First slide

Binomial theorem for positive integral Index

Question

For the expansion ${(xsin p + x^{- 1} \cos p)}^{10}, (p \in R)$

Moderate

A

the greatest value of the term independent of x is 10!/2⁵(5!)²
B

the least value of sum of coefficient is zero
C

the greatest value of sum of coefficient is 32
D

the least value of the term independent of x occurs when $p = (2 n + 1) \frac{π}{4}, n \in Z$

Solution

${(xsin p + x^{- 1} \cos p)}^{10}$
The general term in the expansion is
$T_{r + 1} =^{10} C_{r} (xsin p)^{10 - r} {(x^{- 1} \cos p)}^{r}$
For the term independent of x, we have 10 - 2r = 0 or r = 5.
Hence, the independent term is
$^{10} C_{5} \sin^{5} {pcos}^{5} p =^{10} C_{5} \frac{\sin^{5} 2 p}{32}$
which is the greatest when sin2p = 1.
The least value of $^{10} C_{5} \frac{\sin^{5} 2 p}{32} is - \frac{10!}{2^{5} (5!)^{2}}$ when
$\sin 2 p = - 1 or p = (4 n - 1) \frac{π}{4}, n \in Z$ .
Sum of coefficient is (sin p + cos p)¹⁰, when x = 1
or (1 + sin 2p)⁵, which is least when sin2p = -1.
Hence, least sum of coefficients is zero. Greatest sum of coefficient occurs when sin 2p = 1. Hence, greatest sum is 2⁵ = 32.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App