12th Standard Maths Inverse Trigonometric Functions English Medium Free Online Test One Mark Questions with Answer Key 2020 - 2021

12th Standard

Reg.No. :
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Maths

Time : 00:10:00 Hrs
Total Marks : 10

10 x 1 = 10
1. If sin-1 x+sin-1 y=$\frac{2\pi}{3};$then cos-1x+cos-1 y is equal to

(a)

$\frac{2\pi}{3}$

(b)

$\frac{\pi}{3}$

(c)

$\frac{\pi}{6}$

(d)

$\pi$

2. If sin-1 x+sin-1 y+sin-1 z=$\frac{3\pi}{2}$, the value of x2017+y2018+z2019$-\frac { 9 }{ { x }^{ 101 }+{ y }^{ 101 }+{ z }^{ 101 } }$is

(a)

0

(b)

1

(c)

2

(d)

3

3. The domain of the function defined by f(x)=sin−1$\sqrt{x-1}$ is

(a)

[1,2]

(b)

[-1,1]

(c)

[0,1]

(d)

[-1,0]

4. ${ tan }^{ -1 }\left( \frac { 1 }{ 4 } \right) +{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right)$is equal to

(a)

$\frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 3 }{ 5 } \right)$

(b)

$\frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 3 }{ 5 } \right)$

(c)

$\frac { 1 }{ 2 } {tan }^{ -1 }\left( \frac { 3 }{ 5 } \right)$

(d)

${ tan}^{ -1 }\left( \frac { 1}{ 2 } \right)$

5. sin-1(2cos2x-1)+cos-1(1-2sin2x)=

(a)

$\frac{\pi}{2}$

(b)

$\frac{\pi}{3}$

(c)

$\frac{\pi}{4}$

(d)

$\frac{\pi}{6}$

6. sin(tan-1x), |x|<1 ia equal to

(a)

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

(b)

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

(c)

$\frac{1}{\sqrt{1+x^2}}$

(d)

$\frac{x}{\sqrt{1+x^2}}$

7. If ${ sin }^{ -1 }x-cos^{ -1 }x=\cfrac { \pi }{ 6 }$ then

(a)

$\cfrac { 1 }{ 2 }$

(b)

$\cfrac { \sqrt { 3 } }{ 2 }$

(c)

$\cfrac { -1 }{ 2 }$

(d)

none of these

8. ·If $\alpha ={ tan }^{ -1 }\left( \cfrac { \sqrt { 3 } }{ 2y-x } \right) ,\beta ={ tan }^{ -1 }\left( \cfrac { 2x-y }{ \sqrt { 3y } } \right)$ then $\alpha -\beta$

(a)

$\cfrac { \pi }{ 6 }$

(b)

$\cfrac { \pi }{ 3 }$

(c)

$\cfrac { \pi }{ 2 }$

(d)

$\cfrac { -\pi }{ 3 }$

9. $sin\left\{ 2{ cos }^{ -1 }\left( \cfrac { -3 }{ 5 } \right) \right\} =$

(a)

$\cfrac { 6 }{ 15 }$

(b)

$\cfrac { 24 }{ 25 }$

(c)

$\cfrac { 4 }{ 5 }$

(d)

$\cfrac { -24 }{ 25 }$

10. In a $\Delta ABC$  if C is a right angle, then  ${ tan }^{ -1 }\left( \cfrac { a }{ b+c } \right) +{ tan }^{ -1 }\left( \cfrac { b }{ c+a } \right) =$

(a)

$\cfrac { \pi }{ 3 }$

(b)

$\cfrac { \pi }{ 4 }$

(c)

$\cfrac { 5\pi }{ 2 }$

(d)

$\cfrac { \pi }{ 6 }$