If p is an integral multiple of 4 lying betwe | vedclass

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(A) The general term in the expansion of $\left(x^2+\frac{1}{x}\right)^8$ is given by $T_{r+1} = {}^{8}C_{r} (x^2)^{8-r} (x^{-1})^r = {}^{8}C_{r} x^{1

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$3$

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(A) The general term in the expansion of $\left(x^2+\frac{1}{x}ight)^8$ is given by $T_{r+1} = {}^{8}C_{r} (x^2)^{8-r} (x^{-1})^r = {}^{8}C_{r} x^{16-3r}$.For the coefficient of $x^4$,we set $16-3r = 4$,which gives $3r = 12$,so $r = 4$. The coefficient is ${}^{8}C_{4} = \frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1} = 70$.For the coefficient of $x$,we set $16-3r = 1$,which gives $3r = 15$,so $r = 5$. The coefficient is ${}^{8}C_{5} = {}^{8}C_{3} = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$.We need to find the number of integral multiples of $4$ that lie strictly between $56$ and $70$.The multiples of $4$ are $60, 64, 68$.There are $3$ such values of $p$.Thus,option $A$ is correct.

If $p$ is an integral multiple of $4$ lying between the coefficients of $x^4$ and $x$ in the expansion of $\left(x^2+\frac{1}{x}\right)^8$,then the number of such values of $p$ is

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