Question

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

• A. The greatest value of the term independent of $x$ is $\frac{10!}{2^5(5!{)}^2}$
• B. The least value of sum of coefficient is zero
• C. The greatest value of coefficient is $32$
• D. The least value of the term independent of $x$ occurs when $p=(2n+1)\frac{\pi }{4},n\in Z$

Answer 1

Answer: (A), (B), (C)

The correct options are
A The least value of sum of coefficient is zero
B The greatest value of coefficient is $32$
C The greatest value of the term independent of $x$ is $\frac{10!}{2^5(5!{)}^2}$

$(x\sin p+x^{-1}\cos p{)}^{10}$ by comparing with $(a+b{)}^n$

we get $a=x\sin p,b=x^{-1}\sin p,n=10$

Also, we know that general term of expansion $(a+b{)}^n$ is

$t_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

So, the general term in the expansion is

$t_{r+1}={}^{10}{C}_r\left(x\sin p{)}^{10-r}\right(x^{-1}\cos p{)}^r$

$t_{r+1}={}^{10}{C}_r{x}^{10-2r}\left(\sin p{)}^{10-r}\right(\cos p{)}^r$

For the term independent of $x$, power of $x$ must be zero

So, we have $10-2r=0$ or $r=5$.

Hence, the independent term is

${}^{10}{C}_5{\sin }^{5}p{\cos }^{5}p={}^{10}{C}_5\frac{(2\sin p\cos p{)}^5}{32}$

$={}^{10}{C}_5\frac{{\sin }^{5}2p}{32}$, because $2\sin A\cos A=\sin 2A$

which is the greatest when $\sin 2p=1$

So, The greatest value of ${}^{10}{C}_5\frac{{\sin }^{5}2p}{32}$ is

${}^{10}{C}_5\frac{1}{32}=\frac{10!}{(5!)(5!)}\times \frac{1}{32}=\frac{10!}{2^5(5!{)}^2}$

Hence, option $A$ is correct

The least velue of independent term is when $\sin 2p=-1$

Which means $2p=2n\pi -\frac{\pi }{2}$

or $p=n\pi -\frac{\pi }{4}$

or $p=(4n-1)\frac{\pi }{4},n\in Z$.

For, sum of coefficient we put $x=1$, then given expansion is only with coefficient

So, sum of coefficient$=(\sin p+\cos p{)}^{10}=[(\sin p+\cos p{)}^2{]}^5$

$=({\sin }^{2}p+2\sin p\cos p+{\cos }^{2}p{)}^5$

$=(1+\sin 2p{)}^5$,

which is least when $\sin 2p=-1$.

And this least sum is $(1+\sin 2p{)}^5=(1-1{)}^5=0$

Hence, least sum of coefficient is zero.

hence, option $B$ is correct

Also, the greatest sum of coefficient occurs when $\sin 2p=1$.

Hence, greatest sum is $(1+\sin 2p{)}^5=(1+1{)}^5=2^5=32$

Hence, option $C$ is correct.

So, options $A,B,C$ are correct.