1 Marks Question

MCQ 511 Mark

$\text{If}\ \sin^{-1}\text{x = y},$ then

A
$ 0 \geq\text{y}\geq {\pi}$
✓
$-\frac{\pi}{2}\leq\text{y}\leq\frac{\pi}{2} $
C
$0 < \text{y} < {\pi}$
D
$-\frac{\pi}{2} < \text{y} < \frac{\pi}{2}$

Answer

Correct option: B.

$-\frac{\pi}{2}\leq\text{y}\leq\frac{\pi}{2} $

$\text{y}=\sin^{-1}\text{x}$
$ \therefore\ -\frac{\pi}{2}\leq\text{y}\leq \frac{\pi}{2}$
$\therefore (b)$ is correct answer.

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MCQ 521 Mark

$\sin^{-1}\left(1-x\right)-2\sin^{-1}x=\frac{\pi}{2},$ then $x$ is equal to :

A
$0,\frac{1}{2}$
B
$1,\frac{1}{2}$
✓
$0$
D
$\frac{1}{2}$

Answer

Correct option: C.

$0$

$\sin^{-1}\left(1-x\right)-2\sin^{-1}x=\frac{\pi}{2}$
$\Rightarrow-2\sin^{-1}x=\frac{\pi}{2}-\sin^{-1}\left(1-x\right)$
$ \Rightarrow-2\sin^{-1}x=\cos^{-1}\left(1-x\right)....(\text{i})$
$\text{Let}\sin^{-1}x=\theta$
$\Rightarrow\sin\theta=x$
$\Rightarrow\cos\theta=\sqrt{1-x^2}.$
$ \therefore\theta=\cos^{-1}\left(\sqrt {1-x^2}\right)$
$ \therefore\sin^{-1}x=\cos^{-1}\left(\sqrt{1-x^2}\right)$
Therefore, from equation $(1),$ we have
$-2\cos^{-1}\left(\sqrt{1-x^2}\right)=\cos^{-1}\left(1-x\right)$
Put $x = \sin y,$ Then, we have
$-2\cos^{-1}\bigg(\sqrt{1-\sin^2y}\bigg)=\cos^{-1}\left(1-\sin y\right)$
$\Rightarrow-2\cos^{-1}\left(\cos y\right)=\cos^{-1}\left(1-\sin y\right)$
$\Rightarrow1-\sin y=\cos\left(-2y\right)=\cos2y$
$ \Rightarrow1-\sin y=1-2\sin^2y$
$\Rightarrow2\sin^2 y-\sin y=0$
$\Rightarrow\sin y\left(2\sin y-1\right)=0$
$\Rightarrow\sin y=0$ or $ \frac{1}{2}$
$\therefore x=0$ or $x=\frac{1}{2}$
But, when $x=\frac{1}{2}$, it can be observed that:
$\text{L.H.S.}=\sin^{-1}\bigg(1-\frac{1}{2}\bigg)-2\sin^{-1}\frac{1}{2}$
$=\sin^{-1}\bigg(\frac{1}{2}\bigg)-2\sin^{-1}\frac{1}{2}$
$=-\sin^{-1}\frac{1}{2}$
$=-\frac{\pi}{6}\neq\frac{\pi}{2}\neq\text{R.H.S.}$
$\therefore x=\frac{1}{2}$ is not the solution of the given equation.
Thus, $x=0.$
Hence, the correct answer is $(c).$

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Question 531 Mark

Find the value of $\tan^{-1}\Big(\tan\frac{9\pi}{8}\Big)$

Answer

$\tan^{-1}\Big(\tan\frac{9\pi}{8}\Big)=\tan^{-1}\Big[\tan\Big(\pi+\frac{\pi}{8}\Big)\Big]$
$=\tan^{-1}\Big[\tan\Big(\frac{\pi}{8}\Big)\Big]$
$=\frac{\pi}{8}$

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Question 541 Mark

Find the principal value of the following:
$\sin^{-1}\Big(\cos\frac{2\pi}{3}\Big)$

Answer

$\sin^{-1}\Big(\frac{\cos2\pi}{3}\Big)=\sin^{-1}\Big(-\frac{1}{2}\Big)$
$=\sin^{-1}\Big[\sin\Big(-\frac{\pi}{6}\Big)\Big]=-\frac{\pi}{6}$

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Question 551 Mark

Evaluate the following:
$\sin\Big(\cos^{-1}\frac{5}{13}\Big)$

Answer

$\sin\Big(\cos^{-1}\frac{5}{13}\Big)$
$=\sin\Big(\sin^{-1}\frac{12}{13}\Big)$ $\Big[{\therefore\ \cos^{-1}}\text{x}=\sin^{-1}\sqrt{1-\text{x}^2}\Big]$
$ =\frac{12}{13}$

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Maths 1 Marks Question