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- Question 10, Exercise 1.2 @math-11-nbf:sol:unit01
- .** Given \[z_1 = -3 + 2i, \quad z_2 = 1 - 3i\] Then \begin{align} \frac{z_1}{z_2} &= \frac{-3 + 2i}{1... - 2i, \quad \overline{z_2} = 1 + 3i. \end{align} Then \begin{align} \frac{\overline{z_1}}{\overline{z_2... \\ &= -3 + 11i + 6\\ &= 3 + 11i. \end{align} Then \[ \overline{z_1 z_2}= 3 - 11i. -- (i)\] Now \[ ... z_1} = -3 - 2i, \quad \overline{z_2} = 1 + 3i. \] Then \begin{align} \overline{z_1}\,\, \overline{z_2}
- Question 1,Review Exercise @math-11-nbf:sol:unit09
- and the terminal arm of angle is in III quadrant. Then $\sin \theta=$\\ * $\frac{1}{2}$\\ * (b... {2}cosec \theta \theta $ and $\theta=45^{\circ}$, then the value of the given trigonometric identity is:... al side of an angle $'\theta'$ is in IV quadrant, then $x=$\\ * (a) $\cos \theta$\\ * (b) $-\cos \theta$... If $\tan A=\frac{1}{7}$ and $\tan B=\frac{1}{3}$, then $\cos 2A$ is equal to:\\ * (a) $\sin B$ \\ * (b)
- Question 3, Exercise 1.2 @math-11-nbf:sol:unit01
- bb{R}\, ... (1)$$ First suppose that $z$ is real, then we shall prove $\overline{z}=z$. Since $z$ is real, imaginary part of $z$ is zero. i.e. $b=0$. Then \begin{align} &z=a \\ \implies &\bar{z}=a \end{a... }$. Now conversly suppose that $\overline{z}=z$, then we $z$ is real. As \begin{align}& z=\bar{z}\\ \R... uad & b=0\quad \because \quad 2i\neq 0\end{align} Then (1) becomes $$z=a+i(0)=a$$ This gives $z$ is real
- Question 1, Exercise 2.5 @math-11-nbf:sol:unit02
- First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{cc... First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{ll... First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{cc... First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{cc
- Question 1, Review Exercise @math-11-nbf:sol:unit05
- collapse> ii. Divide $9 y^{2}+9 y-10$ by $3 y-2$, then remainder is:\\ * (a) $ 0$\\ * (b) $1$\... > iv. If $3 x^{3}-2 x^{2}+5$ is divided by $x+1$, then $x+1$ will be its:\\ * (a) divisor as well as... s a zero of the polynomial $x^{3}+5 x^{2}-4 x+k$, then the value of $k$ will be:\\ * (a) $-4$ *... </collapse> vi. If $x-b$ is the factor of $q(x)$, then $\mathrm{q}(\mathrm{b})$ is:\\ * (a) factor\
- Question 7, Exercise 1.1 @math-11-nbf:sol:unit01
- $11+12 i$. **Solution.** Suppose $$z=11+12i$$ Then \begin{align}|z|&= \sqrt{(11)^2+(12)^2}\\ &=\sqrt... +6 i)$. **Solution.** Suppose $z=(2+3i)ā(2+6i)$, then \begin{align}z&=2+3iā2ā6i\\ &=-3i \end{align} Now... i)$. **Solution.** Suppose $$z=(2-i)(6+3 i),$$ then \begin{align} |z|&=|(2-i)(6+3 i)|\\ &=|(2-i)| |(... **Solution.** Suppose $$z=\dfrac{3-2 i}{2+i},$$ then \begin{align} |z|&=\left|\dfrac{3-2 i}{2+i} \righ
- Question 20 and 21, Exercise 4.4 @math-11-nbf:sol:unit04
- $ and $a_5=48$. Assume $r$ be common difference, then by general formula for nth term, we have $$ a_n=a... $ and $a_5=48$. Assume $r$ be common difference, then by general formula for nth term, we have $$ a_n=a... = \pm 2. \end{align*} Thus, if $a_1=3$ and $r=2$, then \begin{align*} & a_2=a_1 r= (3)(2) = 6 \\ & a_3=a... (3)(2)^3=24. \end{align*} If $a_1=3$ and $r=-2$, then \begin{align*} & a_2=a_1 r= (3)(-2) = -6 \\ & a_3
- MCQs: Math 11 NBF
- 0,0)$</collapse> iv. If $z$ is a complex number then its mirror image is * (a) $|z|$ * (b) ... collapse> vi. If $z_{1}=3+2 i$ and $z_{2}=5+6 i$ then * (a) $z_{1}>z_{2}$ * (b) $z_{1}<z_{... %: argand</collapse> viii. If $\mathrm{z}=3+4 i$ then $\mathrm{z}^{-1}$ is * (a) $\left(\frac{1}{3... se> x. If $\left(\dfrac{1+i}{1-i}\right)^{n}=1$ then least positive value of $n$ is * (a) $1$
- Question 1, Exercise 1.3 @math-11-nbf:sol:unit01
- {2})=z+\dfrac{3}{2}$ is the factor of polynomial. Then by using synthetic division: \begin{align} \begin... 0 $$ So $z-(1)=z-1$ is the factor of polynomial. Then by using synthetic division: \begin{align} \begin... ign} So $z-(-3)=z+3$ is the factor of polynomial. Then by using synthetic division: Now, by synthetic di... nd{align} So $z-2$ is the factor of polynomial.\\ Then by using synthetic division: \begin{align} \begin
- Question 3, Exercise 1.3 @math-11-nbf:sol:unit01
- 1,\quad b = 6,\quad \text{and}\quad c = -48.$$ Then \begin{align} z& = \dfrac{{-6 \pm \sqrt{36-4(1)(... $ a = 1, \quad b = -\dfrac{1}{2}, \quad c = 17 $$ Then \begin{align} z &= \dfrac{{-\left(-\dfrac{1}{2}\r... $$ Where $$ a = 1, \quad b = -6, \quad c = 25 $$ Then \begin{align} z &= \dfrac{{-(-6) \pm \sqrt{{(-6)^... $$ Where $$ a = 1, \quad b = -9, \quad c = 11 $$ Then \begin{align} z &= \dfrac{{-(-9) \pm \sqrt{{(-9)^
- Question 1, Review Exercise @math-11-nbf:sol:unit01
- 0,0)$</collapse> iv. If $z$ is a complex number then its mirror image is * (a) $|z|$ * (b) ... collapse> vi. If $z_{1}=3+2 i$ and $z_{2}=5+6 i$ then * (a) $z_{1}>z_{2}$ * (b) $z_{1}<z_{... %: argand</collapse> viii. If $\mathrm{z}=3+4 i$ then $\mathrm{z}^{-1}$ is * (a) $\left(\frac{1}{3... pse> x. If $\left(\frac{1+i}{1-i}\right)^{n}=1$ then least positive value of $n$ is * (a) $1$
- Question 1, Review Exercise @math-11-nbf:sol:unit02
- is $m \times n$ and order of $B$ is $n \times p$ then order of $A B$ is:\\ * (a) $n \times p$\\ ... ii. If $A$ is a row matrix of order $1 \times n$ then order of $A^{t} A$ is:\\ * (a) $1 \times n$\... j}=M_{i j}$</collapse> iv. If $A$ is any matrix then $A$ and $A^{t}$ are always conformable for:\\ ... a square matrix of order $3 \times 3$ and $|A|=3$ then value of $|\operatorname{adj} A|$ is:\\ *
- Definitions: Mathematics 11 NBF
- plex Number:** If $z=x+i y$ is a complex number, then its magnitude, denoted by $|z|$, is defined as $|... ^{2}}$. **Real & Imaginary Parts:** If $z=x+iy$, then * $Re(z)=x$ and $Im(z)=y$. * $Re(z^{-1})= \df... )}{|z|^4}$. If $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$, then * \(\text{Re}\left(\frac{x_{1}+i y_{1}}{x_{2}+i
- Question 6, Exercise 1.1 @math-11-nbf:sol:unit01
- number $4-3 i$. **Solution.** Given: $z=4-3 i$, then $\bar{z}=4+3i$. ====Question 6(ii)==== Find the... ac{-1}{5}}\\ =&2+\sqrt{\dfrac{1}{5}}i,\end{align} Then $$\bar{z}=2-\sqrt{\dfrac{1}{5}}i$$ ====Question ... tion.** Given: $z=\dfrac{5 }{2}i-\dfrac{7}{8},$ then $\bar{z}=-\dfrac{5 }{2}i-\dfrac{7}{8}$. GOOD ==
- Question 9, Exercise 1.2 @math-11-nbf:sol:unit01
- = \sqrt{3^2 + (-2)^2} \\&= \sqrt{13}. \end{align} Then \[|z|^4=169.\] Using in above formulas \begin{al... _2^2\): \[x_2^2 - y_2^2 = 2^2 - 5^2 = -21.\] Then, compute \(x_1^2 - y_1^2\): \[x_1^2 - y_1^2 = 4... _2 + y_1 y_2 = 3 \cdot 2 + (-7) \cdot 5 = -29\] Then, compute \(x_2 y_1 - x_1 y_2\): \[x_2 y_1 - x_1