Please wait...

Inverse Trigonometric Functions Test - 5
Result
Inverse Trigonometric Functions Test - 5

  • /

    Score

  • -

    Rank
Time Taken: -
  • Question 1/15
    1 / -0.25

     

    What is the maximum and minimum value of sin x +cos x?

     

    Solutions

     

     

    Let y= sin x + cos x

    dy/dx=cos x- sin x

    For maximum or minimum dy/dx=0

    Setting cosx- sin x=0

    We get cos x = sin x

    x= π/4, 5 π/4 ———-

    Whether these correspond to maximum or minimum, can be found from the sign of second derivative.

    d^2y/dx^2=-sin x - cos x=-1/√2 –1/√2 (for x=π/4) which is negative. Hence x=π/4 corresponds to maximum.For x=5 π/4

    d^2y/dx^2=-(-1/√2)-(-1/√2)=2/√2 a positive quantity. Hence 5 π/4 corresponds to minimum

    Maximum value of the function

    y= sin π/4 + cos π/4= 2/√2=√2

    Minimum value is

    Sin(5 π/4)+cos (5 π/4)=-2/√2=-√2

     

     

  • Question 2/15
    1 / -0.25

     

    sin (200)0   + cos (200)0 is

     

    Solutions

     

     

    Because both  sin 2000  and cos 2000  lies in 3rd quadrant. In 3rd quadrant values of sin and cos are negative.

     

     

  • Question 3/15
    1 / -0.25

     

    If cos(-1) x + cos(-1) y = 2 π, then the value of sin(-1) x + sin(-1) y is   

     

    Solutions

     

     

    If cos(-1) x + cos(-1)  y = 2 π, then the value of sin(-1) x + sin(-1) y  = π−2 π= −π.

     

     

  • Question 4/15
    1 / -0.25

     

    Domain of f(x) = sin−1 x −sec−1 x is

     

    Solutions

     

     

    Since  sin−1 x  is defined for  |x|⩽1, and sec−1 x is defined for  |x|⩾1,therefore,f(x) is defined only when|x|=1.so, Df  = {−1,1}.

     

     

  • Question 5/15
    1 / -0.25

     

    The value of sin   is  

     

    Solutions

     

     

    Put    therefore the given expressionis  sin2 θ= 2sin θcos θ

     

     

  • Question 6/15
    1 / -0.25

     

    If  5 sin θ= 3, then   is equal to

     

    Solutions

     

     


    = [(5/4) + (3/4)] / [(5/4) - (3/4)]
    =(8/4) / (2/4)
    =4

     

     

  • Question 7/15
    1 / -0.25

     

    If sin  A  + cos  A  = 1, then sin 2A is equal to

     

    Solutions

     

     

    (sinA+cosA)2
    = sin2 A+cos2 A+2sinAcosA
    =11+sin2 A=1sin2 A=0.
    (because Sin 2A = 2sin A cos A)

     

     

  • Question 8/15
    1 / -0.25

     

    If  θ= cos-1, then tan  θis equal to  

     

    Solutions

     

     


    Therefore, tan θ=

     

     

  • Question 9/15
    1 / -0.25

     

    The number of solutions of the equation  cos-1 (1-x) - 2cos-1 x = π/2 is

     

    Solutions

     

     

    As no value of x in (0, 1) can satisfy the given equation.Thus, the given equation has only one solution.

     

     

  • Question 10/15
    1 / -0.25

     

    tan  (sin−1 x) is equal to

     

    Solutions

     

     

     

     

  • Question 11/15
    1 / -0.25

     

    If  x  ∈R, x  ≠0, then the value of sec  θ+ tan  θis  

     

    Solutions

     

     

     

     

     

  • Question 12/15
    1 / -0.25

     

    The value of  cos 1050 is

     

    Solutions

     

     

     

     

  • Question 13/15
    1 / -0.25

     

    If  cos(2sin−1 x) = 1/9  then x =

     

    Solutions

     

     

    Put

    sin-1   x = θ⇒x = sin θ


     

     

  • Question 14/15
    1 / -0.25

     

     

    Solutions

     

     






     

     

  • Question 15/15
    1 / -0.25

     

    cot  (cos−1 x) is equal to

     

    Solutions

     

     

    Put,
    cos-1 x = θ⇒x = cos θ⇒cos θ

     

     

User Profile
-

Correct (-)

Wrong (-)

Skipped (-)

  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
  • 11
  • 12
  • 13
  • 14
  • 15
Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Click on Allow to receive notifications
×
Open Now