Question 5 and 6 Exercise 7.3

Solutions of Question 5 and 6 of Exercise 7.3 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.

Q5 If $x$ is such that $x^2$ ard higher of $x$ may be negleeled. then show that $$\frac{(8+3 x)^{\frac{2}{3}}}{(2+3 x) \sqrt{4-5 x}}=1-\frac{5 x}{8}$$

Solution: Given that: $$\frac{\sqrt[4]{3}-3 x j^{\frac{2}{3}}}{2 \cdot 3 x+4-5 x}$$ $$\begin{aligned} & =\frac{8^{\frac{2}{3}}\left(1+\frac{3 x}{8}\right)^{\frac{2}{3}}}{2\left(1+\frac{3 x}{2}\right) \sqrt{4}\left(1-\frac{5 x}{4}\right)^{\frac{1}{2}}} \\ & =\frac{\left(2^3\right)^{\frac{2}{3}}}{2.2}\left[\left(1+\frac{3 x}{8}\right)^{\frac{2}{3}}\left(1+\frac{3 x}{2}\right)^{-1}\right. \\ & \left.\times\left(1-\frac{5 x}{4}\right)^{-\frac{1}{2}}\right] \end{aligned}$$

Applying binomial expansion and neglecting $x^2$ and higher powers of $x$. $-\left(1+\frac{2}{3} \cdot \frac{3 x}{8}+\right.$ higher powers of $\left.x\right) x$ $\left(1-\frac{3 x}{2}+\right.$ higher powers of $\left.x\right) \times$ $\left(1+\frac{1}{2}, \frac{5 x}{4}+\right.$ higher powers of $\left.x\right)$ $-\left(1+\frac{x}{4}\right)\left(1-\frac{3 x}{2}\right)$ $\times\left(1+\frac{5 x}{8}\right)$ Multiplying and neglecting $x^2$ and higher powers of $x$ $$\begin{aligned} & =\left(1-\frac{3 x}{2}+\frac{x}{4}+\text { higher powers of } \mathrm{x}\right) \\ & \times\left(1+\frac{5 x}{8}\right) \\ & \therefore\left(1-\frac{5 x}{4}\right)\left(1+\frac{5 x}{8}\right) \end{aligned}$$

Multiplying and neglecting $x^2$ and higher powers of $x$, we have $$\begin{aligned} & =1+\frac{5 x}{8}-\frac{5 x}{x} \\ & =1-\frac{5 x-10 x}{8} \\ & =1-\frac{5 x}{8} . \text { Hence } \end{aligned}$$ $$\frac{(8+3 x)^{\frac{2}{3}}}{(2+3 x) \sqrt{4-5 x}}=1-\frac{5 x}{8}$$ Q6 If $x$ is large and $\frac{1}{x^3}$ may be neglected, then find approximate value of: $$\frac{x \sqrt{x^2-2 x}}{(x+1)^2}$$

Solution: We are given $$\begin{aligned} & \frac{x \sqrt{x^2-2 x}}{(x+1)^2}=\frac{x \sqrt{x^2} \sqrt{1-\frac{2 x}{x^2}}}{x^2\left(1+\frac{1}{x}\right)^2} \\ & =\left(1-\frac{2}{x}\right)^{\frac{1}{2}}\left(1+\frac{1}{x}\right)^{-2} \end{aligned}$$

Applying binomial theorem and neglecting $\frac{1}{x^3}$ etc $$\begin{aligned} & =\left[1 \cdots \frac{1}{2} \cdot \frac{2}{x}+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}\left(-\frac{2}{x}\right)^2+\right. \\ & \ldots] \times\left[1 \cdots \frac{2}{x}+\frac{-2(-2-1)}{2 !}\left(\frac{1}{x}\right)^2\right. \\ & +\ldots \ldots] \\ & =\left[1-\frac{1}{x}-\frac{1}{2 x^2}-\ldots\right] \times\left[1-\frac{2}{x}+\frac{3}{x^2}\right. \\ & +\ldots] \end{aligned}$$ Multiplying and neglecting $\frac{1}{x^3}$ and so on we have $$\begin{aligned} & =1 \frac{2}{x}+\frac{3}{x^2}+\frac{1}{x}+\frac{2}{x^2}-\frac{1}{2 x^2} \\ & =1-\frac{x^2+4}{y^2}+\frac{3}{2} \\ & =1-\frac{3}{2}+ \end{aligned}$$

Hence the spproximate value of $\frac{x \sqrt{x^2-2 x}}{(x+1)^2}$ is $1-\frac{3}{x}-\frac{9}{2 x^2}$

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