Search
You can find the results of your search below.
Matching pagenames:
Fulltext results:
- Question 18, Exercise 2.2 @math-11-kpk:sol:unit02
- atrix.\\ $$A=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right]$$ $$|A|=a_{11}a_{22}-a_{12}a_{21}$$ $$AdjA=\left[ \begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix} \right]... 1}{|A|}AdjA$$ $$A^{-1}=\dfrac{1}{a_{11}}a_{22}-a_{12}a_{21}\left[ \begin{matrix} a_{22} & -a_{12}
- Question 13, Exercise 2.1 @math-11-kpk:sol:unit02
- tion==== $$A=\left[ \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ ... gin{matrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33... )$$ $$A+A^t=\left[ \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ ... gin{matrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33
- Question, Exercise 10.1 @math-11-kpk:sol:unit10
- n \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in ... lpha$ is in 3rd quadrant, \\ $\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an ide... quadrant, \begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{144}{169}}... \beta \\ &=\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)+\left(-\frac{3}{5}\right)\left(\frac{
- Question 3 & 4 Exercise 4.3 @math-11-kpk:sol:unit04
- rac{66}{2}(25+350) \\ \Rightarrow S_{66}&=33(375)=12375 .\end{align} =====Question 4===== The sum of ... a+d)&=36 \\ \Rightarrow 3 a&=36 \\ \Rightarrow a&=12\end{align} Now by the second condition, the sum o... \Rightarrow 3 a^3+6 a d^2&=6336 \\ \Rightarrow 3(12)^3+6(12) d^2&=6336 \text { as } a=12 \\ \Rightarrow 3(1728)+72 d^2&=6336 \\ \Rightarrow 72 d^2&=6336-
- Question 9 Exercise 7.2 @math-11-kpk:sol:unit07
- greatest term in the expansion $(x-y)="$ when $x=12$ and $y-4$. Solution: When we put $x=12$, then the give becounes $$ \begin{aligned} & \left(x \quad y=20(12-y)^{20}\right. \\ & =12^{2 n}\left(\begin{array}{ll} 1 & \frac{y}{12} \end{array}\right)^{31} \end{al
- Question 3, Exercise 2.2 @math-11-kpk:sol:unit02
- lution==== Let $$A=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ ... 11} \left( a_{22} a_{33}-a_{23} a_{32} \right)-a_{12}\left( a_{21}a_{33}-a_{23}a_{31} \right)+a_{13}\l... ght)\\ &=a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}\\ &=a_{11}a_{22}a_{33}+a_{12}a_
- Question 7 & 8 Exercise 4.5 @math-11-kpk:sol:unit04
- 28\\ \Rightarrow a^3&=1728\\ \Rightarrow \quad a&=12,\\ \text{putting}\text{in} (1)\\ \dfrac{12}{r}+12+12 r&=38\\ \Rightarrow \dfrac{1}{r}+1+r&=\dfrac{38}{12}=\dfrac{19}{6}\\ \Rightarrow \dfrac{r^2+r+1}{r
- Question 1 Exercise 5.1 @math-11-kpk:sol:unit05
- \dfrac{n(n+1)}{2}+n \\ & =n[\dfrac{4(n+1)(2 n+1)-12(n+1)+6}{6}] \\ & =n[\dfrac{4(2 n^2+3 n+1)-12 n-12+6}{6}] \\ & =n[\dfrac{8 n^2+12n+4-12 n-6}{6}] \\ & =n[\dfrac{8 n^2-2}{6}] \\ & =\dfrac{n(4 n^2-1)}{3}
- Question 2, Exercise 10.1 @math-11-kpk:sol:unit10
- on 2(i)==== Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }{3}-\dfrac{\pi }{4}$ and using the ide... ion 2(iv)=== Evaluate exactly:$\tan \dfrac{5\pi }{12}$ ==Solution== We rewrite $\dfrac{5\pi }{12}$ as $\dfrac{\left( 2+3 \right)\pi }{12}$ or $\dfrac{\pi
- Question 6, Exercise 10.2 @math-11-kpk:sol:unit10
- half angle identities to evaluate exactly $sin{{112.5}^{\circ }}$. ====Solution==== Because ${{112.5}^{\circ }}=\dfrac{{{225}^{\circ }}}{2}$, the $\dfrac... frac{{{225}^{\circ }}}{2}$, so we can find $sin{{112.5}^{\circ }}$by using half angle identity as, \begin{align} sin{{112.5}^{\circ }}&=\sin \dfrac{{{225}^{\circ }}}{2}=\s
- Question 12 & 13 Exercise 4.2 @math-11-kpk:sol:unit04
- ====== Question 12 & 13 Exercise 4.2 ====== Solutions of Question 12 & 13 of Exercise 4.2 of Unit 04: Sequence and Serie... KPTB or KPTBB) Peshawar, Pakistan. =====Question 12===== A man earned dollars 3500 the first year he ... tion 13(i)===== Find the arithmetic mean between $12$ and $18$. GOOD ====Solution==== Here $a=12, b=18
- Question 10 Exercise 7.3 @math-11-kpk:sol:unit07
- t{\frac{2}{3}}$. (ii) $1+\frac{5}{8}+\frac{5.8}{8.12}+\frac{5.8 .11}{8.12 .16}+\ldots$ Solution: The given series is binomial series. Let it be identical... ac{5}{8} \\ & \frac{n(n-1)}{2 !} x^2=\frac{5.8}{8.12} . \end{aligned} $$ Taking square of Eq.(1), we ... $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{5.8}{8.12} \cdot \frac{64}{25}=\frac{16}{15} \\ & \Rightarr
- Question 3, Exercise 10.1 @math-11-kpk:sol:unit10
- 5}.\frac{3}{5}-\frac{3}{5}.\frac{4}{5}\\ &=\dfrac{12}{25}-\frac{12}{25}\\ \cos \left( u+v \right)&=0\end{align} =====Question 3(ii)===== If $\sin u=\df... dfrac{3}{4}\dfrac{4}{3}}\\ &=\dfrac{\dfrac{9-16}{12}}{1+\dfrac{12}{12}}\\ &=\dfrac{\dfrac{-7}{12}}{2}\\ &=\dfrac{-7}{24}\end{align} =====Question 3(iii
- Question 2 and 3 Exercise 3.3 @math-11-kpk:sol:unit03
- }-7 \hat{k}) \\ \Rightarrow &=4 \hat{i}+3 \hat{j}-12 \hat{k}\\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{(4)^2+(3)^2+(-12)^2} \\ \Rightarrow &=\sqrt{16+9+144} \\ \Rightarr... b}}{|\vec{a}+\vec{b}|}=\dfrac{4 \hat{i}+3 \hat{j}-12 k}{13} \\ \Rightarrow &=\dfrac{1}{13}(4 \hat{i}+3 \hat{j}-12 \hat{k})\\ \Rightarrow &=\dfrac{4}{13}\hat{i}+\df
- Question 3 & 4 Exercise 6.1 @math-11-kpk:sol:unit06
- alue of $n$, when $\dfrac{n(n !)}{(n-5) !}=\dfrac{12(n !)}{(n-4) !}$ ====Solution==== We are given: \begin{align}\dfrac{n(n !)}{(n-5) !}&=\dfrac{12(n !)}{(n-4) !} \\ \Rightarrow \dfrac{n}{(n-5) !}&=\dfrac{12}{(n-4)(n-5) !} \\ \Rightarrow n&=\dfrac{12}{(n-4)} \\ \Rightarrow n(n-4)&=12 \\ \rightarrow n^2-4 n-