Domain and range

Correct answer: C. \(x=\frac{1}{2}\)

\(\csc^{-1}x\) 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.

Correct answer: A. \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\)

\(\sin^{-1}(y)\) is defined for real numbers when \(-1\le y\le 1\). Here \(y=2x\), so
\[
-1\le 2x\le 1.
\]
Dividing by \(2\) gives
\[
-\frac{1}{2}\le x\le \frac{1}{2}.
\]
So \(\sin^{-1}(2x)\) is real for \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\).

Correct answer: A. \(x\in[-1,1]\)

\(\sin^{-1}x\) (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]).

Correct answer: B. Only for (x\in[-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.

Correct answer: A. ([-\frac{1}{2},\frac{1}{2}])

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

Correct answer: A. ([0,\frac{2}{3}])

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.

Correct answer: D. ([-2,2])

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.

Correct answer: A. (\mathbb{R})

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.

Correct answer: B. ((-\infty,-1]\cup[1,\infty))

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

Correct answer: C. It is not defined in real numbers

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.

Correct answer: A. Not defined in real numbers

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.

Correct answer: A. \(|2x+1|\ge1\)

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 \(x\ge0\)).

Correct answer: A. \(x\le 1\) or \(x\ge 3\)

For \(\cosec^{-1}(u)\) to be defined, we need \(|u|\ge 1\) because \(\csc\theta = u\) requires \(u\neq 0\) and \(u\) must lie in the domain of inverse cosecant. Here \(u=x-2\), so we need \(|x-2|\ge 1\). This gives \(x-2\ge 1\Rightarrow x\ge 3\) or \(x-2\le -1\Rightarrow x\le 1\). Hence \(x\le 1\) or \(x\ge 3\).