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- Chapter 06: Sequences and Series @fsc:fsc_part_1_solutions
- nd $d$ be common difference of A.P, then $$\begin{align}l=a_1+(p-1)d,\\ m=a_1+(q-1)d,\\ n=a_1+(r-1)d.\end{align}$$ Now $$\begin{align}L.H.S &= l(q-r)+m(r-p)+n(p-q)\\ &= lq-lr+mr-mp+np-nq\\ &=(l-n)q+(m-l)r+(n-m)p\\ &... (r-q)dp\\ &=[pq-qr+qr-pr+pr-pq]d=(0)d=0=R.H.S\end{align}$$ **Exercise 6.4⇒ Question 3(ii)** An equation
- Unit 02: Differentiation @fsc:fsc_part_2_solutions
- }$ with respect to $x$. ===Method 1=== $$ \begin{aligned} \frac{d}{dx}\left(\frac{x+1}{x-1}\right) &= \f... ac{x-1-x-1}{(x-1)^2}\\ &= \frac{-2}{(x-1)^2} \end{aligned} $$ ===Method 2=== By converting improper to pr... }= 1+\frac{2}{x-1}=1+2(x-1)^{-1} $$ Now $$ \begin{aligned} \frac{d}{dx}\left(\frac{x+1}{x-1}\right) &=\fr... &= 0-2(x-1)^{-2}(1)\\ &= \frac{-2}{(x-1)^2} \end{aligned} $$ This was a simple example but try it to fi
- Important Derivatives & Integrals
- cellpadding="2" cellspacing="2"> <tr> <td align="center"><strong><img src="/images/icon_pdf.gif" ... cellpadding="2" cellspacing="2"> <tr> <td align="center"><strong><img src="/images/icon_pdf.gif"
- MCQs-Short Questions by Mr Parvez Khan @fsc:fsc_part_1_mcqs
- the MCQs is given at page 57.</wrap> <HTML> <div align="center"> <iframe src="http://docs.google.com/vie
- MCQs-Short Questions by Mr. Parvez Khan @fsc:fsc_part_2_mcqs
- Answers are given at page 32.</wrap> <HTML> <div align="center"> <iframe src="http://docs.google.com/vi