UPSC NDA 2 2019 Mathematics Solved Paper

Let (r+1)th term in the expansion of (√x−

k

x2

)10 be independent of x
Now, Tr+1=10Cr(√x)10−r⋅(

−k

x2

)r
=10Crx5−

r

2

⋅kr⋅x−2r
=10Crx5−

r

2

−2r⋅kr
⇒Tr+1=10Crx5−

5r

2

⋅kr.....(i)
This will be independent of x

if 5−

5r

2

=0
⇒

10−5r

2

=0
⇒10−5r=0
⇒5r=10
⇒r=

10

5

=2
Putting r=2 in (i), we get
T3=10C2k2
It is given that the term independent of x is equal to 405
∴10C2k2=405⇒

10

2 8

k2=405
⇒

10.9

2

k2=405
⇒k2=

405

45

=9
⇒k2=9
⇒k=√9
⇒k=±3