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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Apr 2007 19:19:01 IST
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Let n be an odd integer then the number of real roots of polynomial equation
Pn(x)= 1+2x+3x2+4x3............(n+1)xn is
(a) 0 (b) n (c) 1 (d) none
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3 Apr 2007 19:22:44 IST
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Hi king,
think the answer is 1.
Is it?
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3 Apr 2007 19:36:49 IST
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ya it has to 1
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light travels the fastest ??? NO
wherever light goes it always finds that darkness has already got there |
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3 Apr 2007 20:57:56 IST
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yes the answer is 1
u see there beimg a n degree poly has n roots atleast one root is real as complex roots occur in conjugate pairs
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khud hi ko kar buland itna ki khuda bhi tujhse puche ...bol bande , teri raza kya hai |
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8 Apr 2007 10:27:21 IST
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Since it is an objective type question, put n=3.
y = 4x3+3x2+2x+1
dy/dx = 12x2+6x+2 > 0 for any x since its discriminant is less than 0.
Hence y is an increasing function and hence it will cut the x-axis only once.
Hence no. of real roots is 1.
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Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
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