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A curve 'C' with negative slope through ...

A curve 'C' with negative slope through the point(0,1) lies in the I Quadrant. The tangent at any point 'P' on it meets the x-axis at 'Q'. Such that `PQ=1`. Then
The area bounded by 'C' and the co-ordinate axes is

A

1

B

`log_(e)2`

C

`pi//4`

D

`pi//2`

Text Solution

Verified by Experts

The correct Answer is:
C
Area, `A = int_(0)^(infty) ydx`
Where `y=sintheta` and `dx=(sintheta-"cosec "theta)(d)theta`
`therefore A=-int_(0)^(pi//2)sin(sintheta-"cosec "theta)(d)theta`
`=-int_(0)^(pi//2)(sin^(2)theta-1)(d)theta=pi/4`
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