Question

if the middle term in the expansion of
$( \frac{1}{x} + x \sin(x) ) {}^{10}$
equals to 63/8 then x is equal to ?​

Answer 1

Question :

if the middle term in the expansion of

$( \frac{1}{x} + x \sin(x) ) {}^{10}$

equals to 63/8 then x is equal to ?

Theory :

In the Binomial expansion of $(x + a) {}^{n}$ . The middle term depends upon the value of n .

Case 1:

When n is even , then only one middle term exists and it is :

$( \frac{n + 2}{2} ) {}^{th} \: term$

Case 2:

When n is odd , there will be two middle terms and they are :

$( \frac{n + 1}{2} ) {}^{th} \: and \: ( \frac{ n + 3 }{2} ) {}^{th} \: term \:$

$\huge{ \underline{ \underline{ \green{ \sf{ Detailed \: Answer :}}}}}$

$( \frac{1}{x} + x \sin(x) ) {}^{10}$

In this binomial expression index n = 10 ( even )

⇒middle term

$( \frac{n + 2}{2} ) {}^{th} \: term$

put n = 10

⇒middle term = 6 th term

middle term given = 63/8

we know Genral term :

${\purple{\boxed{\large{\bold{t _{r + 1} = {}^{n} c _{r}x {}^{n - r}a {}^{r} }}}}}$

⇒sixth term

$t _{5 + 1} = {}^{10} c _{5} \times ( \frac{1}{x} ) {}^{5} \times (x \sin(x) ) {}^{5}$

$\frac{63}{8} = 63 \times 4 \times \frac{1}{x {}^{5} } \times x {}^{5} \times \sin {}^{5} (x)$

Here ; ${}^{10} c _{5} = \frac{10!}{5! \times 5!} = 63 \times 4$

_____________________

$\frac{1}{32} = \sin {}^{5} (x)$

$\sin {}^{5} (x) =( \frac{1}{2} ) {}^{5}$

⇒

$\sin(x) = \frac{1}{2}$

$\sin(x) = \sin( \frac{\pi}{6} )$

⇒

${\purple{\boxed{\large{\bold{x = n\pi + ( - 1) {}^{n} \frac{\pi}{6} }}}}}$