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- Question 6, Exercise 1.3 @math-11-kpk:sol:unit01
- =0$$ $$(z^2+z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{-1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac... 2}$$ $$(z^2-z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac{... t{3}i}{2}$$ The value of $z$ from both equations, we have $$z=\pm \dfrac{1}{2}\pm \dfrac{\sqrt{3}}{2}i... $z^2-2z+4=0$$ According to the quadratic formula, we have $a=1$, $b=-2$ and $c=4$ Thus, we have \begin
- Question 2 Exercise 4.3 @math-11-kpk:sol:unit04
- OOD ====Solution==== Given: $a_1=2, n=17, d=3$ \\ We need to find $a_{17}$ and $S_{17}$. As we know $$a_{n}=a_1+(n-1)d.$$ Thus $$a_{17}=2+(17-1)(3)=50.$$ ... ution==== Given: $a_1=-40$ and $S_{21}=210$.\\ So we have $n=21$ and we have to find $a_{21}$ and $d$. As \begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \
- Question 1 and 2 Exercise 6.5 @math-11-kpk:sol:unit06
- ac{1}{2}$.\\ Find $P(A \cap B)$. ====Solution==== We know by addition law of probability \begin{align}... ign} Substituting $P(A), P(B)$ and $P(A \cup B)$, we get $$P(A \cap B)=\dfrac{2}{5}+\dfrac{2}{5}-\dfra... dfrac{3}{4}$. Find $P(A \cap B)$ ====Solution==== We are given: $$P(A)=\dfrac{1}{2}, P(\bar{B})=\dfrac{5}{8}, P(A \cup B)=\dfrac{3}{4}$$ We know by complementary events $$P(B)=1-P(\bar{B})$
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- }$ and $C=\begin{bmatrix}x\\y\\z\end{bmatrix}$.\\ We have to prove that $$(AB)C=A(BC)$$ First, we take \begin{align} AB&=\begin{bmatrix}x & y & z\end{bmat... fyz+c{{z}^{2}} \right] \ldots (1) \end{align} Now we take \begin{align}BC&=\begin{bmatrix}a & h & g\\h... } \right] \ldots (2)\end{align} From (1) and (2), we have $$(AB)C=A(BC).$$ =====Question 3(ii)(a)===
- Question 1 Exercise 5.3 @math-11-kpk:sol:unit05
- onstants on the both sides of the above equation, we get $$A+B=0 \text{and} A=1$$ Putting $A=1$,then \... 1)}$$ Multiplying both sides by $(2 n-1)(2 n+1)$. we get, \begin{align} & \mathrm{I}=A(2 n+1)+B(2 n-1)... Solving the above two equations for $A$ and $B$, we get \begin{align}A&=\dfrac{1}{2}\\ \text{and} B&=... n+2}$$ Multiplying both sides by $(3 n-1)(3 n+2)$ we get, \begin{align} 1&=A(3 n+2)+B(3 n-1) \\ \Right
- Question 12 Exercise 7.3 @math-11-kpk:sol:unit07
- \ldots$ then show that $4 y^2+4 y-1=0$. Solution: We are given $$ 2 y=\frac{1}{2^2}+\frac{1.3}{2 !} \c... $$ Adding 1 to both sides of the above equation, we get $S=2 y+1=1+\frac{1}{2^2}+\frac{1.3}{2 !} \cdo... dots \end{aligned} $$ Comparing both the series, we have $n x=\frac{1}{2^2}=\frac{1}{4}.... (1)$ and ... \cdot \frac{1}{2^4} $$ Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$ Dividing Eq.(2) by Eq
- Question 3 & 4, Exercise 3.2 @math-11-kpk:sol:unit03
- Solution==== Given $$\vec{r}=p\vec{a}+q\vec{b}.$$ We put the values of $\vec{r},\vec{a}$ and $\vec{b}$ in the given equation. We get $$\hat{i}-9\hat{j}=p(\hat{i}+2\hat{j})+q(5\ha... paring the coeffients of $\hat{i}$ and $\hat{j}$, 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 \\ \mathop
- Question 9 Exercise 3.4 @math-11-kpk:sol:unit03
- -2 \hat{i}+3 \hat{j}+4 \hat{k}$. ====Solution==== We are give the diagonal as shown in figure, instead... t{j}+2 \hat{k}\end{align} From $\triangle A E B$, we have\\ \begin{align}\vec{c}&=\overrightarrow{A E}... dots \ldots(1)\end{align} From $\triangle A E D$. we have\\ \begin{align}\vec{d}&=\overrightarrow{A E}... ext {. }....(2) \end{align} By using (1) and (2), we have,\\ \begin{align}\vec{c} \times \vec{d}&=\lef
- Question 1 Exercise 5.1 @math-11-kpk:sol:unit05
- 5^2+7^2+\ldots$ up to $n$ terms. ====Solution==== We see that each term of the given series is square ... ummation of the both sides of the above equation, we get \begin{align}& \sum_{j=1}^n T_j=\sum_{j=1}^n(... $n$ terms. ====Solution==== In the given series, we see that $T_1=1^2, T_2=1^2+2^2$, $T_3=1^2+2^2+3^2$ and so on we get \begin{align}& T_j=1^2+2^2+3^2+\ldots+j^2 \\
- Question 7 Exercise 6.4 @math-11-kpk:sol:unit06
- number less than 6$\}$, then from sample space, we see that $n(B)=10$. Thus the probability of gett... $ a sum mure than 7$\}$, then from sample space, we see that $n(C)=5$. Thus the probability of getti... sum greater than 10$\}$, then from sample space, we get $n(D)=3$. Thus the probability of getting nu... {a$ sum at least 10$\}$, then from sample space, we see that $n(E)=6$. Thus the probability of getti
- Question 10 Exercise 7.3 @math-11-kpk:sol:unit07
- dots \end{aligned} $$ Comparing both the series, we have $n x=-\frac{1}{4}$ (I) and $\frac{n(n-1)}{2 ... !} \cdot \frac{1}{2^4}$ Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$ Dividing Eq.(2) by Eq.(3), we get $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{3}... {aligned} $$ Putting $n=-\frac{1}{2}$ in Eq.(1), we get $$ \begin{aligned} & -\frac{1}{2} x=-\frac{1}
- Question 12 & 13, Exercise 3.3 @math-11-kpk:sol:unit03
- in a semicircle is right angle. ====Solution==== We are considering a triangle inside a semicircle as shown. We have to show $\overrightarrow{B A} \cdot \overrightarrow{A C}=0$. We see in figure that: $|\vec{a}|=\vec{b}|=| \vec{c}... ut opposile in direction. From $\triangle A B O$, we have \begin{align}\overrightarrow{O B}+\overright
- Question 4 Exercise 4.5 @math-11-kpk:sol:unit04
- mmon fraction $0 . \overline{8}$ ====Solution==== We can write $$0 . \overline{8}=0.888888 \ldots$$\\ ... geometric series with $$a_1=0.8, \quad r=0.1$$\\ We can find the infinite sum as:\\ $$S_{\infty}=\dfr... frac{8}{9}$$\\ Hence putting $S_{\infty}$ in (i), we get $$0 . \overline{8}=\dfrac{8}{9}$$.\\ =====Qu... . \text { (ii) }\end{align} Putting (ii) in (i), we get\\ $$1.63=1+\dfrac{7}{11}=\dfrac{18}{11} \text
- Question 10 Exercise 7.2 @math-11-kpk:sol:unit07
- f even binomial cosficient $s=2^{n-1}$. Solution: We know that $$ \left.(1+x)^n=\left(\begin{array}{l}... ^n \cdot $$ Putting $x=1$ in the above equation, we have $(1 \div 1)^n=\left(\begin{array}{l}n \\ 0\e... ows that the sum of the :nefficiens is $?^n$. Now we know that $$ \begin{aligned} & (1+x)^n=\left(\beg... \end{array}\right) 1^n \\ & \end{aligned} $$ If we put $x=-1$ in the above eyuation, we get $$ \begi
- Question 1, Exercise 1.3 @math-11-kpk:sol:unit01
- z+3w&=11-5i …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{align}2z-8w&=6i …(iii)\end{align} Subtract (iii) from (ii), we get\\ \[\begin{array}{cccc} 2z&-8w&=6i \\ \mat... \ 2z+3w&=2 …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{align}2z+2w&=6i …(iii)\end{align} Subtract (iii) from (ii), we get\\ $$\begin{array}{ccc} 2z & +2w & =6i\\ \ma