Assertion (A) : Domain of $y = \cos^{-1}(x)$ is $[-1, 1]$. Reason (R) : The range of the principal value branch of $y = \cos^{-1}(x)$ is $[0, \pi] - \{\frac{\pi}{2}\}$.
CBSE CLASS 12 • MATHEMATICS • CHAPTER 02
Inverse Trigonometric Functions
153 distinct questions across 205 board paper set appearances (2022–2026).
Evaluate : $\tan^{-1}\left$ -\frac{1}{\sqrt{3}} \right$ + \cot^{-1}\left$ \frac{1}{\sqrt{3}} \right$ + \tan^{-1}\left( \sin\left$ -\frac{\pi}{2} \right$ \right) + \tan^{-1}\left$ \tan \frac{2\pi}{3} \right$$
(a) If $xy = e^{x - y}$, then find $\frac{dy}{dx}$.
OR
(b) Differentiate $\tan^{-1} \left$ \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right$$ with respect to $\cos^{-1} x^2$.
Derivative of $\cos^{-1} \left$ \frac{\sin x + \cos x}{\sqrt{2}} \right$$, $-\frac{\pi}{4} < x < \frac{\pi}{4}$ with respect to x is : (A) $-1$ (B) 1 (C) $\frac{\pi}{4}$ (D) $-\frac{\pi}{4}$
Differential of $\cos^{-1} (e^x)$ with respect to $x$ is :
(A) $\frac{e^x}{\sqrt{1 - e^{2x}}}$
(B) $\frac{-e^x}{\sqrt{1 + e^x}}$
(C) $\frac{-e^x}{\sqrt{1 - e^{2x}}}$
(D) $\frac{e^x}{\sqrt{1 + e^{2x}}}$
The value of $\int_{0}^{1} \frac{dx}{e^x + e^{-x}}$ is :
(A) $-\frac{\pi}{4}$
(B) $\frac{\pi}{4}$
(C) $\tan^{-1} e - \frac{\pi}{4}$
(D) $\tan^{-1} e$
The principal value of $\cot^{-1}\left$-\frac{1}{\sqrt{3}}\right$$ is :
(A) $-\frac{\pi}{3}$
(B) $-\frac{2\pi}{3}$
(C) $\frac{\pi}{3}$
(D) $\frac{2\pi}{3}$
If $\tan^{-1} (x^2 - y^2) = a$, where 'a' is a constant, then $\frac{dy}{dx}$ is : (A) $\frac{x}{y}$ (B) $-\frac{x}{y}$ (C) $\frac{a}{x}$ (D) $\frac{a}{y}$
(a) Differentiate $y = \sin^{-1} (3x - 4x^3)$ w.r.t. $x$, if $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$.
OR
(b) Differentiate $y = \cos^{-1} \left(\frac{1-x^2}{1+x^2}\right)$ with respect to $x$, when $x \in (0, 1)$.
(a) Express $\tan^{-1}\left$ \frac{\cos x}{1 - \sin x} \right$$, where $-\frac{\pi}{2} < x < \frac{\pi}{2}$ in the simplest form. OR (b) Find the principal value of $\tan^{-1}(1) + \cos^{-1}\left$-\frac{1}{2}\right$ + \sin^{-1}\left$-\frac{1}{\sqrt{2}}\right$$.
Evaluate $\sin^{-1} \left$ \sin \frac{3\pi}{4} \right$ + \cos^{-1} $\cos \pi$ + \tan^{-1} (1)$.
Draw the graph of $\cos^{-1} x$, where $x \in [-1, 0]$. Also, write its range.
(a) Find the domain of $y = \sin^{-1} (x^2 - 4)$.
OR
(b) Evaluate :
$$\cos^{-1} \left[ \cos \left$ - \frac{7\pi}{3} \right$ \right]$$
Assertion (A) : The principal value of $\cot^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$.
Reason (R) : Domain of $\cot^{-1} x$ is $\mathbb{R} - \{-1, 1\}$.
(a) Simplify : $\tan^{-1}\left(\frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right)$, $0 < x < \frac{\pi}{4}$.
OR
(b) Evaluate : $\tan\left(\sin^{-1} 1 - \cos^{-1}\left$-\frac{1}{2}\right$\right)$
The principal value of $\sec^{-1}(\sqrt{2}) + 2 \text{cosec}^{-1}(-\sqrt{2})$ is :
(A) $-\frac{\pi}{2}$
(B) $-\frac{\pi}{4}$
(C) $\frac{\pi}{4}$
(D) $\frac{\pi}{2}$
$\int \frac{dx}{\sqrt{25 - 16x^2}}$ is equal to : (A) $\frac{1}{5} \sin^{-1} 4x + C$ (B) $\frac{1}{25} \sin^{-1} 16x + C$ (C) $\frac{1}{4} \sin^{-1} \frac{4x}{5} + C$ (D) $\frac{1}{16} \sin^{-1} \frac{4x}{5} + C$
The value of $\sin^{-1} \left$\sin \frac{3\pi}{5}\right$$ is :
(A) $\frac{3\pi}{5}$
(B) $\frac{\pi}{5}$
(C) $\frac{7\pi}{5}$
(D) $\frac{2\pi}{5}$
$[\tan^{-1} \sqrt{3} - \sec^{-1} (-2)]$ is equal to :
(A) $\pi$
(B) $-\frac{\pi}{3}$
(C) $\frac{\pi}{3}$
(D) $\frac{2\pi}{3}$
(a) Find the principal value of $\cos^{-1}\left$-\frac{1}{2}\right$ + 2\sin^{-1}\left(\frac{1}{2}\right)$.
OR
(b) Prove that :
$\tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1 - x}{1 + x}\right), x \in [0, 1]$
(a) Differentiate $2^{\cos^2 x}$ w.r.t $\cos^2 x$.
OR
(b) If $\tan^{-1} (x^2 + y^2) = a^2$, then find $\frac{dy}{dx}$.
(a) Simplify $\sin^{-1} \left$ \frac{x}{\sqrt{1 + x^2}} \right$$. OR (b) Find domain of $\sin^{-1} \sqrt{x-1}$.
$\left[ \sec^{-1} (-\sqrt{2}) - \tan^{-1} \left$ \frac{1}{\sqrt{3}} \right$ \right]$ is equal to :
(A) $\frac{11\pi}{12}$
(B) $\frac{5\pi}{12}$
(C) $-\frac{5\pi}{12}$
(D) $\frac{7\pi}{12}$
Questions number 19 and 20 are Assertion and Reason based questions. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is...
(a) Find the value of $\tan^{-1}\left$-\frac{1}{\sqrt{3}}\right$ + \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left[\sin\left$-\frac{\pi}{2}\right$\right]$.
Find value of k if
$\sin^{-1}\left[ k \tan\left$ 2 \cos^{-1} \frac{\sqrt{3}}{2} \right$ \right] = \frac{\pi}{3}$.
If $a = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) + \cos^{-1}\left$-\frac{1}{2}\right$$ and $b = \tan^{-1}(\sqrt{3}) - \cot^{-1}\left$-\frac{1}{\sqrt{3}}\right$$ then find the value of $a + b$.
Simplify : $\cos^{-1}x + \cos^{-1} \left[ \frac{x}{2} + \frac{\sqrt{3-3x^2}}{2} \right] ; \frac{1}{2} \leq x \leq 1$
Assertion (A) : $\cos^{-1} \left$ \cos \frac{13\pi}{6} \right$$ is equal to $\frac{\pi}{6}$. Reason (R) : The range of the principal value branch of the function $y = \cos^{-1} x$ is $[0, \pi]$.
Assertion (A) : The range of the function $f(x) = 2 \sin^{-1} x + \frac{3\pi}{2}$, where $x \in [-1, 1]$, is $[\frac{\pi}{2}, \frac{5\pi}{2}]$. Reason (R) : The range of the principal value branch of $\sin^{-1} (x)$ is $[0, \pi]$.
Assertion (A) : Maximum value of $(\cos^{-1} x)^2$ is $\pi^2$.
Reason (R) : Range of the principal value branch of $\cos^{-1} x$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.
Assertion (A) : Range of [$\sin^{-1} x + 2 \cos^{-1} x$] is $[0, \pi]$.
Reason (R) : Principal value branch of $\sin^{-1} x$ has range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
(a) Evaluate : $3 \sin^{-1} \left(\frac{1}{\sqrt{2}}\right) + 2 \cos^{-1} \left(\frac{\sqrt{3}}{2}\right) + \cos^{-1} (0)$
OR
(b) Draw the graph of f(x) = $\sin^{-1} x$, $x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$. Also, write range of f(x).
Assertion (A) : All trigonometric functions have their inverses over their respective domains.
Reason (R) : The inverse of $\tan^{-1} x$ exists for some $x \in \mathbb{R}$.
$\sin \left[ \frac{\pi}{3} + \sin^{-1} \left$ \frac{1}{2} \right$ \right]$ is equal to
(a) 1
(b) $\frac{1}{2}$
(c) $\frac{1}{3}$
(d) $\frac{1}{4}$
(a) Differentiate $\sec^{-1}\left(\frac{1}{\sqrt{1-x^{2}}}\right)$ w.r.t. $\sin^{-1}(2x\sqrt{1-x^{2}})$. OR (b) If $y = \tan x + \sec x$, then prove that $\frac{d^{2}y}{dx^{2}} = \frac{\cos x}{(1-\sin x)^{2}}$.
If $2\cos^{-1}x = y$, then
(A) $0 \le y \le \pi$
(B) $-\pi \le y \le \pi$
(C) $0 \le y \le 2\pi$
(D) $-\pi \le y \le 0$
(a) Simplify : $\tan^{-1}\left$\\frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right$$, $0 < x < \\frac{\pi}{4}$. \
OR \
(b) Evaluate : $\tan \left(\sin^{-1} 1 - \cos^{-1}\left$-\\frac{1}{2}\right$\right)$
If $\tan^{-1} x = 3y$, then
(A) $-\frac{\pi}{2} < y < \frac{\pi}{2}$
(B) $-\frac{3\pi}{2} < y < \frac{3\pi}{2}$
(C) $-\frac{\pi}{6} < y < \frac{\pi}{6}$
(D) $-\frac{\pi}{6} \le y \le \frac{\pi}{6}$
(a) Simplify : $\cot^{-1}\sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}}$, $x \in \left(0, \frac{\pi}{2}\right)$.
OR
(b) Evaluate : $\sin\left(\tan^{-1}(-\sqrt{3}) - \sec^{-1} 2\right)$
If $\sin^{-1} x + \pi = y$, then
(A) $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$
(B) $-\frac{3\pi}{2} \le y \le -\frac{\pi}{2}$
(C) $\frac{\pi}{2} \le y \le \frac{3\pi}{2}$
(D) $0 \le y \le \pi$
(a) Simplify : $\sin^{-1}\sqrt{\frac{1 + \cos 2x}{2}}$, $0 < x < \frac{\pi}{2}$.
OR
(b) Evaluate : $\cos[\sin^{-1}(-1) - \tan^{-1}(-\sqrt{3})]$.
Find the domain of $p(x) = \sin^{-1}(1 - 2x^2)$. Hence, find the value of $x$ for which $p(x) = \frac{\pi}{6}$. Also, write the range of $2p(x) + \frac{\pi}{2}$.
If $x = e^{\sin^{-1} t}, y = e^{\cos^{-1} t}$, find $\frac{dy}{dx}$ at $t = \frac{1}{\sqrt{2}}$.
Find the domain of $\text{q}(x) = \cos^{-1}(4x^2 - 3)$. Hence, find the value of $x$ for which $\text{q}(x) = 0$. Also, write the range of $3\text{q}(x) - \pi$.
If $y = \tan^{-1}\left(\frac{1 - \cos x}{\sin x}\right)$, then $\frac{dy}{dx}$ is equal to :
(A) $1$
(B) $2$
(C) $\frac{1}{2}$
(D) $-\frac{1}{2}$
$\int \frac{dx}{x^2 + 6x + 10}$ is equal to :
(A) $x \tan^{-1} (x + 3) + C$
(B) $\tan^{-1} (x + 3) + C$
(C) $\frac{1}{3} \tan^{-1} (x + 3) + C$
(D) $\tan^{-1} x + C$
The integrating factor of $(1 + y^2) \text{d}x = (\tan^{-1} y - x) \text{d}y$ is :
(A) $\tan y$
(B) $\tan^{-1} y$
(C) $e^{\tan^{-1} y}$
(D) $e^{\tan^{-1} x}$
(a) Find the value of $\sin^{-1}\left$\sin \frac{2\pi}{3}\right$ + \tan^{-1}\left$\tan \frac{5\pi}{6}\right$ + \cos^{-1}\left$\cos \frac{13\pi}{6}\right$$.
OR
(b) Prove that the function $f : \text{N} \to \text{N}$, $f(n) = (n^2 + n + 1)$ is one-one but not onto.
If $\sin^{-1} x = y$, then $\frac{\text{d}y}{\text{d}x}$ is :
(A) $\cos^{-1} x$
(B) $\cos y$
(C) $\frac{1}{1 - x^2}$
(D) $\sec y$
(a) Find the value of $\sin [\cot^{-1} \sqrt{2} \, (\cos \, (\tan^{-1} 1))]$.
If $\tan^{-1} x = y$, then $\frac{\text{dy}}{\text{dx}}$ is equal to :
(A) $(\sec^{-1} x)^2$
(B) $\sec^2 y$
(C) $\frac{1}{\sqrt{1+x^2}}$
(D) $\cos^2 y$
$\int \frac{\text{dx}}{\sqrt{e^{-2x}-1}}$ is equal to :
(A) $\sin^{-1} e^{-x} + C$
(B) $\log \mid e^{-x} + \sqrt{e^{-2x}-1} \mid + C$
(C) $\sin^{-1} e^x + C$
(D) $\log \mid e^{-x} - \sqrt{e^{-2x}-1} \mid + C$
The domain of $\sin^{-1} (1 - 2x)$ is :
(A) $[-1, 1]$
(B) $[-1, 3]$
(C) $[-2, 2]$
(D) $[0, 1]$
The domain of $\cos^{-1} (4x + 1)$ is :
(A) $[-1, 1]$
(B) $[-3, 5]
(C) $[-4, 4]$
(D) $\left[-\frac{1}{2}, 0\right]$
If $2 \cos^{-1} x = y$, then $\frac{dy}{dx}$ is :
(A) $- 2 \sin^{-1} x$
(B) $-\frac{1}{2} \sin \frac{y}{2}$
(C) $\frac{-1}{\sqrt{1 - 2x^2}}$
(D) $- 2 \csc \frac{y}{2}$
$\int \frac{\cos x}{\sqrt{\sin^2 x + 1}} \, dx$ is equal to :
(A) $\tan^{-1}(\sin x) + C$
(B) $\sin^{-1}(\sin x) + C$
(C) $\log | \sin x + \sqrt{1 + \sin^2 x} | + C$
(D) $\log | \sin x - \sqrt{\sin^2 x + 1} | + C$
If $x = e^{\sin^{-1} t}$ and $y = e^{\cos^{-1} t}$, then find $\frac{dy}{dx}$ at $t = \frac{1}{\sqrt{2}}$.
The following graph represents :
[Graph provided in image showing curves asymptotic to y-axis with points labeled at (0, π/2), (1, π/2), (-1, 0), (1, 0), (-1, -π/2), and (0, -π/2)]
(A) $y = \cos^{-1} x$
(B) $y = \sec^{-1} x$
(C) $y = \tan^{-1} x$
(D) $y = \text{cosec}^{-1} x$
The following graph represents : (A) $y = \sec^{-1} x$ (B) $y = \cot^{-1} x$ (C) $y = \tan^{-1} x$ (D) $y = \text{cosec}^{-1} x$
$\int \frac{\text{dx}}{\sqrt{25 - 16x^2}}$ is equal to : (A) $\frac{1}{5} \sin^{-1} 4x + C$ (B) $\frac{1}{25} \sin^{-1} 16x + C$ (C) $\frac{1}{4} \sin^{-1} \frac{4x}{5} + C$ (D) $\frac{1}{16} \sin^{-1} \frac{4x}{5} + C$
The following graph represents : [Graph provided] (A) $y = \cos^{-1} x$ (B) $y = \sec^{-1} x$ (C) $y = \sin^{-1} x$ (D) $y = \tan^{-1} x$
The principal value of $\sec^{-1}(\sqrt{2}) + 2 \csc^{-1}(-\sqrt{2})$ is :
(A) $-\frac{\pi}{2}
(B) $-\frac{\pi}{4}
(C) $\frac{\pi}{4}
(D) $\frac{\pi}{2}$
For the inverse trigonometric functions, which of the following Principal Value Branch is not correctly defined ?
(A) $\tan^{-1} : \mathbb{R} \rightarrow \left$ -\frac{\pi}{2}, \frac{\pi}{2} \right$$
(B) $\sec^{-1} : \mathbb{R} - (-1, 1) \rightarrow [0, \pi] - \left\{ \frac{\pi}{2} \right\}$
(C) $\c...
Evaluate $\sin \left[ \tan^{-1} \tan \left$ \frac{3\pi}{4} \right$ \right]$.
For the inverse trigonometric functions, which of the following Principal Value Branch is not correctly defined ?
(A) $\tan^{-1} : R \rightarrow \left$ -\frac{\pi}{2}, \frac{\pi}{2} \right$$
(B) $\sec^{-1} : R - (-1, 1) \rightarrow [0, \pi] - \left\{ \frac{\pi}{2} \right\}$
(C) $\cot^{-1} : R \right...
Evaluate $\sin \left[ \cos^{-1} \cos \left$ \frac{7\pi}{6} \right$ \right]$.
For the inverse trigonometric functions, which of the following Principal Value Branch is not correctly defined ? (A) $\tan^{-1} : \text{R} \to \left$ -\frac{\pi}{2}, \frac{\pi}{2} \right$$ (B) $\sec^{-1} : \text{R} - (-1, 1) \to [0, \pi] - \left\{ \frac{\pi}{2} \right\}$ (C) $\cot^{-1} : \text{R} \...
Evaluate $\tan\left[\cos^{-1}\left$\tan \frac{3\pi}{4}\right$\right]$.
Evaluate : $\cos^{-1}\left$\cos\frac{2\pi}{3}\right$ + \sin^{-1}\left$\sin\frac{2\pi}{3}\right$ + \tan^{-1}\left$\tan\frac{2\pi}{3}\right$$.
Differentiate $y = e^{\cos^{-1}\sqrt{1 - x^2}}$ with respect to $x$ and hence show that $\frac{\text{dy}}{\text{dx}} = \frac{y}{\sqrt{1 - x^2}}$.
The value of $\cos^{-1}\left[\cos \frac{7\pi}{6}\right]$ is :
(A) $\frac{7\pi}{6}$
(B) $\frac{5\pi}{6}$
(C) $\frac{\pi}{3}$
(D) $\frac{\pi}{6}$
The value of $\left[\tan^{-1}(-1) + \cos^{-1}\left$-\frac{1}{2}\right$ + \sin^{-1}\left(\frac{1}{2}\right)\right]$ is :
(A) $\frac{7\pi}{12}$
(B) $\frac{5\pi}{12}$
(C) $-\frac{5\pi}{12}$
(D) $\frac{\pi}{3}$
Assertion (A): Value of the expression $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) + \tan^{-1}1 - \sec^{-1}(\sqrt{2})$ is $\frac{\pi}{4}$.
Reason (R): Principal value branch of $\sin^{-1}x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and that of $\sec^{-1}x$ is $[0, \pi] - \left\{\frac{\pi}{2}\rig...
Evaluate $\tan \left(\tan^{-1}(-1) + \frac{\pi}{3}\right)$
Find the domain of $\cos^{-1}(3x - 2)$
$\int \frac{1}{9x^2 + 6x + 10} \, dx$ is equal to :
(A) $\frac{1}{3} \tan^{-1} (3x + 1) + C$
(B) $\frac{1}{9} \tan^{-1} (3x + 1) + C$
(C) $\tan^{-1} \frac{3x + 1}{3} + C$
(D) $\frac{1}{9} \tan^{-1} \frac{3x + 1}{3} + C$
The general solution of the differential equation $\frac{\text{dy}}{\text{dx}} = \frac{1 + y^2}{1 + x^2}$ is :
(A) $\tan^{-1}(y - x) = C$
(B) $\tan^{-1} y + \tan^{-1} x = C$
(C) $\tan^{-1} \frac{y}{2} - \tan^{-1} \frac{x}{2} = C$
(D) $\tan^{-1} y = \tan^{-1} x + C$
Evaluate : $\sec^2 \left(\tan^{-1} 3\right) + \text{cosec}^2 \left(\cot^{-1} 2\right)$
(b) Find $\frac{dy}{dx}$, if $y = \tan^{-1} \left(\frac{1 + \sin x}{\cos x}\right)$.
(a) If $y = \tan^{-1}\left[\frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}\right]$, $x^2 \le 1$, then find $\frac{\text{d}y}{\text{d}x}$.
OR
(b) If $y = \left(x + \sqrt{1+x^2}\right)^n$, then show that
$(1+x^2)\frac{\text{d}^2y}{\text{d}x^2} + x\frac{\text{d}y}{\text{d}x} = n^2y$.
The derivative of $\sin^{-1} (2\text{x}^2 - 1)$ with respect to $\sin^{-1} \text{x}$ is :
(A) $\frac{2}{\text{x}}$
(B) $2$
(C) $\frac{\sqrt{1-\text{x}^2}}{\sqrt{1-4\text{x}^2}}$
(D) $1-\text{x}^2$
The principal value of $\cos^{-1}\left$-\frac{1}{\sqrt{2}}\right$$ is :
(A) $\frac{\pi}{4}$
(B) $-\frac{\pi}{4}$
(C) $\frac{3\pi}{4}$
(D) $\frac{2\pi}{3}$
(a) Find the principal value of $\cos^{-1}\left$-\frac{1}{2}\right$ + 2\sin^{-1}\left(\frac{1}{2}\right)$.
OR
(b) Prove that :
$\tan^{-1}\sqrt{\text{x}} = \frac{1}{2}\cos^{-1}\left(\frac{1-\text{x}}{1+\text{x}}\right)$, $\text{x} \in [0, 1]$
If a function $f : X \to Y$ defined as $f(x) = y$ is one-one and onto, then we can define a unique function $g : Y \to X$ such that $g(y) = x$, where $x \in X$ and $y = f(x)$, $y \in Y$. Function $g$ is called the inverse of function $f$.
The domain of sine function is $R$ and function $\text{sine} ...
Evaluate : $\tan^{-1} \left[ 2 \sin \left$ 2 \cos^{-1} \frac{\sqrt{3}}{2} \right$ \right]$
Evaluate: $\sin^{-1} \left$ \sin \frac{3\pi}{5} \right$$
(a) Differentiate $\tan^{-1} \frac{\sqrt{1 - x^2}}{x}$ w.r.t. $\cos^{-1} (2x\sqrt{1 - x^2})$, $x \in \left(\frac{1}{\sqrt{2}}, 1\right)$
OR
(b) Find $\frac{dy}{dx}$, if $y = x^{\tan x} + \frac{\sqrt{x^2 + 1}}{2}$.
Solve for $x$,
$2 \tan^{-1} x + \sin^{-1}\left(\frac{2x}{1 + x^2}\right) = 4\sqrt{3}$
(a) Differentiate $2^{\cos^2 x}$ w.r.t $\cos^2 x$.\nOR\n(b) If $\tan^{-1}(x^2 + y^2) = a^2$, then find $\frac{dy}{dx}$.
If $y = \sin^{-1}x$, $-1 \le x \le 0$, then the range of $y$ is
(A) $\left$\frac{-\pi}{2}, 0
ight$$
(B) $\left[\frac{-\pi}{2}, 0
ight]$
(C) $\left$\frac{-\pi}{2}, 0
ight]$
(D$ $\left[\frac{-\pi}{2}, 0
ight)$
The principal value of $\sin^{-1} \left$ \cos \frac{43\pi}{5} \right$$ is
(A) $\frac{-7\pi}{5}$
(B) $\frac{-\pi}{10}$
(C) $\frac{\pi}{10}$
(D) $\frac{3\pi}{5}$
Value of $4 \cos \left[ \frac{1}{2} \cos^{-1} \left$ \frac{1}{8} \right$ \right]$ is
(A) $3$
(B) $-3$
(C) $1$
(D) $-1$
If $y = \tan^{-1}\left(\frac{1-\cos x}{\sin x}\right)$, then $\frac{\text{d}y}{\text{d}x}$ is
(A) $1$
(B) $\frac{1}{2}$
(C) $-\frac{1}{2}$
(D) $-1$
$\int \frac{2x^3}{4+x^8} \, dx$ is equal to
(A) $\frac{1}{4} \tan^{-1} \frac{x^4}{2} + C$
(B) $\frac{1}{2} \tan^{-1} \frac{x^4}{2} + C$
(C) $\frac{1}{4} \tan^{-1} \frac{x^4}{4} + C$
(D) $\frac{1}{4} \tan^{-1} x^4 + C$
Evaluate : $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$
The principal value of $\sin^{-1} \left( \sin \left$ -\frac{10\pi}{3} \right$ \right)$ is :
(A) $-\frac{2\pi}{3}$
(B) $-\frac{\pi}{3}$
(C) $\frac{\pi}{3}$
(D) $\frac{2\pi}{3}$
(a) Simplify $\sin^{-1} \left$ \frac{\text{x}}{\sqrt{1 + \text{x}^2}} \right$$. OR (b) Find domain of $\sin^{-1} \sqrt{\text{x} - 1}$.
The principal branch of $\cos^{-1} x$ is :
(A) $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$
(B) $[\pi, 2\pi]$
(C) $[0, \pi]$
(D) $[2\pi, 3\pi]$
Domain of $\sin^{-1} (2x^{2} - 3)$ is : (A) $(-1, 0) \cup (1, \sqrt{2})$ (B) $(- \sqrt{2}, - 1) \cup (0, 1)$ (C) $[- \sqrt{2}, - 1] \cup [1, \sqrt{2}]$ (D) $(- \sqrt{2}, - 1) \cup (1, \sqrt{2})$
Find the domain of $\sec^{-1}(2x + 1)$.
Find the domain of $\sin^{-1}(x^2 - 3)$.
The following graph is a combination of :
(A) $y = \sin^{-1} x$ and $y = \cos^{-1} x$
(B) $y = \cos^{-1} x$ and $y = \cos x$
(C) $y = \sin^{-1} x$ and $y = \sin x$
(D) $y = \cos^{-1} x$ and $y = \sin x$
The graph shown below depicts :
(A) $y = \cot x$
(B) $y = \cot^{-1} x$
(C) $y = \tan x$
(D) $y = \tan^{-1} x$
The graph shown below depicts :
(A) $y = \sec^{-1} x$
(B) $y = \sec x$
(C) $y = \text{cosec}^{-1} x$
(D) $y = \text{cosec } x$
$\left[ \sec^{-1} \left(-\sqrt{2}\right) - \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \right]$ is equal to :
(A) $\frac{11\pi}{12}$
(B) $\frac{5\pi}{12}$
(C) $-\frac{5\pi}{12}$
(D) $\frac{7\pi}{12}$
Find the domain of $\text{f(x)} = \sin^{-1} (- x^{2})$.
The given graph illustrates : (A) $y = \tan^{-1} x$ (B) $y = \text{cosec}^{-1} x$ (C) $y = \cot^{-1} x$ (D) $y = \sec^{-1} x$
$\int \frac{\text{e}^x}{\sqrt{4 - \text{e}^{2x}}} \, \text{d}x$ is equal to :
(A) $\frac{1}{2} \cos^{-1}(\text{e}^x) + C$
(B) $\frac{1}{2} \sin^{-1}(\text{e}^x) + C$
(C) $\frac{\text{e}^x}{2} + C$
(D) $\sin^{-1}\left(\frac{\text{e}^x}{2}\right) + C$
Study the given graph. It illustrates : (A) $y = \tan^{-1} x$ (B) $y = \cos^{-1} x$ (C) $y = \sec^{-1} x$ (D) $y = \sin^{-1} x$
$\int \frac{a^x}{\sqrt{1 - a^{2x}}} \, \text{d}x$ is equal to :
(A) $\frac{\sin^{-1}(a^x)}{\log_e a} + C$
(B) $\log_e(1 - a^{2x}) + C$
(C) $\frac{\cos^{-1}(a^x)}{\log_e a} + C$
(D) $\frac{\sin^{-1}(a^x)}{a^x} + C$
28. Show that the derivative of $\tan^{-1}(\sec x + \tan x)$, $\left[-\frac{\pi}{2} < x < \frac{\pi}{2}\right]$ with respect to x is equal to $\frac{1}{2}$.
The given graph illustrates : (A) $y = \sec^{-1} x$ (B) $y = \cot^{-1} x$ (C) $y = \tan^{-1} x$ (D) $y = \text{cosec}^{-1} x$
$\int \frac{e^{-x}}{16 + 9e^{-2x}} \, dx$ is equal to :
(A) $\frac{16}{9} \tan^{-1} (e^{-x}) + C$
(B) $-\frac{1}{12} \tan^{-1} \left(\frac{3e^{-x}}{4}
ight) + C$
(C) $\tan^{-1} \left(\frac{e^{-x}}{4}
ight) + C$
(D) $-\frac{1}{3} \tan^{-1} \left(\frac{e^{-x}}{4}
ight) + C$
Assertion (A) : Set of values of $\sec^{-1}\left(\frac{\sqrt{3}}{2}\right)$ is a null set.
Reason (R) : $\sec^{-1} x$ is defined for $x \in \mathbb{R} - (-1, 1)$.
Solve the following differential equation :
$$(\tan^{-1} y - x)\ \mathrm{d}y = (1 + y^2)\ \mathrm{d}x$$
If A is the interval other than principal value branch, give an example of one such interval.
If $\sin^{-1} (x)$ is defined from $[-1, 1]$ to its principal value branch, find the value of $\sin^{-1} \left$ -\frac{1}{2} \right$ - \sin^{-1} (1)$.
Draw the graph of $\sin^{-1} x$ from $[-1, 1]$ to its principal value branch.
Q.15. The graph drawn below depicts
(A) $y = \sin^{-1} x$
(B) $y = \cos^{-1} x$
(C) $y = \operatorname{cosec}^{-1} x$
(D) $y = \cot^{-1} x$
If $\cot^{-1}(3x + 5) > \frac{\pi}{4}$, then find the range of the values of $x$.
(a) Find the derivative of $\tan^{-1} x$ with respect to $\log x$; (where $x \in (1, \infty)$).
(b) If $y = \text{cosec}(\cot^{-1} x)$, then prove that $\sqrt{1+x^2} \frac{dy}{dx} - x = 0$.
If $y = \cos^{-1}(e^x)$, then $\frac{dy}{dx}$ is :
(A) $\frac{1}{\sqrt{e^{-2x} + 1}}$
(B) $-\frac{1}{\sqrt{e^{-2x} + 1}}$
(C) $\frac{1}{\sqrt{e^{-2x} - 1}}$
(D) $-\frac{1}{\sqrt{e^{-2x} - 1}}$
(a) Find the value of $\sin^{-1}\left$-\frac{1}{2}\right$ + \cos^{-1}\left$-\frac{\sqrt{3}}{2}\right$ + \cot^{-1}\left$\tan\frac{4\pi}{3}\right$$. OR (b) Find the domain of $f(x) = \cos^{-1}(1 - x^{2})$. Also, find its range.
(a) If y = $(\log x)^2$, prove that $x^2y'' + xy' = 2$. OR (b) If y = $\sin(\tan^{-1} e^x)$, then find $\frac{dy}{dx}$ at x = 0.
Assertion (A) : $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$
Reason (R) : $\cos\left$\frac{\pi}{6}\right$ = \frac{\sqrt{3}}{2}$
(a) Find the value of $\tan^{-1}(1) + \cos^{-1}\left$-\frac{1}{2}\right$ + \sin^{-1}\left$-\frac{1}{2}\right$$.
OR
(b) Find the domain of the function $y = \cos^{-1}(x^2 - 4)$.
(a) Differentiate $\cot^{-1}(\sqrt{1 + x^2} + x)$ w.r.t. x.
OR
(b) If $(\cos x)^y = (\cos y)^x$, find $\frac{dy}{dx}$.
Evaluate : $\sec^2 \left$\tan^{-1} \frac{1}{2}\right$ + \operatorname{cosec}^2 \left$\cot^{-1} \frac{1}{3}\right$$
Find the value of $\left[ \sin^2 \left\{ \cos^{-1} \left$ \frac{3}{5} \right$ \right\} + \tan^2 \left\{ \sec^{-1} (3) \right\} \right]$.
If $y = \sin^{-1} x$, then $\frac{d^2y}{dx^2}$ is :
(A) $\sec y$
(B) $\sec y \tan y$
(C) $\sec^2 y \tan y$
(D) $\tan^2 y \sec y$
Evaluate :
$\cot^2 \{ \text{cosec}^{-1} 3 \} + \sin^2 \{ \cos^{-1} \left(\frac{1}{3}\right) \}$
Find $\frac{dy}{dx}$, if $y = (\cos x)^x + \cos^{-1} \sqrt{x}$ is given.
The integral $\int \frac{dx}{\sqrt{9-4x^{2}}}$ is equal to :
(A) $\frac{1}{6}\sin^{-1}\left(\frac{2x}{3}\right)+c$
(B) $\frac{1}{2}\sin^{-1}\left(\frac{2x}{3}\right)+c$
(C) $\sin^{-1}\left(\frac{2x}{3}\right)+c$
(D) $\frac{3}{2}\sin^{-1}\left(\frac{2x}{3}\right)+c$
The integral $\int \frac{\text{dx}}{\sqrt{9-4x^2}}$ is equal to :
(A) $\frac{1}{6}\sin^{-1}\left(\frac{2x}{3}\right)+c$
(B) $\frac{1}{2}\sin^{-1}\left(\frac{2x}{3}\right)+c$
(C) $\sin^{-1}\left(\frac{2x}{3}\right)+c$
(D) $\frac{3}{2}\sin^{-1}\left(\frac{2x}{3}\right)+c$
Assertion (A) : Domain of $y = \cos^{-1}(x)$ is $[-1, 1]$.
Reason (R) : The range of the principal value branch of $y = \cos^{-1}(x)$ is $[0, \pi] - \left\{ \frac{\pi}{2} \right\}$.
Find the value of $\cos^{-1} \left$ \frac{1}{2} \right$ - \tan^{-1} \left$ -\frac{1}{\sqrt{3}} \right$ + \text{cosec}^{-1} (-2)$.
Find the value of \tan^{-1} \left$ \tan \frac{3\pi}{5} \right$ + \cos^{-1} \left$ \cos \frac{13\pi}{6} \right$ + \sin^{-1} \left$ -\frac{1}{2} \right$.
If $y = e^{a \cos^{-1} x}$, then show that $(1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} - a^2y = 0$.
Evaluate: $\sin^{-1} \left$ \sin \frac{3\pi}{4} \right$ + \cos^{-1} \left$ \cos \frac{3\pi}{4} \right$ + \tan^{-1} (1)$.
If $x = \sqrt{a^{\tan^{-1} t}}$, $y = \sqrt{a^{\cot^{-1} t}}$, then show that $x \frac{dy}{dx} + y = 0$.
Find the value of $\sin^{-1}\left(\cos\left$\frac{33\pi}{5}\right$\right)$. OR Find the domain of $\sin^{-1}(x^2-4)$.
Assertion (A) : Maximum value of $(\cos^{-1} x)^2$ is $\pi^2$.
Reason (R) : Range of the principal value branch of $\cos^{-1} x$ is $[\frac{-\pi}{2}, \frac{\pi}{2}]$.
If $y = \sec(\tan^{-1} x)$, then $\frac{dy}{dx}$ at $x = 1$ is equal to :
(a) $\sqrt{2}$
(b) $\frac{1}{\sqrt{2}}$
(c) $1$
(d) $\frac{1}{2}$
(a) Find the value of $\tan^{-1} \sqrt{3} - \sec^{-1} (-2)$. OR (b) Check the injectivity and surjectivity of the function f : N $\rightarrow$ N, given by f(x) = $x^3$.
Questions number 19 and 20 are Assertion and Reason based questions carrying 1 mark each. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (a), (b), (c) and (d) as given below.
(a) Both Assertion (A) and Reason (R) are t...
Derivative of $\tan^{-1}\left(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\right)$ with respect to $\sin^{-1}\left(2\text{x}\sqrt{1-\text{x}^2}\right)$ is :
(a) $-\frac{1}{4}$
(b) $\frac{1}{2}$
(c) $2$
(d) $-\frac{1}{2}$
Solution of the differential equation $(1 + y^2) (1 + \log x) dx + x dy = 0$ is :
(a) $\tan^{-1} y + \log |x| + \frac{(\log |x|)^2}{2} = C$
(b) $\tan^{-1} y - \log |x| + \frac{(\log |x|)^2}{2} = C$
(c) $\tan^{-1} y - \log |x| - \frac{(\log |x|)^2}{2} = C$
(d) $\tan^{-1} y + \log |x| - \frac{(\log |x...
The angle between the lines $\frac{x + 1}{1} = \frac{4 - y}{-1} = \frac{z - 5}{2}$ and $\frac{x + 3}{-3} = \frac{y - 2}{5} = \frac{z + 5}{4}$ is :
(a) $\cos^{-1} \left$ \frac{2}{3} \right$$
(b) $\cos^{-1} \left$ \frac{1}{\sqrt{3}} \right$$
(c) $\frac{\pi}{2}$
(d) $\frac{\pi}{4}$
If $y = \sin^{-1} (2x \sqrt{1 - x^2})$, $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$, then $\frac{dy}{dx}$ is :
(a) $2$
(b) $\frac{2}{\sqrt{1 - x^2}}$
(c) $-\frac{2}{\sqrt{1 - x^2}}$
(d) $\sqrt{1 - x^2}$
If $y = \sin^{-1}x$, then $(1 - x^2)y_2$ is equal to
(a) $xy_1$ (b) $xy$ (c) $xy_2$ (d) $x^2$
Assertion (A): The domain of the function $\sec^{-1} 2x$ is $(-\infty, -\frac{1}{2}] \cup [\frac{1}{2}, \infty)$
Reason (R): $\sec^{-1}(-2) = -\frac{\pi}{4}$