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12th Maths Weekly Test -1 (Inverse Trigonometric Functions)

St. Britto Hr. Sec. School - Madurai

12th Maths Weekly Test -1 (Inverse Trigonometric Functions)-Aug 2020

Time : 1.30 Hours

Part - A

Multiple Choice Question - Answer All Question

The value of sin^-1 (cos x),0\(\le x\le\pi\) is

\(\pi-x\)
\(x-\frac{\pi}{2}\)
\(\frac{\pi}{2}-x\)
\(\pi-x\)

\({ sin }^{ -1 }\frac { 3 }{ 5 } -{ cos }^{ -1 }\frac { 12 }{ 13 } +{ sec }^{ -1 }\frac { 5 }{ 3 } { -cosec }^{ 1- }\frac { 13 }{1 2 } \)is equal to

2\(\pi\)
\(\pi\)
0
tan^-1\(\frac{12}{65}\)

sin⁻¹(cos x)\(=\frac{\pi}{2}-x \) is valid for

\(-\pi \le x\le 0\)
\(0 \le x\le \pi\)
\(-\frac { \pi }{ 2 } \le x\le \frac { \pi }{ 2 } \)
\(-\frac { \pi }{ 4 } \le x\le \frac { 3\pi }{ 4 } \)

If x=\(\frac{1}{5}\), the valur of cos (cos^-1x+2sin^-1x) is
\(-\sqrt { \frac { 24 }{ 25 } } \)

\(\sqrt { \frac { 24 }{ 25 } } \)

\(\frac{1}{5}\)

-\(\frac{1}{5}\)
If cot⁻¹x=\(\frac{2\pi}{5}\) for some x\(\in\)R, the value of tan^-1 x is
\(-\frac{\pi}{10}\)

\(\frac{\pi}{5}\)

\(\frac{\pi}{10}\)

\(-\frac{\pi}{5}\)

If the function f(x)sin^-1(x²-3), then x belongs to

[-1,1]
[\(\sqrt2\),2]
\(\\ \\ \\ \left[ -2,-\sqrt { 2 } \right] \cup \left[ \sqrt { 2 } ,2 \right] \)
\(\left[ -2,-\sqrt { 2 } \right] \cap \left[ \sqrt { 2 } ,2 \right] \)

\({ sin }^{ -1 }\left[ tan\frac { \pi }{ 4 } \right] -{ sin }^{ -1 }\left[ \sqrt { \frac { 3 }{ x } } \right] =\frac { \pi }{ 6 } \).Then x is a root of the equation

x²−x−6=0
x²−x−12=0
x²+x−12=0
x²+x−6=0

If \(\\ \\ \\ { cot }^{ -1 }\left( \sqrt { sin\alpha } \right) +{ tan }^{ -1 }\left( \sqrt { sin\alpha } \right) =u\), then cos2u is equal to

tan²\(\alpha\)
0
-1
tan2\(\alpha\)

The equation tan^-1x-cot^-1x=tan^-1\(\left( \frac { 1 }{ \sqrt { 3 } } \right) \)has

no solution
unique solution
two solutions
infinite number of solutions

If sin^-1 \(\frac{x}{5}+ cot^{-1}\frac{5}{4}=\frac{\pi}{2}\), then the value of x is

4
5
2
3

If \(\alpha ={ tan }^{ -1 }\left( tan\cfrac { 5\pi }{ 4 } \right) \) and \(\beta ={ tan }^{ -1 }\left( -tan\cfrac { 2\pi }{ 3 } \right) \) then

\(4\alpha =3\beta \quad \)
\(3\alpha =4\beta \)
\(\alpha -\beta =\cfrac { 7\pi }{ 12 } \)
none

Part - B

Generic 2 - Answer All Question

Find all the values of x such that
-10\(\pi\)\(\le x\le\)10\(\pi\) and sin x=0

Find the value of
\({ sin }^{ -1 }\left( sin\left( \frac { 2\pi }{ 3 } \right) \right) \)

Find the value of sin^-1\(\left( sin\frac { 5\pi }{ 9 } cos\frac { \pi }{ 9 } +cos\frac { 5\pi }{ 9 } sin\frac { \pi }{ 9 } \right) \).

Is cos^-1(-x)=\(\pi\)-cos⁻¹(x) true? Justify your answer.
Find the domain of the following functions :
\(tan^{-1}(\sqrt{9-x^{2}})\)

Find the domain of
f(x)=sin^-1 \((\frac{|x|-2}{3})+cos^{-1}(\frac{1-|x|}{4})\)

Part - C

Generic 3 - Answer All Question

Prove that \({ cos }^{ -1 }\left( \cfrac { 4 }{ 5 } \right) +{ tan }^{ -1 }\left( \cfrac { 3 }{ 5 } \right) ={ tan }^{ -1 }\left( \cfrac { 27 }{ 11 } \right) \)

Solve: \({ tan }^{ -1 }\left( \cfrac { x-1 }{ x-2 } \right) +{ tan }^{ -1 }\left( \cfrac { x+1 }{ x+2 } \right) =\cfrac { \pi }{ 4 } \)

Solve: cos(tan^-1x) = \(sin\left( { cot }^{ -1 }\cfrac { 3 }{ 4 } \right) \)

Find the domain of the following

sin⁻¹\(\left(\frac{x^2 + 1}{2x} \right)\)

For what value of x , the inequality \(\frac{\pi}{2}\)< cos⁻¹(3x − 1) < π
Prove that \({ tan }^{ -1 }\sqrt { x } =\cfrac { 1 }{ 2 } { cos }^{ -1 }={ \cfrac { 1 }{ 2 } { cos }^{ -1 }\left( \cfrac { 1-x }{ 1+x } \right) ,x\epsilon \left| 0,1 \right| }\)

Part - D

Generic 5 - Answer All Question

Find the domain of the following functions
(i) f(x) = sin^-1(2x - 3)
(ii) f(x) = sin^-1x + cos x

Find (i) cos^-1\((-\frac{1}{\sqrt2})\)

ii) cos^-1\((cos(-\frac{\pi}{3}))\)

iii) cos^-1\((cos(-\frac{7\pi}{6}))\)

Solve \(tan^{ -1 }\left( \frac { x-1 }{ x-2 } \right) +tan^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =\frac { \pi }{ 4 }\)
If \({ tan }^{ -1 }\left( \cfrac { \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } } \right) =a\) than prove that x²=sin2a

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