1 Marks Question

Question 11 Mark

Find the value of $3 \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer

$
3 \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\sin \left(\frac{\sqrt{3}}{2}\right)=3 \times \frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}+\frac{\pi}{3}=\frac{5 \pi}{6}
$

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Question 21 Mark

Find the principal value of $\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Answer

Suppose that $\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=y, \Rightarrow \cos y=\frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}$ We know that the range of principal value branch of $\cos ^{-1} x$ is $[0, \pi]$ and $\cos y=\cos \frac{\pi}{4}$. So, the principal value of $\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)$ will be $\frac{\pi}{4}$.

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Question 31 Mark

Find the domain of $f(x)=\cos ^{-1}\left(x^2-4\right)$.

Answer

$f(x)=\cos ^{-1}\left(x^2-4\right)$ will be defined if $-1 \leq x^2-4 \leq 1$$
\begin{array}{ll}
\Rightarrow & -1+4 \leq x^2-4+4 \leq 1+4 \\
\Rightarrow & 3 \leq x^2 \leq 5 \\
\Rightarrow & x \in[-\sqrt{5},-\sqrt{3}] \cup[\sqrt{3}, \sqrt{5}] \\
& \left\{\because a^2 \leq x^2 \leq b^2 \Leftrightarrow x \in[-b,-a] \cup[a, b\right.
\end{array}
$

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Question 41 Mark

If $\sin ^{-1} x=\frac{\pi}{3}$ then write the value of $\cos ^{-1} x$

Answer

$\sin ^{-1}\left\{2 x \sqrt{1-x^2}\right\}=\sin ^{-1}\left\{2 \sin \left(\frac{1}{2}\right) \sqrt{1-\sin ^2\left(\frac{1}{2}\right)}\right\}$${\left[\because \text { given } \sin ^{-1}(x)=\frac{1}{2}\right]}$
$=\sin ^{-1}\left\{2 \sin \left(\frac{1}{2}\right) \sqrt{\cos ^2\left(\frac{1}{2}\right)}\right\}$
$=\sin ^{-1}\left\{2 \sin \left(\frac{1}{2}\right) \cos \left(\frac{1}{2}\right)\right\}$
$=\sin ^{-1}\left\{\sin \left(2 \times \frac{1}{2}\right)\right\}$
$=\sin ^{-1}(\sin (1))$
$=1 \text {}$

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Question 51 Mark

Write the value of $\cos \left[\left(\frac{\pi}{2}\right)+\sin ^{-1}\left(\frac{1}{3}\right)\right]$

Answer

suppose $\sin ^{-1} \frac{1}{3}=\theta$
$
\begin{aligned}
\therefore \quad \cos \left(\frac{\pi}{2}+\theta\right)=-\sin \theta & =-\sin \left(\sin ^{-1}\left(\frac{1}{3}\right)\right) \\
& =-\left(\frac{1}{3}\right)=-\frac{1}{3} \text { }
\end{aligned}
$

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Question 61 Mark

Find the value of expression $2 \sec ^{-1} 2+\sin ^{-1}\left(\frac{1}{2}\right)$

Answer

$2 \sec ^{-1}(2)+\sin ^{-1}\left(\frac{1}{2}\right)$
$=2 \times \frac{\pi}{3}+\frac{\pi}{6}$
$=\frac{2 \pi}{3}+\frac{\pi}{6}$
$=\frac{5 \pi}{6} \text { }$

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