Inverse Trigonometric Functions

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Question
CBSEENMA12036032

straight l im with straight x rightwards arrow 0 below space fraction numerator left parenthesis 1 minus cos 2 straight x right parenthesis left parenthesis 3 space plus cos space straight x right parenthesis over denominator straight x space tan space 4 straight x end fraction space is equal to 
  • 4

  • 3

  • 2

  • 1/2

Solution

C.

2

We have,
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Question
CBSEENMA12036248

A particle is projected from a point O with velocity u at an angle of 60o with the horizontal. When it is moving in a direction at right angles to its direction at O, its velocity then is given by

  • u/3

  • u/2

  • 2u/3

  • fraction numerator u over denominator square root of 3 end fraction

Solution

D.

fraction numerator u over denominator square root of 3 end fraction WiredFaculty
u cos 60o = v cos 30o
straight v equals space fraction numerator 4 over denominator square root of 3 end fraction

Question
CBSEENMA12036008

A value of θ for which fraction numerator 2 space plus space 3 straight i space sin space straight theta over denominator 1 minus 2 straight i space sin space straight theta end fraction is purely imaginary is

  • π/3

  • π/6

  • sin to the power of negative 1 end exponent open parentheses fraction numerator square root of 3 over denominator 4 end fraction close parentheses
  • sin to the power of negative 1 end exponent space open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses

Solution

D.

sin to the power of negative 1 end exponent space open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses

Let z = straight z space equals space fraction numerator 2 space plus space 3 straight i space sin space straight theta over denominator 1 minus 2 straight i space sin space straight theta end fraction is purely imaginary. Then, we have Re (z) = 0
We have Re (z) = 0
Now, consider z = 
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Question
CBSEENMA12036012

Consider f(x) = tan-1 open parentheses square root of fraction numerator 1 space plus sin space straight x over denominator 1 minus sin space straight x end fraction end root close parentheses comma space straight x space element of space open parentheses 0 comma space straight pi over 2 close parentheses. A normal to y = f (x) at x = π/6 also passes through the point

  • (0,0)

  • (0, 2π/3)

  • (π/6 ,0)

  • (π/4, 0)

Solution

B.

(0, 2π/3)


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Question
CBSEENMA12036028

If 0≤x<2π, then the number of real values of x, which satisfy the equation cosx+cos2x+cos3x+cos4x=0, is :

  • 3

  • 5

  • 7

  • 9

Solution

C.

7

5