JEE Main 2019 — Mathematics Algebra

To find the coefficient of term independent of $x$x in given expression, first let's find constant term and coefficient of ${x}^{-8}$x−8 in the expansion of ${(2{x}^{2}-\frac{3}{{x}^{2}})}^{6}$(2x2−x23​)6

$${T}_{r+1}{= }^{6}{C}_{r} {(2{x}^{2})}^{6-r}\times {(-\frac{3}{{x}^{2}})}^{r}$$Tr+1​=6Cr​(2x2)6−r×(−x23​)r

$${T}_{r+1}{= }^{6}{C}_{r} {2}^{6-r}\times {(-3)}^{r} \times {x}^{12-4r}$$Tr+1​=6Cr​26−r×(−3)r×x12−4r

For constant term: $r=3$r=3 , hence constant term = ${ }^{6}{C}_{3} {2}^{3}\times {(-3)}^{3}\times {x}^{0}$6C3​23×(−3)3×x0

For coefficient of ${x}^{-8} : r=5$x−8:r=5 , hence coefficient of ${x}^{-8}{= }^{6}{C}_{5}\times {2}^{1}{(-3)}^{5}\times {x}^{-8}$x−8=6C5​×21(−3)5×x−8

Given expression can be written as

$$(\frac{1}{60}-\frac{{x}^{8}}{81})({ }^{6}{C}_{3}\times {2}^{3}\times {(-3)}^{3}\times {x}^{0}+\ldots + { }^{6}{C}_{5}\times {2}^{1}\times {(-3)}^{5}\times {x}^{-8}+\ldots )$$(601​−81x8​)(6C3​×23×(−3)3×x0+…+6C5​×21×(−3)5×x−8+…)

Hence, coefficient of term independent of $x$x

$$=\frac{1}{60}{\times }^{6}{C}_{3}\times {2}^{3}{(-3)}^{3}-\frac{1}{81}{\times }^{ 6}{C}_{5}\times {2}^{1}\times {(-3)}^{5}$$=601​×6C3​×23(−3)3−811​×6C5​×21×(−3)5

$$= -72+36= -36$$=−72+36=−36