A) 4
B) 5
C) 6
D) 3
Correct Answer: B
Solution :
[b]We have, \[{{a}^{2}}+{{b}^{2}}=7\] and \[{{a}^{3}}+{{b}^{3}}=10\] |
\[\Rightarrow \]\[{{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}+{{b}^{2}}-ab)\]\[\Rightarrow \]\[10=(a+b)(7-ab)\] |
\[\Rightarrow \]\[10=(a+b)\left( 7-\frac{{{(a+b)}^{2}}-7}{2} \right)\] \[[\because {{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab]\] |
\[\Rightarrow \]\[20=(a+b)(21-{{(a+b)}^{2}})\]\[\Rightarrow \]\[{{(a+b)}^{3}}-21(a+b)+20=0\] |
Let \[a+b=x\] |
\[\therefore \]\[{{x}^{3}}-21x+20=0\] |
\[(x-1)(x-4)(x+5)=0\] |
\[\therefore \]\[x=1,4,-\,5\] |
\[\left| a+b \right|=1,4,5\] |
\[\therefore \]Maximum value of \[\left| a+b \right|=5\] |
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