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- Question 10, Exercise 1.2
- Question 9, Exercise 1.4
- Question 10, Exercise 2.2
- Question 19 and 20, Exercise 4.1
- Question 16 and 17, Exercise 4.2
- Question 20, 21 and 22, Exercise 4.3
- Question 20 and 21, Exercise 4.4
- Question 19 and 20, Exercise 4.7
- Question 11, Exercise 8.1
- Question 8(xiii, xiv & xv) Exercise 8.2
- Question 9, Exercise 9.1
- Question 10(xi-xv), Review Exercise
Fulltext results:
- Question 4, Exercise 1.3 @math-11-nbf:sol:unit01
- i}\\ &=\dfrac{9-5-5i-9i}{81+25}\\ &=\dfrac{4-14i}{106}\\ &=\dfrac{2}{53}-\dfrac{7}{53}i\end{align} Put... =\dfrac{1}{53}\dfrac{155+145i}{2}\\ &=\dfrac{155}{106}+\dfrac{145}{106}i\end{align} Thus, we have $$z=\dfrac{155}{106}+\dfrac{145}{106}i, \omega=\dfrac{2}{53}-\dfrac{7}{53
- Question 29 and 30, Exercise 4.7 @math-11-nbf:sol:unit04
- nd sum to infinity of the series: $$1+4 x+7 x^{2}+10 x^{3}+\ldots$$ ** Solution. ** The given arithmetic-geometric series is:\\ \[ 1 + 4x + 7x^2 + 10x^3 + \ldots \] It can be rewritten as:\\ \[ 1 \times 1 + 4 \times x + 7 \times x^2 + 10 \times x^3 + \ldots \] The numbers \(1, 4, 7, 10, \ldots\) are in AP with \(a = 1\) and \(d = 4 - 1
- Question 10, Exercise 1.2 @math-11-nbf:sol:unit01
- ====== Question 10, Exercise 1.2 ====== Solutions of Question 10 of Exercise 1.2 of Unit 01: Complex Numbers. This is un... extbook Board, Islamabad, Pakistan. ====Question 10(i)==== For $z_{1}=-3+2 i$, verify: $$\left|z_{1}\... ht| = \sqrt{13}.$$ As required. GOOD ====Question 10(ii)==== For $z_{1}=-3+2 i$ and $z_{2}=1-3 i$ veri
- Question 5 and 6, Exercise 4.1 @math-11-nbf:sol:unit04
- he sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$.$a_{n}=n^{2}-2 n$ ** Solution. ** Given $$a_n = n^2 - ... 16 - 8 = 8\\ \end{align*} Now \begin{align*} a_{10} &= (10)^2 - 2(10) = 100 - 20 = 80\\ a_{15} &= (15)^2 - 2(15) = 225 - 30 = 195 \end{align*} So, $a_1
- Question 1(i, ii, iii & iv) Exercise 8.3 @math-11-nbf:sol:unit08
- change the sum or difference: $$4 \sin 16x \cos 10x $$ ** Solution. ** \begin{align*} &4 \sin 16x \cos 10x \\ & = 2 (2\sin 16x \cos 10x) \\ &= 2[\sin(16x+10x)+\sin(16x-10x)]\\ &= 2[\sin (26x)+\sin(6x)] \end{align*} GOOD =====Question 1(
- Question 10, Exercise 1.4 @math-11-nbf:sol:unit01
- ====== Question 10, Exercise 1.4 ====== Solutions of Question 10 of Exercise 1.4 of Unit 01: Complex Numbers. This is un... tbook Board, Islamabad, Pakistan. =====Question 10(i)===== Find the impedance $Z$ for the following values: $E=(-50+100 i)$ volts, $I=(-6-2 i)$ amps. ** Solution. **
- Question 4, Exercise 2.6 @math-11-nbf:sol:unit02
- {1}-4 x_{2}+3 x_{3}=7$\\ $4 x_{1}+2 x_{2}-5 x_{3}=10$\\ ** Solution. ** \begin{align*} 2x_1 - x_2 -... 3x_1 - 4x_2 + 3x_3 &= 7, \\ 4x_1 + 2x_2 - 5x_3 &= 10, \end{align*} The associative augment matrix: \b... & : & 2 \\ 3 & -4 & 3 & : & 7 \\ 4 & 2 & -5 & : & 10 \end{bmatrix}\\ &\sim \text{R}\begin{bmatrix} 1 &... \\ 3 & -4 & 3 & \vert & 7 \\ 4 & 2 & -5 & \vert & 10 \end{bmatrix}\quad \dfrac{1}{2}\\ &\sim \text{R}\
- Question 8, Exercise 1.2 @math-11-nbf:sol:unit01
- OD ====Question 8(iii)==== Write $|z-4 i|+|z+4 i|=10$ in terms of $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$|z-4i| + |z+4i| = 10.$$ Put $z = x + iy$, we have \begin{align} & |(x + iy) - 4i| + |(x + iy) + 4i| = 10 \\ \implies & |x + i(y - 4)| + |x + i(y + 4)| = 10 \\ \implies & \sqrt{x^2 + (y - 4)^2} + \sqrt{x^2 +
- Question 6, Exercise 2.6 @math-11-nbf:sol:unit02
- \right|\\ &=5(4 - 6) - 3(8 - 3) + 1(4 - 1)\\ &= -10 - 15 + 3\\ &=-22 \end{align*} This system is cons... 3 & 1 \\ 2 & 4 \end{array} \right| = -(12 - 2) = -10\\ A_{22} &= (-1)^{2+2} \left| \begin{array}{cc} 5... array}{cc} 5 & 3 \\ 1 & 2 \end{array} \right| = -(10 - 3) = -7\\ A_{31} &= (-1)^{3+1} \left| \begin{ar... nd{bmatrix}\\ A &=\begin{bmatrix} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -18 & -1 \end{bmatrix}\\ \text{a
- Question 1 and 2, Exercise 4.1 @math-11-nbf:sol:unit04
- he sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$. $$a_{n}=3 n+1$$ ** Solution. ** Given $$a_{n}=3 n+1$... 3(2) + 1 = 6 + 1 = 7\\ a_3 &= 3(3) + 1 = 9 + 1 = 10\\ a_4 &= 3(4) + 1 = 12 + 1 = 13\\ \end{align*} Now \begin{align*} a_{10} &= 3(10) + 1 = 30 + 1 = 31\\ a_{15} &= 3(15) + 1
- Question 5, Exercise 2.6 @math-11-nbf:sol:unit02
- {1}-2 x_{2}+3 x_{3}=1$\\ $3 x_{1}-7 x_{2}+4 x_{3}=10$\\ ** Solution. ** The above system may be writt... end{bmatrix}, \quad B = \begin{bmatrix} 8 \\ 1 \\ 10 \end{bmatrix}\\ |A| &= \begin{vmatrix} 1 & 1 & 2 ... \frac{ \begin{vmatrix} 8 & 1 & 2 \\ 1 & -2 & 3 \\ 10 & -7 & 4 \end{vmatrix}}{52}\\ &=\frac{8(-8 + 21) - 1(4 - 30) + 2(-7 + 20)}{52}\\ &=\frac{104 + 26 + 26}{52}\\ & = \frac{156}{52} = 3 \end{ali
- Question 9 and 10, Exercise 4.1 @math-11-nbf:sol:unit04
- ====== Question 9 and 10, Exercise 4.1 ====== Solutions of Question 9 and 10 of Exercise 4.1 of Unit 04: Sequence and Series. ... he sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term, $a_{15}$: $a_{n}=(-1)^{n}(n+3)$ ** Solution. ** Do yourself.
- Question 1 and 2, Exercise 4.7 @math-11-nbf:sol:unit04
- frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10}\\ &= \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10}\\ &= \frac{30}{60} + \frac{15}{60} + \frac{10}{60} + \frac{7.5}{60} + \frac{6}{60}\\ &= \frac{68.5}{... }{9} + \frac{1}{11} + \frac{1}{13}\\ &= \frac{35}{105} + \frac{21}{105} + \frac{15}{105} + \frac{11.67
- Question 3 and 4, Exercise 4.1 @math-11-nbf:sol:unit04
- he sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{n}=\frac{n}{n+1}$ ** Solution. ** Given $$a_n = \frac{n}{n+... frac{4}{5}\\ \end{align*} Now \begin{align*} a_{10} &= \frac{10}{10+1} = \frac{10}{11}\\ a_{15} &= \frac{15}{15+1} = \frac{15}{16} \end{align*} So, $a_
- Question 1 and 2, Exercise 4.3 @math-11-nbf:sol:unit04
- =====Question 1===== Find the sum of series: $4+7+10+13+16+19+22+25$ ** Solution. ** Given: $4+7+10+13+16+19+22+25$. As the given series is arithmetic s... ind the sum of series. $a_{1}=2$, $a_{n}=200$, $n=100$ FIXME Statement is logically incorrenct. Find... ithematic series with: $a_{1}=2$, $a_{n}=200$, $n=100$ ** Solution. ** Given $a_{1}=2$, $a_{n}=200$,