Inverse Trigonometric Functions

More Topic from Mathematics

Question 1

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,
limit as straight x space rightwards arrow 0 of space fraction numerator left parenthesis 1 minus cos space 2 straight x right parenthesis left parenthesis 3 plus cosx over denominator straight x space tan space 4 straight x end fraction space
equals limit as straight x space rightwards arrow 0 of fraction numerator 2 sin squared space straight x space left parenthesis 3 plus cosx right parenthesis over denominator straight X space straight x space begin display style fraction numerator tanx over denominator 4 straight x end fraction end style space straight x space 4 straight x end fraction
equals limit as straight x space rightwards arrow 0 of fraction numerator 2 sin squared straight x over denominator straight x squared end fraction space straight x space limit as straight x space rightwards arrow 0 of fraction numerator 3 plus cos space straight x over denominator 4 end fraction space straight x limit as straight x space rightwards arrow 0 of fraction numerator 1 over denominator begin display style fraction numerator tan begin display style space end style begin display style 4 end style begin display style straight x end style over denominator 4 straight x end fraction end style end fraction
space equals space 2 space straight x space 4 over 4 space straight x space 1 space equals space 2

Question 2

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

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 = 
straight z 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
space equals space fraction numerator left parenthesis 2 plus space 3 straight i space sin space straight theta right parenthesis left parenthesis 1 space plus space 2 straight i space sin space straight theta right parenthesis over denominator left parenthesis 1 minus 2 straight i space sin space straight theta right parenthesis left parenthesis 1 plus space 2 straight i space sin space straight theta right parenthesis end fraction
equals space fraction numerator 2 space plus space 4 straight i space sin space straight theta space plus space 3 straight i space sin space straight theta space plus space 6 space straight i squared space sin space squared space straight theta over denominator 1 squared minus space left parenthesis 2 straight i space sin space straight theta right parenthesis squared end fraction
equals space fraction numerator 2 space plus space 7 space straight i space sin space straight theta space minus space 6 space sin squared space straight theta over denominator 1 space plus space 4 space sin squared space straight theta end fraction
equals space fraction numerator 2 minus 6 space sin squared space straight theta over denominator 1 plus 4 space sin squared space straight theta end fraction space plus space straight i fraction numerator 7 space sin space straight theta over denominator 1 space plus space 4 space sin squared space straight theta end fraction
therefore space Re space left parenthesis straight z right parenthesis space equals space 0
therefore space fraction numerator 2 minus 6 space sin squared space straight theta over denominator 1 plus space 4 space sin squared space straight theta end fraction space equals space 0
rightwards double arrow space 2 space equals space 6 space sin squared straight theta
rightwards double arrow space sin squared straight theta space equals space 1 third
rightwards double arrow space sin space straight theta space equals space plus-or-minus space fraction numerator 1 over denominator square root of 3 end fraction
rightwards double arrow space straight theta space equals space sin to the power of negative 1 end exponent space open parentheses plus-or-minus fraction numerator 1 over denominator square root of 3 end fraction close parentheses space equals space plus-or-minus space sin to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses

Question 4

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)


We space have comma space straight f left parenthesis straight x right parenthesis space equals space tan to the power of negative 1 end exponent square root of fraction numerator 1 plus space sin space straight x over denominator 1 minus space sin space straight x end fraction end root comma space straight x element of space open parentheses 0 comma space straight pi over 2 close parentheses
rightwards double arrow space straight f left parenthesis straight x right parenthesis space equals space tan to the power of negative 1 end exponent space square root of open parentheses cos begin display style straight x over 2 end style plus space sin begin display style straight x over 2 end style close parentheses squared over open parentheses cos straight x over 2 minus space sin straight x over 2 close parentheses squared end root
space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator cos space begin display style straight x over 2 end style space plus space sin space begin display style straight x over 2 end style over denominator cos space begin display style straight x over 2 end style space minus space sin space begin display style straight x over 2 end style end fraction close parentheses
open square brackets therefore space cos space straight x over 2 greater than space sin space straight x over 2 space for space 0 space less than thin space straight x over 2 less than straight x over 4 close square brackets
equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator 1 plus tan space begin display style straight x over 2 end style over denominator 1 minus space tan space begin display style straight x over 2 end style end fraction close parentheses
equals space tan to the power of negative 1 end exponent space open square brackets tan space open parentheses straight pi over 4 plus straight pi over 2 close parentheses close square brackets

space equals space straight pi over 4 plus straight pi over 2
rightwards double arrow space straight f apostrophe left parenthesis straight x right parenthesis space equals space 1 half
rightwards double arrow space straight f apostrophe open parentheses straight pi over 6 close parentheses equals 1 half
Now comma space equation space of space normal space at space straight x space equals space straight pi over 6 space is space given space by
open square brackets straight y minus straight f open parentheses straight pi over 6 close parentheses close square brackets space equals space minus 2 open parentheses straight x minus straight pi over 6 close parentheses
rightwards double arrow open parentheses straight y minus straight pi over 3 close parentheses space equals space minus space 2 open parentheses straight x minus straight pi over 6 close parentheses
open square brackets therefore comma space straight f open parentheses straight pi over 6 close parentheses space equals space straight pi over 4 plus straight pi over 12 equals fraction numerator 4 straight pi over denominator 12 end fraction space equals straight pi over 3 close square brackets
Whcih space passes space through space the space point space open parentheses 0 comma space fraction numerator 2 straight pi over denominator 3 end fraction close parentheses
Question 5

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