A) 104
B) 108
C) 122
D) 5
Correct Answer: B
Solution :
[b] Given, \[f(t)={{t}^{2}}-4t+3\] |
\[\therefore \] Sum of zeroes \[=\alpha +\beta =4\] |
and product of zeroes \[=\alpha \beta =3\] |
Now, \[{{\alpha }^{4}}{{\beta }^{3}}+{{\alpha }^{3}}{{\beta }^{4}}={{\alpha }^{3}}{{\beta }^{3}}(\alpha +\beta )={{(\alpha \beta )}^{3}}(\alpha +\beta )\] |
\[={{(3)}^{3}}\times 4=108\] |
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