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- Question 7, Exercise 1.2 @math-11-kpk:sol:unit01
- estion 7(i)===== Separate into real and imaginary parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{... =&\dfrac{4}{29}+\dfrac{19}{29}i \end{align} Real part $=\dfrac{4}{29}$\\ Imaginary part $=\dfrac{19}{29}$ =====Question 7(ii)===== Separate into real and imaginary parts $\dfrac{{{\left( 1+2i \right)}^{2}}}{1-3i}$. ====Solution==== \begin{align}&\dfrac
- Unit 02: Matrices and Determinants (Solutions) @math-11-kpk:sol
- ===== Unit 02: Matrices and Determinants (Solutions) ===== This is a second unit of the book Mathema... bers and $i=\sqrt{-1}$. * Recognize $a$ as real part of $z$ and $b$ as imaginary part of $z$. * Know condition for equality of complex numbers. * Carry ... conjugate of $z=a+ib$. * Define $|z| = \sqrt{a^2+b^2}$ as the absolute value or modulus of a compl
- Question 1 Exercise 4.5 @math-11-kpk:sol:unit04
- {align} ====Go To==== <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit04:ex4-5-p2|Question 2 >]]</btn></text>
- Question 1 Exercise 5.3 @math-11-kpk:sol:unit05
- $$u_n=\dfrac{1}{(3 n-1)(3 n+2)}$$ Resolving into partial fractions: $$\dfrac{1}{(3 n-1)(3 n+2)}=\dfrac{A}{3 n-1}-\dfrac{B}{3 n+2}$$ Multiplying both sides by $(3 n-1)(3 n+2)$ we get, \begin{align} 1&=A(3 n+2)+B(3 n-1) \\ \Rightarrow(3 A+3 B) n+2 A-B&=1\end{align} Comparing the coefficients of $n$ and consta
- Question 7 & 8 Exercise 4.5 @math-11-kpk:sol:unit04
- the first $n$ terms of the sequence $\{(\dfrac{1}{2})^n\}$. ====Solution==== The sequence is:\\ $$\{(\dfrac{1}{2})^n\}=\dfrac{1}{2}, \dfrac{1}{2^2}, \dfrac{1}{2^3}, \ldots$$\\ where $$a_1=\dfrac{1}{2}$$\\ and $$r=\dfrac{\dfrac{1}{2
- Question 12, 13 & 14, Exercise 3.2 @math-11-kpk:sol:unit03
- . Find the value of $z.$ ====Solution==== {{ :fsc-part1-kpk:sol:unit03:math-11-kpk-3-2-q13.svg?nolink |Question 13}} By head to tail rul... have\\ \begin{align}\vec{u}+\vec{v}&=\vec{w}\\ (2\hat{i}+3\hat{j}+4\hat{k})+(-\hat{i}+3\hat{j}-\hat... ition vectors of the points $A,B,C$ and $D$ are $2\hat{i}-\hat{j}+\hat{k}$,$3\hat{i}+\hat{j},$ $2\hat{i}+4\hat{j}-2\hat{k}$ and $-\hat{i}-2\hat{j}+\hat
- Question 5 & 6 Exercise 4.5 @math-11-kpk:sol:unit04
- =====Question 5===== Find $r$ such that $S_{10}=244 S_5$. ====Solution==== We know that $$S_n=\dfra... we get\\ \begin{align}\dfrac{a_1(r^{10}-1)}{r-1}&=244 \dfrac{a_1(r^5-1)}{r-1} \\ \Rightarrow r^{10}-1&=244(r^5-1) \\ \Rightarrow r^{10}-244 r^5 \cdots 1+244&=0 \\ \Rightarrow r^{10}-244 r^5+243&=0 \\ \Righ
- Question 1, Exercise 10.3 @math-11-kpk:sol:unit10
- KPTB or KPTBB) Peshawar, Pakistan. There are four parts in Question 1. =====Question 1(i)===== Express the product as sum or difference $2\sin 6x\sin x$. ====Solution==== We have an identity: $$-2\sin \alpha \sin \beta =\cos (\alpha +\beta )-\cos... ut $\alpha =6x$ and $\beta =x$ \begin{align}-\,2\sin 6x\sin x&=\cos (6x+x)-\cos (6x-x)\\ &=\cos 7x
- Question 1, Exercise 10.2 @math-11-kpk:sol:unit10
- drawing the reference triangle as shown: {{ :fsc-part1-kpk:sol:unit10:fsc-part1-kpk-ex10-2-q1.png?nolink |reference triangle}} we find $\sin \theta =\dfrac{1}{\sqrt{26}}$ and $\cos \theta =\dfrac{-5}{\sqrt{26}}$ Thus, we have the following by using double angle identities. \begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\ &=2\left( \d
- Question 2 Exercise 4.5 @math-11-kpk:sol:unit04
- ====== Question 2 Exercise 4.5 ====== Solutions of Question 2 of Exercise 4.5 of Unit 04: Sequence and Series. This is... KPTB or KPTBB) Peshawar, Pakistan. =====Question 2(i)===== Some of the components $a_1, a_n, n_2 r$ and $S_n$ of a geometric sequence are given. Find t
- Question 4 Review Exercise @math-11-kpk:sol:unit05
- neral term of the series is: $$a_n=\dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}$$ Resolving into partial fractions \begin{align} \dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}&=\dfrac{A}{3 n-2}+\dfrac{B}{3 n+1}+\dfrac{C}{3 n+4}\end{align} Multiplying both side
- Question 2 & 3 Exercise 5.4 @math-11-kpk:sol:unit05
- ====== Question 2 & 3 Exercise 5.4 ====== Solutions of Question 2 & 3 of Exercise 5.4 of Unit 05: Miscullaneous Series... KPTB or KPTBB) Peshawar, Pakistan. =====Question 2===== Find sum of the series: $\sum_{k=1}^n \dfrac{1}{9 k^2+3 k-2}$ ====Solution==== \begin{align}\text { Let
- Question 8, Exercise 1.2 @math-11-kpk:sol:unit01
- ====== Question 8, Exercise 1.2 ====== Solutions of Question 8 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbo... ====Question 8(i)===== Show that $z+\overline{z}=2\operatorname{Re}\left( z \right)$. ====Solution==... +ib \right)+\left( a-ib \right)\\ &=a+ib+a-ib\\ &=2a\\ z+\overline{z}&=2\operatorname{Re}\left( z \ri
- Question 6 Exercise 3.4 @math-11-kpk:sol:unit03
- ====Question 6(i)===== A force $\vec{F}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ acts on a particle at $(1,-2,2)$. Find the moment or torque of t... ===Question 6(ii)===== A force $\vec{F}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ acts on a particle at $(1,-2,2)$. Find the moment or torque of the force about the point $(1,2,1)$ ====Solution==== Let $\vec{r}$
- Question 7, Exercise 3.2 @math-11-kpk:sol:unit03
- ==== <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit03:ex3-2-p4 |< Question 5 & 6 ]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit03:ex3-2-p6|Question 8 >]]</btn></text>