Search
You can find the results of your search below.
Matching pagenames:
Fulltext results:
- Question 1 Exercise 5.2 @math-11-kpk:sol:unit05
- n=1.2^2+2.2^3+3.2^4+4.2^5+\ldots +n \cdot 2^n....(ii)\end{align} Suburacting the (ii) from (i), we get \begin{align} (1-2) S_n&=1 \cdot 2+(2-1) 2^2+(3-2) ... S_n& =2+(n-1) 2^{n+1}\end{align} =====Question 1(ii)===== Sum up to $n$ terms the series $1+4 x+7 x^2... =x+4 x^2+7 x^3+10 x^4+\ldots +(3 n-2) x^{4 t}....(ii)\end{align} Subtracting the (ii) from (i), we get
- Question 1, Exercise 1.3 @math-11-kpk:sol:unit01
- hat \begin{align}z-4w&=3i …(i)\\ 2z+3w&=11-5i …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{... 2z-8w&=6i …(iii)\end{align} Subtract (iii) from (ii), we get\\ \[\begin{array}{cccc} 2z&-8w&=6i \\ ... n} Hence $$z=4-i, \quad w=1-i.$$ =====Question 1(ii)===== Solve the simultaneous linear equation with... en that \begin{align}z+w&=3i …(i)\\ 2z+3w&=2 …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{
- Question 4 Exercise 4.5 @math-11-kpk:sol:unit04
- . \overline{8}=\dfrac{8}{9}$$.\\ =====Question 4(ii)===== Convert each decimal to common fraction $1 ... 7}{11} \ldots \ldots \ldots \ldots . . . \text { (ii) }\end{align} Putting (ii) in (i), we get\\ $$1.63=1+\dfrac{7}{11}=\dfrac{18}{11} \text {. }$$\\ =====Question 4(ii)===== Convert each decimal to common fraction $2
- Question 4 & 5 Exercise 5.2 @math-11-kpk:sol:unit05
- ac{7}{3^2}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots.(ii) \end{align} Subtracting the (ii) from (i), we get \begin{align} & \dfrac{2}{3} S_{\infty}=5+\dfrac{2... $$r S_{\infty}=3 r+5 r^2+7 r^3+\ldots \infty....(ii)$$ Subtracsing the (ii) from (i), we get \begin{align} (1-r) S_{\infty}&=3+2 r+2 r^2+2 r^3+\ldots \
- Question, Exercise 10.1 @math-11-kpk:sol:unit10
- $\alpha $in Quadrant III and $\beta $in Quadrant II, find the exact value of $\sin \left( \alpha -\be... ght)&=\frac{33}{65}.\end{align} =====Question 4(ii)===== If $\sin \alpha =-\dfrac{4}{5}$ and $\cos \... $\alpha $in Quadrant III and $\beta $in Quadrant II, find the exact value of $\cos \left( \alpha +\be... $\alpha $in Quadrant III and $\beta $in Quadrant II, find the exact value of $\tan \left( \alpha +\b
- Question 1, Exercise 3.2 @math-11-kpk:sol:unit03
- \\ &=-\hat{i}+\hat{j}\end{align} =====Question.1(ii)===== If $\vec{a}=3\hat{i}-5\hat{j}$ and $\vec{b}... } …(i)\\ |\vec{b}|&=\sqrt(-2)^2+(3)^2=\sqrt{13} …(ii)\end{align} Subtracting (i) from (ii). We get $$|\hat{a}|-|\hat{b}|=\sqrt{34}-\sqrt{13}$$ =====Quest
- Question 5(i) & 5(ii) Exercise 3.5 @math-11-kpk:sol:unit03
- ====== Question 5(i) & 5(ii) Exercise 3.5 ====== Solutions of Question 5(i) & 5(ii) of Exercise 3.5 of Unit 03: Vectors. This is uni... a} \times \vec{b}\perp \vec{b}$. =====Question 5(ii)===== Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \h
- Question 2 & 3 Exercise 4.4 @math-11-kpk:sol:unit04
- 1 r^2=27\\ a_5&=a_1 r^4=243.\end{align} Dividing (ii) by (i), we get\\ \begin{align}\dfrac{a_1 r^4}{a_... text{the third term is}=a_3&=a_1 r^2=-\sqrt{2}...(ii)\end{align} Dividing (ii) by (i), we get\\ \begin{align}\dfrac{a_1 r^2}{a_1 r}&=-\dfrac{\sqrt{2}}{2}=
- Question 6 & 7 Exercise 4.4 @math-11-kpk:sol:unit04
- \quad \ln &=m^2 \because m=a_1 r^{12} \text { by (ii) }\end{align} Hence showed that $\ln =m^2$. ====... , \dfrac{1}{a_1 r^{n-1}}$,\\ General term of the (ii) sequence is:\\ \begin{align}a_n&=\dfrac{1}{a_1 r... endent of $n$, \\ which implies that sequence in (ii) is also geometric sequence with common ratio $\d
- Question 5 & 6 Exercise 4.5 @math-11-kpk:sol:unit04
- _{2 n}&=\dfrac{a_1(r^{2 n}-1)}{r-1} \ldots . . . (ii)\\ \text { and } S_{3 n}&=\dfrac{a_1(r^{3 n}-1)}{r-1}...(iii)\end{align}\\ Puting (i),(ii) and (iii) in L.H.S of the given, we get \begin{a... -1)}{r-1}]^2 \ldots .(2)\end{align} From (i) and (ii), we see that \\ $$S_n(S_{3 n}-S_{2 n})=(S_n-S_{2
- Question 2 Exercise 6.3 @math-11-kpk:sol:unit06
- ....(i)\\ &^n C_r=\dfrac{n !}{(n-r) ! r !}=35....(ii)\end{align} Dividing Eq.(i) by Eq.(ii) \begin{align}\dfrac{n !}{(n-r) !} \cdot \dfrac{(n-r) ! r !}{n... xt{or}\quad r &=4\end{align} Putting $r=4$ in Eq.(ii), we get \begin{align} & { }^n C_4=\dfrac{n !}{(n
- Question 1, Exercise 1.2 @math-11-kpk:sol:unit01
- \begin{align} z_2+z_1&=(1-i)+(2+i)\\ &=3 \ldots (ii)\end{align} From (i) and (ii), we get required result. Now, we prove commutative property under mult
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- d (2), we have $$(AB)C=A(BC).$$ =====Question 3(ii)(a)===== If $A=\begin{bmatrix}1 & 3 \\ -1 & 4 \e... (2), we have $$A(B+C)=AB+AC.$$ =====Question 3(ii)(b)===== If $A=\begin{bmatrix} 1 & 3 \\ -1 & 4 \
- Question 3 & 4, Exercise 3.2 @math-11-kpk:sol:unit03
- nd $\hat{j}$, we have, $$p+5q=1…(i)$$ $$2p-q=-9 …(ii)$$ Multiply $2$ by (i) and subtract (ii) from (i). We have \[\begin{array}{ccc} 2p&+10q&=2 \\ \math
- Question 5(iii) & 5(iv) Exercise 3.5 @math-11-kpk:sol:unit03
- olution==== We have already calculated L.H.S in (ii) that\\ \begin{align}|\vec{a} \times \vec{b}|^2&=... h-11-kpk:sol:unit03:ex3-5-p3 |< Question 5(i) & 5(ii)]]</btn></text> <text align="right"><btn type="su