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MATHS M.C.Q (1 Marks)

Inverse Trigonometric Functions · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 15 questions.

Questions

M.C.Q (1 Marks)

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15 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark
The domain of function $\cos ^{-1}(2 x-3)$ is :
Answer
(D)
The domain of $\cos ^{-1}(\theta)$ is $-1 \leq \theta \leq 1$$
\begin{array}{ll}
\text { So, } & -1 \leq 2 x-3 \leq 1 \\
\Rightarrow & -1+3 \leq 2 x-3+3 \leq 1+3 \\
\Rightarrow & 2 \leq 2 x \leq 4 \\
\Rightarrow & 1 \leq x \leq 2 \\
\Rightarrow & x \in[1,2]
\end{array}
$
Hence correct option is (D).
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Question 21 Mark
The principal value of $\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)$ :
Answer
$\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)=\tan ^{-1}\left[\tan \left(\pi+\frac{\pi}{8}\right)\right]$
$=\tan ^{-1}\left[\tan \frac{\pi}{8}\right]=\frac{\pi}{8}$
Hence, correct option is $(A)$
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Question 31 Mark
The principal value of $\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$
Answer
$\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$
$=\frac{\pi}{3}+\left(\frac{-\pi}{4}\right)=\frac{\pi}{12}$
Hence correct opiton is $(A).$
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Question 41 Mark
Principal value of $\tan ^{-1}(-1)$ :
Answer
(C)
Suppose $\tan ^{-1}(-1)=x$
$\therefore$ tan x = -1 = -tan $45^{\circ}$
tan x =tan $\left(-45^{\circ}\right)$ $\therefore$ x = $-45^{\circ}$
$\because$The range of principal value branch of $\tan ^{-1} x$ is
$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Hence correct option is (C).
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Question 61 Mark
$\cot ^{-1} \frac{\sqrt{1-x^2}}{x}$ equal to :
Answer
(B)
$\cot ^{-1} \frac{\sqrt{1-x^2}}{x}$
taking $x=\sin \theta$
$\Rightarrow \quad \cot ^{-1} \frac{\sqrt{1-\sin ^2 \theta}}{\sin \theta}=\cot ^{-1} \frac{\cos \theta}{\sin \theta}$
$\Rightarrow \quad \cot ^{-1}(\cot \theta)=\theta=\sin ^{-1} x=\operatorname{cosec}^{-1}\left(\frac{1}{x}\right)$
Hence correct option is (B).
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Question 71 Mark
The domain of function $\sin ^{-1} 2 x$ is :
Answer
(C)
$\quad \sin ^{-1} 2 x$ will be defined if $-1 \leq 2 x \leq 1 \Rightarrow-\frac{1}{2} \leq$$
x \leq \frac{1}{2} \Rightarrow x \in\left[-\frac{1}{2}, \frac{1}{2}\right]
$
Hence correct option is (C)
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Question 81 Mark
If $\tan ^{-1}(1)+\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\sin ^{-1} x$ the value of $x$ is :
Answer
(C)
$\tan ^{-1}(1)+\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\sin ^{-1} x$
$\Rightarrow \quad \frac{\pi}{4}+\frac{\pi}{4}=\sin ^{-1} x$
$\Rightarrow \quad \frac{\pi}{2}=\sin ^{-1} x \therefore x=\sin \frac{\pi}{2}=1$
Hence correct option is (C).
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Question 91 Mark
The value of $\sin \cot ^{-1} x$ :
Answer
(D)
$\sin \cot ^{-1} x$
Suppose $\cot ^{-1} x=\theta \therefore x=\cot \theta$
$\therefore \quad \sin \theta=\frac{1}{\sqrt{1+x^2}}$
or $\sin \theta=\left(1+x^2\right)^{\frac{-1}{2}}$ Hence correct option is (D).

Image
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Question 101 Mark
If $\sin ^{-1}\left(\frac{1}{2}\right)=x$ then find the general value of $x$ :
Answer
(D)
If $\sin ^{-1}\left(\frac{1}{2}\right)=x$ then $\sin x=\frac{1}{2}$
$\sin x=\sin \frac{\pi}{6}$ then general value of $x=n \pi+(-1)^n \frac{\pi}{6}$
Hence correct option is (D).
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Question 111 Mark
The domain of $\cos ^{-1}(2 x-1)$
Answer
$\cos ^{-1}(2 x-1)$ will be defined if $-1 \leq 2 x-1 \leq 1$
$\Rightarrow 0 \leq 2 x \leq 2$
$\Rightarrow 0 \leq x \leq 1$
$\Rightarrow x \in[0,1]$
Hence domain of $\cos ^{-1}(2 x-1)[0,1]$
Hence correct option is $(A)$
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Question 121 Mark
The domain of function $f(x)=\sin ^{-1} x+\cos x$ is-
Answer
(A)
$[-1,1]$
Function $\sin ^{-1} x$ has domain $D _1=[-1,1]$ and function $\cos x$ has domain $D _2= R$
$\therefore$ Domain of $f(x)=\sin ^{-1} x+\cos x$ is $= D _1 \cap D _2=$ $[-1,1] \cap R =[-1,1]$
Hence correct option is (A).
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Question 131 Mark
Domain of $f(x)=\sin ^{-1}\left(-x^2\right)$ is :
Answer
(C)
$f(x)=\sin ^{-1}(-x)^2$ will be defined if $-1 \leq-x^2 \leq 1$$
\begin{array}{ll}
\Rightarrow & 1 \geq x^2 \geq-1 \\
\Rightarrow & 0 \leq x^2 \leq 1 \\
\Rightarrow & x^2-1 \leq 0
\end{array}
$
$
\begin{array}{ll}
\Rightarrow & -1 \leq x \leq 1 \\
\Rightarrow & x \in[-1,1]
\end{array}
$
Hence function $f(x)=\sin ^{-1}\left(-x^2\right)$ has domain $[-1,1]$
Hence correct option is (C)
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Question 141 Mark
Domain of $f(x)=\sin ^{-1} \sqrt{x-1}$ is :
Answer
(A)
$f(x)=\sin ^{-1} \sqrt{x-1}$ will be defined if$
\begin{array}{ll} 
& x-1 \geq 0 \text { and }-1 \leq \sqrt{x-1} \leq 1 \\
\Rightarrow & x \geq 1 \text { and } 0 \leq \sqrt{x-1} \leq 1 \quad[\because \sqrt{x-1} \geq 0] \\
\Rightarrow & x \geq 1 \text { and } 1 \leq x \leq 2 \\
\Rightarrow & x \in[1,2]
\end{array}
$
So, domain of $f(x)$ is $[1,2]$
Hence correct option is (A)
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Question 151 Mark
The value of $\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)+2 \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ :
Answer
(C)
The range of principal value branch of $\sin ^{-1} x$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ and the range of principal value branch of $\cos ^{-1} x$ is $[0, \pi]$.
$\therefore \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$ and $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}$
Hence, $\frac{\pi}{3}+2 \times \frac{\pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}$
Hence correct option is (C)
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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip