A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
\[{{T}_{r+1}}={}^{5}{{C}_{r}}{{({{x}^{2}})}^{5-r}}{{\left( \frac{k}{x} \right)}^{r}}\] For coefficient of x, \[10-2r-r=1\,\,\,\,\Rightarrow r=3\] Hence, \[{{T}_{3+1}}={}^{5}{{C}_{3}}{{({{x}^{2}})}^{5-3}}{{\left( \frac{k}{x} \right)}^{3}}\] According to question, \[10\,{{k}^{3}}=270\,\,\Rightarrow \,\,k=3\].
You need to login to perform this action.
You will be redirected in
3 sec
