Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

If $$x$$ takes negative permissible value, then $$\sin^{-1}x=$$

A
$$\cos^{-1}\sqrt{1-x^{2}}$$

B
$${-\cos^{-1}}\sqrt{1-x^{2}}$$

C
$${\cos^{-1}}\sqrt{x^{2}-1}$$

D
$${\pi-\cos^{-1}}\sqrt{1-x^{2}}$$

Solution

If $$-1\leq x <0$$, then $$ \dfrac{-\pi}{2} \leq sin^{-1}(x) < 0$$
Let, $$sin \theta = x$$
Hence, $$ \dfrac{-\pi}{2} \leq \theta < 0$$
Hence, $$cos \theta = \sqrt{1-x^2}$$
Since, the range of $$cos^{-1}(x)$$ is $$[0,\pi]$$,
$$\theta = - cos^{-1} ( \sqrt{1-x^2} )$$
Question 2
1 / -0

There is flag-staff at the top of $$10$$ metres high tower. lf the flag-staff makes an angle $$\tan ^{ -1 }{ \left( 1/8 \right) }$$ at a point $$24$$ metres away from the tower, then the height of the flag staff in metres is

A
$$26/7$$

B
$$27/8$$

C
$$27/6$$

D
$$26/3$$

Solution

$$tan\theta =\frac{1}{8}, tan\alpha =\frac{10}{24}$$
$$=\frac{5}{12}$$
$$tan(\alpha +\theta )= \frac{h+10}{24}$$
$$\frac{tan\alpha +tan\theta} {1-tan\alpha tan\theta }=\frac{h+10}{24}$$
$$\frac{1}{8}+\frac{5}{12}1-\frac{5}{12} \times\frac{1}{8}=\frac{h+10}{24}$$
$$\frac{12+40}{(69-5)}=\frac{h+10}{24}$$
$$\frac{25\times24}{91}=4+10$$
$$h=\frac{338}{91}= \frac{26}{7}$$
Question 3
1 / -0

If $$\tan(cos^{-1}x)=\sin (\sec^{-1} (\sqrt{5}) )$$, then $$x=$$

A
$$\sqrt{\dfrac{5}{3}}$$

B
$$\displaystyle \dfrac{\sqrt{5}}{3}$$

C
$$\sqrt{\dfrac{5}{2}}$$

D
$$\displaystyle \dfrac{\sqrt{5}}{2}$$

Solution

If $$tan (cos ^{-1}x)=sin (sec^{-1}\sqrt{5})$$
$$\therefore Let \space \alpha =cos ^{-1}x$$
$$cos \alpha =x$$
$$\displaystyle \therefore tan \alpha =\frac{\sqrt{1-x^{2}}}{x}$$
$$\displaystyle \therefore tan (cos ^{-1}x)=\frac{\sqrt{1-x^{2}}}{x}=\dfrac{2}{\sqrt{5}}$$
$$\Rightarrow 5(1-x^{2})=4x^{2}$$
$$5=9x^{2}$$
$$\displaystyle x=\pm\frac{\sqrt{5}}{3}$$
Question 4
1 / -0

A tower stands at the top of a hill whose height is three times the height of the tower. The tower is found to subtend an angle of\$$\tan ^{ -1 }{ \left( { 1 }/{ 7 } \right) } $$ at a point $$2km$$ away on the horizontal throught the foot of the hill. Then the height of the tower is

A
$$\displaystyle \frac{1}{2}km$$ or $$\displaystyle \frac{1}{3}km$$

B
$$\displaystyle \frac{1}{3}km \space or \space \frac{2}{3}km$$

C
$$\displaystyle \frac{2}{3}km \space or \space\frac{1}{2}km$$

D
$$\displaystyle \frac{3}{4}km \space or \space \frac{1}{2}km$$

Solution

$$tan\theta =\frac{1}{7}$$
$$tan\theta =\frac{3h}{2}$$
$$(tan\theta+\alpha)=\frac{4h}{2}$$
$$\frac{4tan\alpha}{3}=\frac{tan\theta+tan\alpha}{1-tan\alpha tan\theta}$$
$$\frac{4tan\alpha}{3}=\frac{1+7tan\alpha}{7-tan\alpha}$$
$$28tan\alpha-4tan^{2}\theta=3+21tan\theta$$
$$4tan^{2}\theta-7tan\theta =3+21tan\theta$$
$$tan\theta=1,\frac{3}{4}$$
$$h=\frac{2}{3},1/2$$
Question 5
1 / -0

If $$\theta = sin^{-1}x+cos^{-1}x+tan^{-1}x,\ 0\leq x\leq 1$$, then the smallest interval in which $$\theta$$ lies is given by

A
$$\displaystyle \frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}$$

B
$$-\displaystyle \frac{\pi}{4}\leq\theta\leq 0$$

C
$$0\displaystyle \leq\theta\leq\frac{\pi}{4}$$

D
$$\displaystyle \frac{\pi}{2}\leq\theta\leq\frac{3\pi}{4}$$

Solution

$$\theta = sin^{-1}x + cos^{-1} x +tan^{-1} x$$
Now, $$sin ^{-1}x + cos ^{-1} x = \dfrac{\pi}{2} $$
$$\Rightarrow \theta = \dfrac{\pi}{2} + tan^{-1}{x} $$
For ,$$0 \leq x \leq 1 $$
$$0 \leq tan^{-1} x \leq \dfrac{\pi}{4}$$
$$\Rightarrow \dfrac{\pi}{2} \leq \theta \leq \dfrac{3\pi}{4} $$
Hence, option D is correct.
Question 6
1 / -0

The domain of $$\sin^{-1}[\log_{2}(x^{2}/2)]$$ is

A
$$[2, 1]$$

B
$$[1, 2]$$

C
$$[-2,-1]\cup [1,2]$$

D
$$[-2,0]$$

Solution

$$\sin ^{-1}(x) domain : x\in [-1,1]$$

$$\therefore $$domain of $$\displaystyle \sin ^{-1}[\log _{2}\dfrac{x^{2}}{2}]$$

$$ \therefore \log_{2}\dfrac{x^{2}}{2} \in [-1,1]$$

$$\displaystyle -1 \leq \log_{2}\frac{x^{2}}{2} \leq 1$$ and $$\displaystyle \frac{1}{2} \leq \frac{x^{2}}{2} \leq 2$$

$$1\leq x^{2} \leq 4$$

For $$x^{2} \geq 1$$

$$x\in [-\infty ,-1] \cup [1, \infty ]$$...........(i)

For $$x^{2}\leq 4$$

$$x\in [-2,2]$$.................(ii)

By taking intersection of (i) & (ii)

$$\therefore x\epsilon [-2,-1] \cup [1,2]$$
Question 7
1 / -0

$$A$$ vertical pole subtends an angle $$\displaystyle \tan^{-1}(\frac{1}{2})$$ at a point $$P$$ on the ground. The angle subtended by the upper half of the pole at $$P$$ is

A
$$\displaystyle \tan^{-1}(\frac{1}{4})$$

B
$$\displaystyle \tan^{-1}(\frac{1}{8})$$

C
$$\displaystyle \tan^{-1}(\frac{2}{3})$$

D
$$\displaystyle \tan^{-1}(\frac{2}{9})$$

Solution

as shown in the figure
$$tan(tan^{-1}(\dfrac{1}{2}))=\dfrac{1}{2}=\dfrac{2h}{b}$$ .......................(1)

$$tan(tan^{-1}(\dfrac{1}{2})-\phi)=\dfrac{h}{b}$$ .......................(2)

(1) divided by (2) we get
$$2= \dfrac{\dfrac{1}{2}}{\dfrac{1-\dfrac{tan\phi}{2}}{1+\dfrac{tan\phi}{2}}}$$

solving the above equation we get

$$tan\phi= \dfrac{2}{9}$$

$$\phi= tan^{-1}\dfrac{2}{9}$$
Question 8
1 / -0

Range of $$\sin^{-1}x-\cos^{-1}x$$ is

A
$$[\displaystyle \frac{-3\pi}{2}, \displaystyle \frac{\pi}{2}]$$

B
$$[\displaystyle \frac{-5\pi}{3}, \displaystyle \frac{\pi}{3}]$$

C
$$[\displaystyle \frac{-3\pi}{2}, \pi]$$

D
$$[0, \pi]$$

Solution

range of $$\displaystyle sin ^{-1}x \epsilon [-\frac{\pi }{2} , \frac{\pi }{2}]$$
range of $$cos ^{-1}x \epsilon [0, \pi ]$$

$$\therefore f(x)=sin ^{-1}x-cos ^{-1}x=sin ^{-1}x+cos ^{-1}x-2 cos ^{-1}x$$
$$\displaystyle \quad\quad\quad=\frac{\pi }{2}-2cos ^{-1}x\quad\quad [\because sin ^{-1}x+cos ^{-1}x=\frac{\pi }{2}] identity$$
range of $$\displaystyle f(x) \epsilon [\frac{\pi }{2} -2(\pi ) , \frac{\pi }{2}-2(0)]$$
$$\therefore f(x)\displaystyle \epsilon [-\frac{3\pi }{2} ,\frac{\pi }{2}]$$
Question 9
1 / -0

$$ABC$$ is a triangular park with $$AB=AC=100$$ cm. $$A$$ clock tower is situated at the midpoint of $$BC$$. The angles of elevation of the top of the tower from $$A$$ and $$B$$ are $$cot ^{-1}3.2$$ and $$cosec ^{-1} 2.6$$. The height of the tower is

A
$$25$$ mt

B
$$50$$ mt

C
$$100$$ mt

D
$$50\sqrt{2}$$ mt

Solution

$$y^{2}=AD^{2}=100^{2}-BD^{2}=10000-x^{2}$$
$$\cot^{-1}(3.2)=\alpha,cosec^{-1}(2.6)=\beta)$$
$$\cot \alpha =3.2$$ & $$ cosec \beta =2.6$$
$$\cot \alpha =\dfrac{y}{h}$$
$$\Rightarrow y=3.2 \times h$$
Now,
$$\cot \beta =\dfrac{x}{h}$$
$$x=\cot \beta \times h$$
$$x=h\sqrt{2.6^{2}-1}$$
$$x= 2.4h$$
$$x^{2}+y^{2}=100000$$
$$\Rightarrow (2.4h)^{2}+(3.2h)^{2}=100^{2}$$
$$\Rightarrow h^{2}(16)=100^{2}$$
$$\Rightarrow h=25$$
Question 10
1 / -0

A vertical pole more than $$100ft$$ high consists of two portions, the lower being $$ 1/3$$ of the whole. lf the upper portion subtends an angle $$\tan^{-1}(1/2)$$ at a point distant 40 ft. from the foot of the pole, the height of the pole is

A
$$ 105 $$

B
$$ 120$$

C
$$135$$

D
$$150$$

Solution

$$tan=1/2$$
$$tan\alpha =\frac{h}{3+40}$$
$$tan(\alpha +\theta )=\frac{h}{40}$$
$$\dfrac{tan\alpha }{3}=\dfrac{tan\alpha +tan\theta }{1-tan\alpha tan\theta }$$
$$\dfrac{1+2tan\alpha }{2-tan\alpha }=\dfrac{tan\alpha }{3}$$
$$3+6tan\alpha =2tan\alpha -tan\alpha $$
$$tan^{2}\alpha +4tan\alpha +3 =0$$
$$tan\alpha=-1,tan\alpha=-3$$
$$1=\frac{h}{120}$$
$$h=120$$

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Inverse Trigonometric Functions Test - 24

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Inverse Trigonometric Functions Test - 24

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Question 1

$$\cos^{-1}\sqrt{1-x^{2}}$$

$${-\cos^{-1}}\sqrt{1-x^{2}}$$

$${\cos^{-1}}\sqrt{x^{2}-1}$$

$${\pi-\cos^{-1}}\sqrt{1-x^{2}}$$

Solution

Question 2

$$26/7$$

$$27/8$$

$$27/6$$

$$26/3$$

Solution

Question 3

$$\sqrt{\dfrac{5}{3}}$$

$$\displaystyle \dfrac{\sqrt{5}}{3}$$

$$\sqrt{\dfrac{5}{2}}$$

$$\displaystyle \dfrac{\sqrt{5}}{2}$$

Solution

Question 4

$$\displaystyle \frac{1}{2}km$$ or $$\displaystyle \frac{1}{3}km$$

$$\displaystyle \frac{1}{3}km \space or \space \frac{2}{3}km$$

$$\displaystyle \frac{2}{3}km \space or \space\frac{1}{2}km$$

$$\displaystyle \frac{3}{4}km \space or \space \frac{1}{2}km$$

Solution

Question 5

$$\displaystyle \frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}$$

$$-\displaystyle \frac{\pi}{4}\leq\theta\leq 0$$

$$0\displaystyle \leq\theta\leq\frac{\pi}{4}$$

$$\displaystyle \frac{\pi}{2}\leq\theta\leq\frac{3\pi}{4}$$

Solution

Question 6

$$[2, 1]$$

$$[1, 2]$$

$$[-2,-1]\cup [1,2]$$

$$[-2,0]$$

Solution

Question 7

$$\displaystyle \tan^{-1}(\frac{1}{4})$$

$$\displaystyle \tan^{-1}(\frac{1}{8})$$

$$\displaystyle \tan^{-1}(\frac{2}{3})$$

$$\displaystyle \tan^{-1}(\frac{2}{9})$$

Solution

Question 8

$$[\displaystyle \frac{-3\pi}{2}, \displaystyle \frac{\pi}{2}]$$

$$[\displaystyle \frac{-5\pi}{3}, \displaystyle \frac{\pi}{3}]$$

$$[\displaystyle \frac{-3\pi}{2}, \pi]$$

$$[0, \pi]$$

Solution

Question 9

$$25$$ mt

$$50$$ mt

$$100$$ mt

$$50\sqrt{2}$$ mt

Solution

Question 10

$$ 105 $$

$$ 120$$

$$135$$

$$150$$

Solution

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