If P(x1,y1), Q(x2,y2), R(x3,y3) be the points of inflection of the curve x2y−x+y−1=0, then number of rational vertices of the polygon PQR is
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Solution
y=x+1x2+1 dydx=(−x2−2x+1)(x2+1)2 d2ydx2=2(x−1)(x2+4x+1)(x2+1)3⇒x=1,x=−2+√3,−2−√3
For these three points f′′(x) vanishes andf′(x) is defined so only one rational point.