The greatest value of the term independent of | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

${\left( {x\sin 	heta  + \frac{{\cos 	heta }}{x}} ight)^{10}}$ ;$ T_{r + 1} = ^{10}C_r(x sin 	heta )^{10 - r} &middot;{\left( {\frac{{\cos x

Vedclass · Accepted Answer

$\frac{{^{10}{C_5}}}{{{2^5}}}$

Vedclass · Answer

${\left( {x\sin 	heta  + \frac{{\cos 	heta }}{x}} ight)^{10}}$ ;$ T_{r + 1} = ^{10}C_r(x sin 	heta )^{10 - r} &middot;{\left( {\frac{{\cos x}}{x}} ight)^r} = ^{10}C_r (sin 	heta )^{10 - r} &middot; x^{10 - 2r} (cos	heta )^r$

hence for independent of $x, r = 5$

$T_6 = ^{10}C_5 (sin 	heta )^5 &middot; (cos	heta )^5$.

$=\frac{{^{10}{C_5}{{(\sin 2	heta )}^5}}}{{{2^5}}}$

The greatest value of the term independent of $x$ in the expansion of ${\left( {x\sin \theta + \frac{{\cos \theta }}{x}} \right)^{10}}$ is

Similar Questions

The constant term in the expansion of $\left(2 x+\frac{1}{x^7}+3 x^2\right)^5 \text { is }........$.

The coefficient of the term independent of $x$ in the expansion of $(1 + x + 2{x^3}){\left( {\frac{3}{2}{x^2} - \frac{1}{{3x}}} \right)^9}$ is

The term independent of $x$ in the expression of $\left(1-x^{2}+3 x^{3}\right)\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}, x \neq 0$ is

The middle term in the expression of ${\left( {x - \frac{1}{x}} \right)^{18}}$ is

The coefficient of ${x^{100}}$ in the expansion of $\sum\limits_{j = 0}^{200} {{{(1 + x)}^j}} $ is