TS EAMCET 04 May 2019 Shift 1 Solved Paper

It is given that kth term in the expansion (‌

3

2

x2−‌

1

3x

)6 is
independent of x so,
Tr+1‌‌=6Cr(‌

3

2

)6−r(x)12−2r×(−‌

1

3

)r(x−r)
‌‌=6Cr(‌

3

2

)6−r(x)12−3r×(−‌

1

3

)
since, Tr+1 is independent of x so
12−3r=0
r=4
And
k‌‌=r+1
‌‌=4+1
‌‌=5
Now, the greatest term of the expansion (‌

3

2

x2−‌

1

3x

)6 is
r‌‌=[‌

n+1

1+|‌ 9x3 2 ]

]
‌‌=[‌

6

1+‌ 9 2 (‌ 8 27 )

]
‌‌=2
This implies
T3‌‌=5C2(‌

3

2

x2)3(−‌

1

3x

)2
‌‌=10×‌

27

8

×(‌

2

3

)4×‌

1

9

‌‌=‌

20

27