JEE Main 2021 — Mathematics Algebra

The general term in the binomial expansion of ${(a+b)}^{n}$(a+b)n is $CRn{a}^{n-R}{b}^{R}.$CRnan−RbR.

Thus, the general term in the expansion of ${(2{x}^{r}+\frac{1}{{x}^{2}})}^{10}$(2xr+x21​)10 is $=CR10{(2{x}^{r})}^{10-R}{(\frac{1}{{x}^{2}})}^{R}$=CR10(2xr)10−R(x21​)R

$$=CR10({2}^{10-R}{x}^{r(10-R)}){x}^{-2R}$$=CR10(210−Rxr(10−R))x−2R

Given, the constant term in the expansion is $180$180

$$\Rightarrow CR10\cdot {2}^{10-R}=180\ldots (1)$$⇒CR10⋅210−R=180…(1)

And $(10-R)r-2R=0$(10−R)r−2R=0

$$\Rightarrow r=\frac{2R}{10-R}$$⇒r=10−R2R​

$$\Rightarrow r=\frac{2(R-10)}{10-R}+\frac{20}{10-R}$$⇒r=10−R2(R−10)​+10−R20​

$$\Rightarrow r=-2+\frac{20}{10-R}\ldots (2)$$⇒r=−2+10−R20​…(2)

Since $R$R is positive integer less than or equal to $10$10 and $r$r is also an integer, hence, the value of $R$R is such that, $10-R$10−R must divide $20.$20.

This is possible, when the value of $10-R$10−R can be one from the divisors of $20$20 i.e. $10-R$10−R can be from $1,2,4$1,2,4 or $5$5 and consequently, $R$R can be $9,8,6$9,8,6 or $5$5 respectively.

But for R=9,6&5 equation $(1)$(1) is not satisfied, hence $R=8$R=8

as $CR10{2}^{10-R}=180$CR10210−R=180

$$\Rightarrow r=\frac{2\times 8}{10-8}=8.$$⇒r=10−82×8​=8.