(cosec y\) satisfy \(|x|\ge1\). Use the reciprocal nature to remember the domain.