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Consider a curve C : y=cos^(-1)(2x-1) an...

Consider a curve `C : y=cos^(-1)(2x-1)` and a straight line `L :2p x-4y+2pi-p=0.` Statement 1: The set of values of `p` for which the line `L` intersects the curve at three distinct points is `[-2pi,-4]dot` Statement 2: The line `L` is always passing through point of inflection of the curve `Cdot`

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We have curve `y =f(x)=cos^(-1)(2x-1)`
and line 2px-4y+2pi-p=0
Domain of y =`f(x)=cos^(-1)(2x-1) is [0,1)`
`f(0)=cos^(-1)=pi and f(1)=cos^(-1)=0`
Graph of the function =f(x) is as shown in the following figure.

Clearly curve has point of inflection at `(1//2,pi//2)`
Examining the line , we find that it passes through `(1//2,pi//2)`

This line must intersect the curve at three distinct points.
Now `(dy)/(dx)=(-2)/sqrt(1-(2x-1))^(2)=(-1)/sqrt(x-x^(2))=-(x-x^(2))^(-1//2)`
`therefore ((dy)/(dx))_(x=0.5)=-2`
=slope of tangent to the curve at `(1//2,pi//2)`
It can be verified that that points `(0,pi)` and `(1//2,pi//2) and (1,0)` are collinear
slope of line joining these points is `(pi-(pi)/(2))/(0-(1)/(2))=-pi`
Hence given line intersect the curve at three distinct point if its slope is less than '-2' but more than or equal, to `-pi`
`therefore (p)/(2)in[(-pi,-2) rarr p in [-2pi,-4)`
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