# Question 14 Exercise 7.3

Solutions of Question 14 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.

Q14 If $x$ is nearly equal to unity, then show that $p x^p-q x^q=(p-q) x^{p+q}$. Solution: Since, $x$ is nearly equal to unity, so let say $x=1+h$. where $h \longrightarrow 0$. Then $$ p x^p-q x^q=p(1+h)^p-q(1+h)^q $$

Applying binomial theorem on the R.H.S of the above last equation, $$ \begin{aligned} & p x^p-q x^q \\ & =p(1+p h+\text { higher powers h) } \\ & -q(1+q h+\text { higher powcrs } h) \\ & \Rightarrow p x^p-q x^q=p+p^2 h-q-q^2 h \\ & \Rightarrow p x^p-q x^q=(p-q)+\left(p^2-q^2\right) h \\ & \Rightarrow p x^p-q x^q \\ & -(p-q)+[(p-q)(p+q)] h \\ & \Rightarrow p x^p-q x^q=(p-q)[1+(p+q) h] \\ & \Rightarrow p x^p-q x^q=(p-q)[1+h]^{p+q} . \end{aligned} $$

Replacing $1+h$ by $x$ then we have

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