A) 21, 24 or \[-\]24, \[-\]21
B) 30, 31 or \[-\]30, \[-\]31
C) 40, 41 or \[-\]40, \[-\]41
D) none of these
Correct Answer: A
Solution :
Let the two numbers are a and b, such that a>b, by the hypothesis, we get \[2({{b}^{3}}+3{{a}^{2}}b)\] ?..(1) \[2({{a}^{3}}+3a{{b}^{2}})\] ?..(2) From (1), \[0<a<1\] ..?(3) From (2) and (3), \[a+\frac{1}{a}\] \[\Rightarrow \] \[{{b}^{2}}+3b-504=0\] \[\Rightarrow \] \[{{b}^{2}}+24b-21b-504=0\] \[\Rightarrow \] \[\left( 3+\frac{5}{x} \right)\left( 9-\frac{15}{x}+\frac{25}{{{x}^{2}}} \right)\] \[\Rightarrow \] \[\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca}=\_\_\_\_\_\_\_\_\] \[\Rightarrow \] \[11x+12y=58\] Substituting these values for b separately in eq. (1), we have, \[12x+11y=57,\] \[4(x+y)\] \[\frac{2}{x}+3y=15\] \[12x+11y=57,\] Thus, \[a=24,\,b=21\]or \[a=-21,\,\,b=-24.\]
You need to login to perform this action.
You will be redirected in
3 sec
