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Find the set of values of p for which line 2px-4y+2pi-9=0 intersect the curve y=cos^(-1)at three distinct points. |
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Answer» Solution :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 atthree 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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