# Question 8 Exercise 7.2

Solutions of Question 8 of Exercise 7.2 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Q8 Find numerically greatest term in $(3-2 x)^{10}$. when $x=\frac{3}{4}$. Solution: We first write in form $\left(3-2,1^{10}=3^{10}\left(1-\frac{3 x}{2}\right)^{10}\right.$. The numerically greatest term in $\left(1-\frac{3 x}{2}\right)^{10}$ is: \begin{aligned} & \frac{(1+n) i}{1+i} \quad \frac{1+10}{1}{ }^4 \text { here }
& x^2=\frac{3}{2} \frac{3}{2}: 8
& \therefore \frac{11}{1} \cdot i \quad 7
& \end{aligned} Hence by perpe:ts of like binomial theorem, we hisu that: $p+1$ - 5.. 1 - 6 icrm is numerically greates in $: 3-\mathbf{2}_1 1^{10}$ Now $T_{5} !=\left(\begin{array}{c}10 \\ 5\end{array}\right) 3^{10} 5-2 \gamma^{15}$ we compute at $x=\frac{3}{4}$, so \begin{aligned} & T_{51}=\left(\begin{array}{c} 0 \\ 5 \end{array}\right) 3^{10 i-5} \quad 2, \frac{3}{4} y \\ & =\frac{10 !}{10-5, ! 5 !} 3^{\prime}\left(\frac{3}{2}\right)^2 \\ & \frac{25213^{16}}{2^5}-46.5016475 \\ & \end{aligned}
Hence $T$ is mumerically greatest lerm all $x-\frac{3}{4}$ and is 19375453125