Inverse Trigonometric Functions — JEE MCQs with answers

Answer: A — Option A. Both $\operatorname{cosec} x$ and $\cot x$ are built on $\sin x$ in the denominator, so both fail exactly where $\sin x = 0$, that is at the integral multiples of $\pi$. Option D names the set shared by $\sec x$ and $\tan x$ instead. Option B names positions that matter on the $X$-axis of the companion panels, not here. Option C denies an agreement that both drawings show.

Answer: A — The obstruction named on the page is that the functions are not one-one and onto over their natural domains and ranges, and the Chapter 1 condition for an inverse is exactly being one-one and onto. Restrictions that ensure the existence of the inverses must therefore make the restricted functions one-one and onto, option A. Option C thinks the rule changes; a restriction keeps the rule and limits the inputs. Option B thinks the condition is waived, when it is the very condition being met. Option D misses that repeated outputs come from too many inputs, so the domain has to be restricted as well; the page names domains and ranges together.

Answer: C — Both terms use the same angle $\frac{\pi}{6}$ here, because $1 - x$ happens to equal $x$ at $x = \frac{1}{2}$. Option C is correct: $\sin^{-1}\frac{1}{2} - 2\sin^{-1}\frac{1}{2} = -\sin^{-1}\frac{1}{2} = -\frac{\pi}{6}$, which is not $\frac{\pi}{2}$; this is why $\frac{1}{2}$ is not a root. Option A assumes the value satisfies the equation instead of computing it. Option B drops the sign produced by subtracting twice the same term. Option D would need $\sin^{-1}\frac{1}{2} = 2\sin^{-1}\frac{1}{2}$, which is false for a non-zero value.

Answer: B — The branches are $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ -- the SAME two endpoints, closed in one row and open in the other. So the difference is precisely the endpoints, and option B, $\frac{\pi}{2}$, is one of them. Options C and A are interior and lie in both. Option D exceeds both branches and lies in neither.

Answer: A — The page's words are that they play an important role in calculus for they serve to define many integrals, option A. Option B swaps integrals for derivatives, a different part of calculus from the one named. Option C reverses the direction: the trigonometric functions come first, and the inverses are built from them. Option D inflates 'many integrals' into a claim that no other function can be integrated, which the page does not make.

Answer: D — $\frac{4}{5} = 0.8$ and $\frac{1}{\sqrt{2}} \approx 0.7071$, so $\frac{4}{5}$ lies in $\left[\frac{1}{\sqrt{2}}, 1\right]$, the interval of Example 3 (ii): option D. Option A quotes (i), whose interval stops at $\frac{1}{\sqrt{2}}$. Option B is wrong because the two intervals meet at a single point, and $0.8$ is not it. Option C overlooks that $0.8 \le 1$.

Answer: A — Option A is correct. The Note prints $(\sin x)^{-1} = \frac{1}{\sin x}$ and adds that the same holds for the other trigonometric functions. Option B: this is another inverse function, not a reciprocal of a value. Option C: here the reciprocal has been taken of the angle instead of the sine. Option D: this is the inverse function, and the Note exists precisely to say the two are different.

Answer: A — $\frac{2}{3}$ lies in $[-1, 1]$, so the first term is $\frac{2}{3}$; $\frac{\pi}{9}$ lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so the second term is $\frac{\pi}{9}$. Option A. Option B reads $\sin^{-1}$ as a reciprocal in the first term. Option C reduces the second term by $\pi - \frac{\pi}{9}$ although it already lies inside the branch. Option D turns the first term into an angle, as if the answer to $\sin\left(\sin^{-1}\frac{2}{3}\right)$ had to be a multiple of $\pi$; the two compositions return different kinds of thing — a number and an angle.

Answer: C — The page recalls that the inverse of a function $f$, denoted by $f^{-1}$, exists if $f$ is one-one and onto -- both conditions together, option C. Option D keeps only the one-one half and option A keeps only the onto half; the very next sentence of Section 2.1 says a function that is not one-one, onto or both has no inverse to talk of, so neither half alone is enough. Option B swaps the condition for a requirement on the domain, which the recalled condition never mentions: sine is defined at every real number and still has no inverse over its natural domain.

Answer: C — The $\sec^{-1}$ row prints $[0, \pi] - \left\{\frac{\pi}{2}\right\}$, option C. Option B omits the deleted point; but $\sec\frac{\pi}{2}$ is undefined, so $\frac{\pi}{2}$ cannot be an output. Option D is the branch the table prints for $\operatorname{cosec}^{-1}$, which removes a different point from a different interval. Option A is the branch of $\sin^{-1}$, which removes nothing at all.