Question 7 and 8 Exercise 7.3

Solutions of Question 7 and 8 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.

Q7 II $x^4$ and higher powers are neglected and $(1-x)^{\frac{1}{4}}+(1-x)^{\frac{1}{4}}=a-b x^2$ then find $a$, and $b$. Solution: We are taking L.H.S of the above given equation and apply the binomial theorem $$\begin{aligned} & (1+x)^{\frac{1}{4}}+(1-x)^{\frac{1}{4}} \\ & =\left[1+\frac{x}{4}+\frac{\frac{1}{4}\left(\frac{1}{4}-1\right)}{2 !} x^2+\right. \\ & \left.\frac{\frac{1}{4}\left(\frac{1}{4}-1\right)\left(\frac{1}{4}-2\right)}{3 !} x^3+\ldots\right] \times \\ & 1-\frac{x}{4}+\frac{\frac{1}{4}\left(\frac{1}{4}-1\right)}{2 !}(-x)^2+ \\ & \left.\frac{\frac{1}{4}\left(\frac{1}{4} 1\right)\left(\frac{1}{4}-2\right)}{3 !}(-x)^3+\ldots\right] \\ & =\left[1+\frac{x}{4}-\frac{3 x^2}{32}+\frac{5 x^3}{128}-\ldots\right]+ \\ & {\left[1-\frac{x}{4}-\frac{3 x^2}{32}-\frac{5 x^3}{128}+\ldots\right]} \\ & \end{aligned}$$

Adding and neglecting $x^4$ and higher powers, we get $-2-\frac{3 x^2}{16}$ Hence $(1-x) \frac{1}{4}-(1-x) \frac{1}{4}=a-b x^2$ and tve oltain: $\left.(1+x) \frac{1}{4}+11-x\right) \frac{1}{4}=2-\frac{3 x^2}{16}$. From the above two equations, we get that $a \cdot b x^2=2-\frac{3 x^2}{16}$ $$\Rightarrow a=2 \text { and } b=-\frac{3}{16} .$$

Q8 If $x$ is of such a size that its values are considered up to $x^3$. Show that $$\frac{\left(1+\frac{x}{2}\right)^3-(1+3 x)^{\frac{1}{2}}}{1-\frac{5 x}{6}}=\frac{15 x^2}{8} .$$

Solution: We are taking numerator in the L.H.S of the above given equation

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