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  • If the coefficient of  in  <img alt="\left(\mathrm{ax}-\frac{1}{\mathrm{bx}^2}\right)^{13}"

If the coefficient of x^{7} in  \left(\mathrm{ax}-\frac{1}{\mathrm{bx}^2}\right)^{13}  and the coefficient of  x^{-5}  in \left(\mathrm{ax}+\frac{1}{\mathrm{bx}^2}\right)^{13}  are equal, then  a^{4}b^{4}  

is equal to :

 

 

Option: 1

22 


Option: 2

44


Option: 3

11


Option: 4

33


Answers (1)

best_answer

\begin{aligned} & \left(a x-\frac{1}{b x^2}\right)^{13} \\ & \text { We have, } \\ & T_{r+1}={ }^n C_r(p)^{n-1}(q)^r \\ & T_{r+1}={ }^{13} \mathrm{C}_{\mathrm{r}}(\mathrm{ax})^{13-r}\left(-\frac{1}{\mathrm{bx}^2}\right)^r \\ \end{aligned}

\begin{aligned} & ={ }^{13} C_r(a)^{13-r}\left(-\frac{1}{b}\right)^t(x)^{13-r} \cdot(x)^{-2 r} \\ & ={ }^{13} C_r(a)^{13-r}\left(-\frac{1}{b}\right)^r(x)^{13-3 r} \\ & \end{aligned}                   ......(1)

Coefficient of  x^{7} 

\begin{aligned} & \Rightarrow 13-3 r=7 \\ & r=2 \end{aligned}

r in equation (1)

\begin{aligned} & \mathrm{T}_3={ }^{13} \mathrm{C}_2(\mathrm{a})^{13-2}\left(-\frac{1}{b}\right)^2(\mathrm{x})^{13-6} \\ & ={ }^{13} \mathrm{C}_2(\mathrm{a})^{11}\left(\frac{1}{\mathrm{~b}}\right)^2(\mathrm{x})^7 \end{aligned}

Coefficient of x^{7}  is   ^{13}C_2  \frac{(a)^{11}}{b^{2}} 

\begin{aligned} & \text { Now, }\left(a x+\frac{1}{b^2}\right)^{13} \\ & \mathrm{~T}_{\mathrm{r}+1}={ }^{13} \mathrm{C}_{\mathrm{r}}(\mathrm{ax})^{13-\mathrm{r}}\left(\frac{1}{\mathrm{bx}^2}\right)^{\mathrm{r}} \end{aligned}

\begin{aligned} & ={ }^{13} C_t(a)^{13-r}\left(\frac{1}{b}\right)^r(x)^{13-r}(x)^{-2 r} \\ & ={ }^{13} C_r(a)^{13-r}\left(\frac{1}{b}\right)^r(x)^{13-3 r} \end{aligned}                   .......(2)

Coefficient of x^{-5}

\begin{aligned} & \Rightarrow 13-3 r=-5 \\ & r=6 \end{aligned}

r in equation

$$ \begin{aligned} & \mathrm{T}_7={ }^{13} \mathrm{C}_2(\mathrm{a})^{13-6}\left(-\frac{1}{b}\right)^6(\mathrm{x})^{13-18} \\ \end{aligned}

$$ \begin{aligned} T_7 & ={ }^{13} \mathrm{C}_6(\mathrm{a})^{7}\left(\frac{1}{\mathrm{~b}}\right)^6(\mathrm{x})^-5 \end{aligned}

Coefficient of x^{-5}  is    ^{13}C_6\: \: (a)^{7}\left ( \frac{1}{b} \right )^{6}

ATQ
Coefficient of  x^{7}  = coefficient of x^{-5}

T_3=T_7

\begin{aligned} & { }^{13} \mathrm{C}_2\left(\frac{\mathrm{a}^{11}}{\mathrm{~b}^2}\right)={ }^{13} \mathrm{C}_6(\mathrm{a})^7\left(\frac{1}{\mathrm{~b}}\right)^6 \\ & \mathrm{a}^4 \cdot \mathrm{b}^4=\frac{{ }^{13} \mathrm{C}_6}{{ }^{13} \mathrm{C}_2} \\ & =\frac{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 1}{13 \times 12 \times 6 \times 5 \times 4 \times 3}=22 \end{aligned}

 

 

Posted by

Shailly goel

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