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#1 Posted 11:25pm 02-01-10 FIITJEE: cubic functionf(x)=ax[p]3[/p]+bx[p]2[/p]+cx+d g(y)=ay[p]3[/p]+f'''(m)y[p]2[/p]/2+cf''(m)/1+f(m)=0. f(x) has roots α[ss]1[/ss], α[ss]2[/ss], α[ss]3[/ss] and g(y) has corresponding β roots. PT difference between the corresponding roots are equal.
I'm not dwelling on the past; I don't think you should either. |
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#2 Posted 00:10am 03-01-10 Re: FIITJEE: cubic functionwhich corressponding roots are talking about?? Can u explain ur ques a lil bit more....
Nikhil Kaushik
IIIrd yr BTech, IIT kharagpur |
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#3 Posted 07:03am 03-01-10 Re: FIITJEE: cubic functionseems to me the qn has been typed wrongly. I guess it should have been: [im]http://codecogs.izyba.com/gif.latex?g%28y%29%20%3D%20ay%5E3+%5Cfrac%7Bf%22%28m%29%7D%7B2%7Dy%5E2%20+%20f%27%28m%29%20+%20f%28m%29[/im] Then you can write g(y) as [im]http://codecogs.izyba.com/gif.latex?g%28y%29%20%3D%20a%28y+m%29%5E3+b%28y+m%29%5E2+c%28y+m%29+d[/im] Its now obvious that if the roots of f are [im]http://codecogs.izyba.com/gif.latex?%5Calpha_1%2C%20%5Calpha_2%2C%20%5Calpha_3[/im], the roots of g are correspondingly [im]http://codecogs.izyba.com/gif.latex?%5Calpha_1%20-m%2C%20%5Calpha_2-m%2C%20%5Calpha_3-m[/im] So that the difference of corresponding roots is m |
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#4 Posted 5:57pm 03-01-10 Re: FIITJEE: cubic functionno i meant that g(y) has roots β[ss]1[/ss], β[ss]2[/ss], β[ss]3[/ss]. So PT α[ss]1[/ss]-β[ss]2[/ss]=α[ss]2[/ss]-β[ss]2[/ss]=α[ss]3[/ss]-β[ss]3[/ss] and yes g(y) is typed wrong. Its ay[p]3[/p]+f''(m)y[p]2[/p]/2+f'(m)y+f(m)=0 and f(x)=0 sorry again 4 da inconvenience!
I'm not dwelling on the past; I don't think you should either. Edited on 6:12pm 03-01-10 |
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