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23 Sep 2021, 01:00
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B
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Difficulty:
15%
(low)
Question Stats:
86% (01:58) correct
14%
(02:32)
wrong
based on 113
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History
Date
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Not Attempted Yet
If \(\frac{5x^2+ax+b}{3x^2+7x+5}=x\) and \(3x^3+2x^2-b=0\), where \(x \neq 0\), what is the value of a ?
A. 4
B. 5
C. 6
D. 7
E. 8
A. 4
B. 5
C. 6
D. 7
E. 8
23 Sep 2021, 01:20
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3x^3+2x^2-b=0 => 3x^3+2x^2=b--- eq-(1)
now, 5x^2+ax+b=x(3x^2+7x+5)=> 5x^2+ax+b=3x^3+7x^2+5x=> ax+b=3x^3+2x^2+5---eq-2
Using eq-1 in eq-2
we got, ax+b=b+5=> a=5
now, 5x^2+ax+b=x(3x^2+7x+5)=> 5x^2+ax+b=3x^3+7x^2+5x=> ax+b=3x^3+2x^2+5---eq-2
Using eq-1 in eq-2
we got, ax+b=b+5=> a=5
Bunuel
\(\frac{5x^2+ax+b}{3x^2+7x+5}=x\)
\(5x^2+ax+b = 3x^3+7x^2+5x\)
\(ax = 3x^3+7x^2 -5x^2 - b +5x\)
\(ax =(3x^3+2x^2-b) +5x\)
\(ax =5x\)
a = 5, since \(x \neq 0\)
IMO, (B)!
