If the greatest value of the term independent | vedclass

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

Question

$T_{r+1}={ }^{10} C_{r}(x \sin \alpha)^{10-t}\left(\frac{a \cos \alpha}{x}\right)^{r}$

$r=0,1,2, \ldots, 10$

$T_{r+1}$ will be independent of $x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$T_{r+1}={ }^{10} C_{r}(x \sin \alpha)^{10-t}\left(\frac{a \cos \alpha}{x}ight)^{r}$

$r=0,1,2, \ldots, 10$

$T_{r+1}$ will be independent of $x$

When $10-2 r=0 \Rightarrow r=5$

$T_{6}={ }^{10} C_{5}(x \sin \alpha)^{5} 	imes\left(\frac{a \cos \alpha}{x}ight)^{5}$

$=^{10} C_{5} 	imes a^{5} 	imes \frac{1}{2^{5}}(\sin 2 \alpha)^{5}$

will be greatest when $\sin 2 \alpha=1$

$\Rightarrow^{10} C_{5} \frac{a^{5}}{2^{5}}={ }^{10} C_{5} \Rightarrow a=2$

If the greatest value of the term independent of $^{\prime}x^{\prime}$ in the expansion of $\left(x \sin \alpha+a \frac{\cos \alpha}{x}\right)^{10}$ is $\frac{10 !}{(5 !)^{2}}$, then the value of $' a^{\prime}$ is equal to:

Similar Questions

If the coefficients of ${r^{th}}$ term and ${(r + 4)^{th}}$ term are equal in the expansion of ${(1 + x)^{20}}$, then the value of r will be

The middle term in the expansion of ${(1 + x)^{2n}}$ is

Coefficient of ${t^{12}}$ in ${\left( {1 + {t^2}} \right)^6}\left( {1 + {t^6}} \right)\left( {1 + {t^{12}}} \right)$ is-

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

The coefficient of ${x^{32}}$ in the expansion of ${\left( {{x^4} - \frac{1}{{{x^3}}}} \right)^{15}}$ is