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
& =5.82 \quad \because \quad i
&
\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
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