Search
You can find the results of your search below.
Fulltext results:
- 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... }}{2}.\end{align} Thus the solutions of the given equations are $\dfrac{1\pm\sqrt{5}}{2}$. =====Question 5(i
- 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==== $$z... c{1\pm \sqrt{3}i}{2}$$ The value of $z$ from both equations, we have $$z=\pm \dfrac{1}{2}\pm \dfrac{\sqrt{3}... ====Question 6(ii)===== Find the solutions of the equation ${{z}^{3}}=-8$. ====Solution==== Given: $$z^3=-8... 1\pm \sqrt{3}i$ are the solutions of the required equations. =====Question 6(iii)===== Find the solutions o
- 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... B=0 \text { and } A-B=1 $$ Solving the above two equations for $A$ and $B$, we get \begin{align}A&=\dfrac{1... $n$ and constants on the both sides of the above equation, we get $$3 A+3 B=0 \quad \text{and} \quad 2 A-B
- 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 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
- 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 2 & 3 Exercise 5.4 @math-11-kpk:sol:unit05
- $k$ and constants on the both sides of the above equation, we get $$3 A+3 B=0\quad \text{and}\quad 2 A-B=1$$ Solving the above two equations for $A$ and $B$ we get $$A=\dfrac{1}{3}\quad\te... }=\dfrac{A}{n}+\dfrac{B}{n-1}$$ Solving the above equation for $A$ and $B$ we get $A=1$ and $B=-1$. So, \be
- Question 7 and 8 Exercise 7.3 @math-11-kpk:sol:unit07
- Solution: We are taking L.H.S of the above given equation and apply the binomial theorem $$ \begin{aligned}... rac{1}{4}=2-\frac{3 x^2}{16}$. From the above two equations, we get that $a \cdot b x^2=2-\frac{3 x^2}{16}$ ... taking numerator in the L.H.S of the above given equation ====Go To==== <text align="left"><btn type="
- Unit 01: Complex Numbers (Solutions)
- complex numbers. * Solve simultaneous linear equations with complex coefficients. * Write the polyn... product of linear factors. * Solve quadratic equation of the form $pz^2+ qz+ r = 0$ by completing squar
- Unit 02: Matrices and Determinants (Solutions)
- of complex numbers. * Solve simultaneous linear equations with complex coefficients. * Write the polynom... a product of linear factors. * Solve quadratic equation of the form $pz^2+ qz+ r = 0$ by completing squar
- Question 5 & 6, Exercise 2.1 @math-11-kpk:sol:unit02
- }{2}.$$ =====Question 6(i)===== Solve the matrix equations for $X.$ Find $X-3A=2B$, if $A=\begin{bmatrix} 1... lign} =====Question 6(ii)===== Solve the matrix equations for $X.$ Find $2( X-A )=B$, if $A=\begin{bmatrix
- 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_{