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- Question 5, Exercise 1.3 @math-11-kpk:sol:unit01
- =====Question 5(i)===== Find the solutions of the equation ${{z}^{2}}+z+3=0$. ====Solution==== Given: $${{z... }}{2}i\end{align} Thus the solutions of the given equation are $-\dfrac{1}{2}\pm\dfrac{\sqrt{11}}{2}i$. =====Question 5(ii)===== Find the solutions of the equation ${{z}^{2}}-1=z$.\\ ====Solution==== Given: $${{z... ===Question 5(iii)===== Find the solutions of the equation ${{z}^{2}}-2z+i=0$\\ ====Solution==== Given: $$
- Question 6, Exercise 1.3 @math-11-kpk:sol:unit01
- =====Question 6(i)===== Find the solutions of the equation ${{z}^{4}}+{{z}^{2}}+1=0$. ====Solution==== $$z... ====Question 6(ii)===== Find the solutions of the equation ${{z}^{3}}=-8$. ====Solution==== Given: $$z^3=-8... ===Question 6(iii)===== Find the solutions of the equation ${{\left( z-1 \right)}^{3}}=-1$.\\ ====Solution==... {3}i}{2}.\end{align} Hence solutions of the given equation are $0$, $\dfrac{3}{2}\pm \dfrac{\sqrt{3}i}{2}$.
- Definitions: FSc Part1 KPK
- * **Trigonometry equation:** The equation, containing at least one trigonometry function are called Trigonometry equation. \\ e.g. $sin x=\frac{2}{7}$, $cos x-tan x=0$. ... or e-circle. * **Trigonometric function:** The equation, containing at least one trigonometric function,
- Question 3 & 4, Exercise 1.3 @math-11-kpk:sol:unit01
- {z}_{1}}=-1+i$ and ${{z}_{2}}=-1-i$ satisfied the equation ${{z}^{2}}+2z+2=0$\\ ====Solution==== Given: $... ign} This implies $z_1=-1+i$ satisfied the given equation.\\ Now put $z_2=-1-i$ in (i) \begin{align} L.H.S&... \end{align} This implies $z_2=-1-i$ satisfied the equation. =====Question 4===== Determine weather $1+2i$ i... ign} This implies $1+2i$ is solution of the given equation. ====Go To==== <text align="left"><btn type="prim
- Question 1 Exercise 5.3 @math-11-kpk:sol:unit05
- $n$ and constants on the both sides of the above equation, we get $$A+B=0 \text{and} A=1$$ Putting $A=1$,th... } Taking summation of the both sides of the above equation \begin{align} \sum_{k=1}^n T_k&=\sum_{k=1}^n(\dfr... $n$ and constants on the both sides of the above equation, we get $$3 A+3 B=0 \quad \text{and} \quad 2 A-B... $n$ and constants on the both sides of the above equation, we get $$9 A+9 B=0\quad \text{ and} \quad 4 A-5
- Question 5 Exercise 6.1 @math-11-kpk:sol:unit06
- ====Solution==== We are taking L.H.S of the above equation \begin{align}\dfrac{(2 n) !}{n !}&=\dfrac{1}{n !}... n-(2 n-1))]\end{align} In the L.H.S of the above equation are total $2 n$ terms \begin{align}\dfrac{(2 n) !... ====Solution==== We are taking L.H.S of the above equation \begin{align}\dfrac{(2 n+1) !}{n !}&=\dfrac{1}{n ... n+1-(2 n))]\end{align} In the L.H.S of the above equation. are total $2 n+1$ terms \begin{align}\dfrac{(2 n
- Multiple Choice Questions (MCQs)
- -\omega, -\omega^2$ * (D) $1, -1, 2$ * The equation $ax^2+bx+c=0$ will be quadratic if * (A) $a=0... e formed by using a sign of '=' is a/an * (A) equation * (B) formula * (C) rational fraction ... on * (C) series * (D) permutation * The equation $ax^2+bx+c=0$ will be quadratic if * (A) $a=0
- Question 1, Exercise 1.3 @math-11-kpk:sol:unit01
- =Question 1(i)===== Solve the simultaneous linear equation with complex coefficient. \begin{align}&z-4w=3i\\... Question 1(ii)===== Solve the simultaneous linear equation with complex coefficient. \begin{align}&z+w=3i\\ ... uestion 1(iii)===== Solve the simultaneous linear equation with complex coefficient. \begin{align} &3z+(2+i)
- Question 3 & 4, Exercise 3.2 @math-11-kpk:sol:unit03
- s of $\vec{r},\vec{a}$ and $\vec{b}$ in the given equation. We get $$\hat{i}-9\hat{j}=p(\hat{i}+2\hat{j})+q(... mplies & x^2+4x-17=0\end{align} This is quadratic equation with $a=1$, $b=4$ and $c=-17$, so \begin{align}x&
- Question 1 Exercise 5.1 @math-11-kpk:sol:unit05
- Taking summation of the both sides of the above equation, we get \begin{align}& \sum_{j=1}^n T_j=\sum_{j=1... {align} Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=64 \sum_
- Question 4 & 5 Exercise 5.1 @math-11-kpk:sol:unit05
- {align} Taking sum of the both sides of the above equation, we get \begin{align}& \sum_{j=1}^n T_i=\dfrac{1}... 1+j^2$ Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^{j=n} T_j=\sum
- Question 9 Exercise 5.1 @math-11-kpk:sol:unit05
- {align} Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=2 \sum_{... 4 n^3$$ Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=3 \sum_{
- Question 2 & 3 Exercise 5.4 @math-11-kpk:sol:unit05
- $k$ and constants on the both sides of the above equation, we get $$3 A+3 B=0\quad \text{and}\quad 2 A-B=1... }=\dfrac{A}{n}+\dfrac{B}{n-1}$$ Solving the above equation for $A$ and $B$ we get $A=1$ and $B=-1$. So, \be
- Question 5 & 6 Review Exercise @math-11-kpk:sol:unit05
- $n$ and constants on the both sides of the above equation, we get $$A+B=0\quad \text{and}\quad A=1$$ Puttin... } Taking summation of the both sides of the above equation \begin{align} \sum_{k=1}^n T_k&=\sum_{k=1}^n(\dfr
- Question 3 & 4 Exercise 6.1 @math-11-kpk:sol:unit06
- on==== We are taking the L.H.S of the above given equation. \begin{align}\dfrac{1}{6 !}+\dfrac{2}{7 !}+\dfra... on==== We are taking the L.H.S of the above given equation. \begin{align}\dfrac{(n+5) !}{(n+3) !}&=\dfrac{(n