The term independent of x in the expansion of | vedclass

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

Question

$\left(\frac{1}{60}-\frac{x^{8}}{81}ight)\left(2 x^{2}-\frac{3}{x^{2}}ight)^{6}$

Term independent of $x$ will be $\frac{1}{60} 	imes$

indep

Vedclass · Accepted Answer

$-36$

Vedclass · Answer

$\left(\frac{1}{60}-\frac{x^{8}}{81}ight)\left(2 x^{2}-\frac{3}{x^{2}}ight)^{6}$

Term independent of $x$ will be $\frac{1}{60} 	imes$

independent of $x$ in $\left(2 x^{2}-\frac{3}{x^{2}}ight)^{6}-\frac{1}{8} 	imes$

Termof $x^{-8}$ in $\left(2 x^{2} \frac{3}{x^{3}}ight)^{6}$

$\mathrm{T}_{r+1}$ in $\left(2 \mathrm{x}^{2}-\frac{3}{\mathrm{x}^{2}}ight)^{6}$ will be

$\mathrm{T}_{r+1}=^{6} \mathrm{C}_{r}\left(2 \mathrm{x}^{2}ight)^{6-\mathrm{f}}\left(-\frac{3}{\mathrm{x}^{2}}ight)^{r}$

$=^{6} \mathrm{C}_{r} 2^{6-5}(-1)^{r} 	imes 3^{r} 	imes \mathrm{x}^{12-2 r-2 r}$

Case $I:$ For term independent of $x, 12-4 r=0$ $\Rightarrow r=3$

$T_{4}=^{6} C_{3} 	imes 2^{3} 	imes 3^{3} x^{6}=-20 	imes 2^{3} 	imes 3^{3}$

case $II :$ For term of $x^{-8}$

${12-4 r=-8}$

${4 r=20 \Rightarrow r=5}$

$T_{6}=^{6} C_{5} \cdot 2^{1}(-1) \cdot 3^{5} \cdot x^{-8}$

Required Answer $=\frac{1}{60} 	imes(-20) 2^{3} 	imes 3^{3}-\frac{1}{81} 	imes 6 	imes 2 	imes(-1) 	imes 3^{5}$

$=-72+36=-36$

Hence the correct answer is option $(B).$

The term independent of $x$ in the expansion of $\left( {\frac{1}{{60}} - \frac{{{x^8}}}{{81}}} \right).{\left( {2{x^2} - \frac{3}{{{x^2}}}} \right)^6}$ is equal to

Similar Questions

Coefficient of $t^{20}$ in the expansion of $(1 + t^2)^{10}(1 + t^{10})(1 + t^{20})$ is

If the expansion of ${\left( {{y^2} + \frac{c}{y}} \right)^5}$, the coefficient of $y$ will be

Coefficient of $x$ in the expansion of ${\left( {{x^2} + \frac{a}{x}} \right)^5}$ is

If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$, is $\alpha \times 2^8 \times \sqrt[5]{3}$, then $25 \alpha$ is equal to :

If ${\left( {2 + \frac{x}{3}} \right)^{55}}$ is expanded in the ascending powers of $x$ and the coefficients of powers of $x$ in two consecutive terms of the expansion are equal, then these terms are