Trigonometric Functions

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Question 1

If space sin to the power of negative 1 end exponent space open parentheses straight x over 5 close parentheses space plus space cosec to the power of negative 1 end exponent space open parentheses 5 over 4 close parentheses space equals space straight pi over 2 then the value of x
  • 1

  • 3

  • 4

  • 5

Solution

B.

3

sin to the power of negative 1 end exponent space straight x over 5 space plus space sin to the power of negative 1 end exponent space 4 over 5 space equals space straight pi over 2
rightwards double arrow space sin to the power of negative 1 end exponent space straight x over 5 space equals space cos to the power of negative 1 end exponent space 4 over 5
rightwards double arrow space sin to the power of negative 1 end exponent space straight x over 5
space equals space sin to the power of negative 1 end exponent space 3 over 5
therefore space straight x space equals space 3
Question 2

limit as straight n rightwards arrow infinity of space open square brackets 1 over straight n squared sec squared space 1 over straight n squared plus 2 over straight n squared space plus 2 over straight n squared sec squared 4 over straight n squared plus.....1 over straight n squared sec squared 1 close square brackets equal
  • 1 half sec space 1
  • 1 half cosec space 1
  • tan 1

  • 1 half tan space 1

Solution

D.

1 half tan space 1 limit as straight n rightwards arrow infinity of space open square brackets 1 over straight n squared sec squared 1 over straight n squared plus 2 over straight n squared sec squared 4 over straight n squared plus 3 over straight n squared sec squared 9 over straight n squared plus..... plus 1 over straight n sec squared 1 close square brackets is equals
limit as straight n rightwards arrow infinity of space straight r over straight n squared space sec squared space straight r squared over straight n squared space equals space limit as straight n rightwards arrow infinity of 1 over straight n. straight r over straight n space sec squared straight r squared over straight n squared
rightwards double arrow space Given space limit space is space equal space to space value space of space integral space integral subscript 0 superscript 1 straight x space sec squared space straight x squared space dx
or space 1 half integral subscript 0 superscript 1 space 2 straight x space sec space straight x squared space dx space equals space 1 half integral subscript 0 superscript 1 space sec squared space tdt
space equals space 1 half left parenthesis tan space straight t right parenthesis subscript 0 superscript 1 space equals space 1 half space tan space 1
Question 3

If space straight l subscript 1 space equals space integral subscript 0 superscript 1 space 2 to the power of straight x squared 1 end exponent space dx comma space straight I subscript 2 space equals space integral subscript 0 superscript 1 space 2 to the power of straight x cubed end exponent space dx space comma space straight I subscript 3 space equals space integral subscript 1 superscript 2 space 2 to the power of straight x squared space end exponent dx space and space straight I subscript 4 space equals space integral subscript 1 superscript 2 2 to the power of straight x cubed space end exponent space dx space then
  • I2 > I1

  • I1 > I2

  • I3 = I4

  • I3 > I4

Solution

B.

I1 > I2

straight I subscript 1 space equals space integral subscript 0 superscript 1 2 to the power of straight x squared end exponent space dx comma space straight I subscript 2 space equals space integral subscript 0 superscript 1 space 2 to the power of straight x cubed end exponent space dx space comma space straight I subscript 3 space equals integral subscript 1 superscript 2 2 to the power of straight x squared end exponent space dx space straight I subscript 4 space equals integral subscript 0 superscript 1 2 to the power of straight x cubed space dx end exponent
for all space 0 space less than space straight x space less than space 1 comma space straight x to the power of 2 space end exponent space greater than space straight x cubed
rightwards double arrow space integral subscript 0 superscript 1 2 to the power of straight x squared space end exponent dx space greater than integral subscript 0 superscript 1 space 2 to the power of straight x cubed space end exponent space dx
rightwards double arrow space straight I subscript 1 space greater than straight I subscript 2
Question 4

A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30o. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60o. Then the time taken (in minutes) by him, from B to reach the pillar, is:

  • 6

  • 10

  • 20

  • 5

Solution

D.

5

According to given information, we have the following figure:

Now, from ΔACD and ΔBCD, we have
tan space 30 to the power of straight o space equals space fraction numerator straight h over denominator straight x plus straight y end fraction
and
tan space 60 to the power of straight o space equals space straight h over straight y
rightwards double arrow space straight h space equals space fraction numerator straight x plus straight y over denominator square root of 3 end fraction space space space space... space left parenthesis straight i right parenthesis
and space space straight h equals square root of 3 straight y end root space space space space space.. space left parenthesis ii right parenthesis
From space eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
fraction numerator straight x plus straight y over denominator square root of 3 end fraction space equals space square root of 3 straight y space
rightwards double arrow space straight x plus straight y space equals space 3 straight y
rightwards double arrow space straight x minus 2 straight y space equals space 0
rightwards double arrow space straight y space equals space straight x over 2
∵ speed is uniform.
∵ distance y will be cover in 5 min
∵ distance x covered in 10 min
∵ Distance x/2 will be cover in 5 min

Question 5

A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is 60o and when he retires 40 meters away from the tree the angle of elevation becomes 30o. The breadth of the river is

  • 20 m

  • 30 m

  • 40 m

  • 60 m

Solution

A.

20 m


tan to the power of straight o space 30 space space equals space fraction numerator straight h over denominator 40 plus straight b end fraction
rightwards double arrow space square root of 3 straight h end root space equals space 40 plus straight b space.... space left parenthesis straight i right parenthesis
tan space 60 to the power of straight o space equals space straight h divided by straight b
rightwards double arrow space straight h square root of 3 straight b space... left parenthesis ii right parenthesis
rightwards double arrow space straight b space equals space 20 space straight m