TS EAMCET 5-Aug-2021 Shift 2 Question Paper

(1+x)2k‌‌=‌2kC0+‌2kC1x+‌2kC2x2+...+‌2kC2kx2k
Tr+1‌‌=‌2kCrxr‌‌Tr=‌2kCr−1xr−1
If Tr+1 is greatest
‌

Tr+1

Tr

>1
⇒‌‌(‌

2k−r+1

r

)x>1
⇒‌‌x>(‌

r

2k−r+1

) . . . (i)
Similarly, ‌

Tr+1

Tr+2

>1
⇒‌‌‌

‌2kCrxr

‌2kCr+1xr+1

⇒‌‌‌

r+1

2k−(r+1)+1

⋅‌

1

x

>1
⇒‌‌‌

r+1

2k−r

>x . . . (i)
In the expansion of (1+x)2k, the middle term is
‌

2k

2

+1=(k+1)‌th ‌ term.
So, r=k and from Eq. (ii), we get
x>‌

k

k+1

And from Eq. (ii), we have
‌‌

k+1

k

>x⇒x<‌

k+1

k

∴‌‌‌

k

k+1

<x<‌

k+1

k

' x ' can be positive or negative, so it would be better to say that
‌

k

k+1

<|x|<‌

k+1

k

So, x∈(−‌

k+1

k

,‌

−k

k+1

)∪(‌

k

k+1

,‌

k+1

k

)
According to the given options. None is correct.