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Inverse Trigonometric Functions Model Question Paper 1

12th Standard EM

    Reg.No. :
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Maths

Time : 01:00:00 Hrs
Total Marks : 50
    5 x 1 = 5
  1. sin−1(cos x)\(=\frac{\pi}{2}-x \) is valid for

    (a)

    \(-\pi \le x\le 0\)

    (b)

    \(0\pi \le x\le 0\)

    (c)

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

    (d)

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

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

    (a)

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

    (b)

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

    (c)

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

    (d)

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

  3. The number of solutions of the equation \({ tan }^{ -1 }2x+{ tan }^{ -1 }3x=\cfrac { \pi }{ 4 } \) 

    (a)

    2

    (b)

    3

    (c)

    1

    (d)

    none

  4. \({ tan }^{ -1 }\left( \cfrac { 1 }{ 4 } \right) +{ tan }^{ -1 }\left( \cfrac { 2 }{ 11 } \right) \) =

    (a)

    0

    (b)

    \(\cfrac { 1 }{ 2 } \)

    (c)

    -1

    (d)

    none

  5. If \({ cos }^{ -1 }x>x>{ sin }^{ -1 }x\) then

    (a)

    \(\cfrac { 1 }{ \sqrt { 2 } } <x\le 1\)

    (b)

    \(0\le x<\cfrac { 1 }{ \sqrt { 2 } } \)

    (c)

    \(-1\le x<\cfrac { 1 }{ \sqrt { 2 } } \)

    (d)

    x>0

  6. 5 x 2 = 10
  7. For what value of x does sinx=sin−1x?

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

  9. Show that cot−1\(\left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) ={ sec }^{ -1 }x,|x|>1\)

  10. Prove that \({ tan }^{ -1 }\left( \cfrac { 1 }{ 7 } \right) +{ tan }^{ -1 }\left( \cfrac { 1 }{ 13 } \right) ={ tan }^{ -1 }\left( \cfrac { 2 }{ 9 } \right) \)

  11. Ecalute \(sin\left( { cos }^{ -1 }\left( \cfrac { 3 }{ 5 } \right) \right) \)
     

  12. 5 x 3 = 15
  13. For what value of x , the inequality\(\cfrac { \pi }{ 2 } <{ cos }^{ -1 }(3x-1)<\pi \)

  14. Find the value of
     \(cos\left( { cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) +{ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)

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

  16. Solve \({ tan }^{ -1 }\left( \cfrac { 2x }{ 1-{ x }^{ 2 } } \right) +{ cot }^{ -1 }\left( \cfrac { 1-{ x }^{ 2 } }{ 2x } \right) =\cfrac { \pi }{ 3 } ,x>0\)

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

  18. 4 x 5 = 20
  19. Solve \(tan^{ -1 }\left( \frac { x-1 }{ x-2 } \right) +tan^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =\frac { \pi }{ 4 } \)

  20. Solve \(cos\left( sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right) =sin\left\{ cot^{ -1 }\left( \frac { 3 }{ 4 } \right) \right\} \)

  21. 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

  22. Provethat \({ tan }^{ -1 }\left( \cfrac { 1-x }{ 1+x } \right) -{ tan }^{ -1 }\left( \cfrac { 1-y }{ 1+y } \right) ={ sin }^{ -1 }\left( \cfrac { y-x }{ \sqrt { 1+{ x }^{ 2 } } .\sqrt { 1+{ y }^{ 2 } } } \right) \\ \)

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