The number of all possible values of k for wh | vedclass

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(A) The general term in the expansion of $(\sqrt{x}+\sqrt[k]{y})^{10}$ is given by $T_{r+1} = \binom{10}{r} (x^{1/2})^{10-r} (y^{1/k})^r = \binom{10}{

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$3$

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(A) The general term in the expansion of $(\sqrt{x}+\sqrt[k]{y})^{10}$ is given by $T_{r+1} = \binom{10}{r} (x^{1/2})^{10-r} (y^{1/k})^r = \binom{10}{r} x^{(10-r)/2} y^{r/k}$ for $r = 0, 1, 2, \dots, 10$.For the term to be rational,both exponents $\frac{10-r}{2}$ and $\frac{r}{k}$ must be integers.Since there are $11$ terms in total and exactly $9$ terms are irrational,there must be $11 - 9 = 2$ rational terms.The term $T_{r+1}$ is rational if $r$ is even (so that $\frac{10-r}{2}$ is an integer) and $r$ is a multiple of $k$ (so that $\frac{r}{k}$ is an integer).For $r=0$,$T_1 = \binom{10}{0} x^5 y^0 = x^5$ is always rational.For $r=10$,$T_{11} = \binom{10}{10} x^0 y^{10/k} = y^{10/k}$ is rational if $k$ divides $10$.If $k$ divides $10$,then $r=0$ and $r=10$ are rational,giving $2$ rational terms. The divisors of $10$ are $1, 2, 5, 10$.If $k=1$,$r$ can be any value,so all terms are rational. This is rejected.If $k=2$,rational terms occur when $r$ is even and $r$ is a multiple of $2$. $r \in \{0, 2, 4, 6, 8, 10\}$,which gives $6$ rational terms. Rejected.If $k=5$,rational terms occur when $r$ is even and $r$ is a multiple of $5$. $r \in \{0, 10\}$,which gives $2$ rational terms. Accepted.If $k=10$,rational terms occur when $r$ is even and $r$ is a multiple of $10$. $r \in \{0, 10\}$,which gives $2$ rational terms. Accepted.Checking other values of $k$ where $r=0$ is the only rational term is not possible as $r=10$ is always a candidate for rationality if $k|10$. Thus,$k=5$ and $k=10$ are the solutions.

The number of all possible values of $k$ for which the expansion $(\sqrt{x}+\sqrt[k]{y})^{10}$ will have exactly nine irrational terms is

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