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- Question 1 and 2 Exercise 4.1 @math-11-kpk:sol:unit04
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... Given: $$a_n=\dfrac{n(n+1)}{2}$$ For first term, put $n=1,$ $$a_1=\dfrac{1(1+1)}{2}=1$$ For second term, put $n=2$. $$a_2=\dfrac{2(2+1)}{2}=3$$ For third term, put $n=3,$ $$a_3=\dfrac{3(3+1)}{2}=6$$ For forth ter
- Question 4 Exercise 4.5 @math-11-kpk:sol:unit04
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... 8}{1-0.1}=\dfrac{0.8}{0.9}=\dfrac{8}{9}$$\\ Hence putting $S_{\infty}$ in (i), we get $$0 . \overline{... ots \ldots \ldots . . . \text { (ii) }\end{align} Putting (ii) in (i), we get\\ $$1.63=1+\dfrac{7}{11}... s and given by $$S_{\infty}=\dfrac{a_1}{1-r}$$,\\ putting $a_1, r$ we get\\ $$S_{\infty}=\dfrac{0.15}{
- Question 1 and 2 Exercise 6.5 @math-11-kpk:sol:unit06
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... now by complementary events $$P(B)=1-P(\bar{B})$$ Putting \begin{align}P(\bar{B})&=\dfrac{5}{8}\\ P(B)... frac{3}{8}\end{align} Also $$P(\bar{A})=1-P(A)$$ Putting $P(A)=\dfrac{1}{2}$, we get \begin{align}P(\... now by complementary events $$P(B)=1-P(\bar{B})$$ Putting $P(\bar{B})=\dfrac{5}{8}$, we get $$P(B)=1-\
- Question 5 Exercise 7.2 @math-11-kpk:sol:unit07
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... }{x})^{8-r}(b x)^r$$ To get middle term $T_5$, we put $r=4$ \begin{align}T_5&=\dfrac{8 !}{(8-4) ! 4 !}... 9 !}{(9-r) ! r !}(3 x)^{9-r}(-\dfrac{x^2}{2})^r$$ Putting $r=4$ to get first middle term \begin{align}... x^{13}\\ & T_5=\dfrac{15309}{8} x^{13}\end{align} Putting $r=5$ to get the $2^{\text {nd }}$ middle te
- Question 1, Exercise 10.3 @math-11-kpk:sol:unit10
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... a =\cos (\alpha +\beta )-\cos (\alpha -\beta ).$$ Put $\alpha =6x$ and $\beta =x$ \begin{align}-\,2... a =\sin (\alpha +\beta )+\sin (\alpha -\beta )$$ Put $\alpha ={{55}^{\circ }}$ and $\beta ={{123}^... ta =\sin (\alpha +\beta )+\sin (\alpha -\beta )$$ Put $\alpha =\dfrac{A+B}{2}$ and $\beta =\dfrac{A
- Question 2, Exercise 10.3 @math-11-kpk:sol:unit10
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... )\cos \left( \dfrac{\alpha -\beta }{2} \right).$$ Put $\alpha ={{37}^{\circ }}$, $\beta ={{43}^{\circ ... t)\sin \left( \dfrac{\alpha -\beta }{2} \right)$$ Put $\alpha =36^\circ$, adn $\beta =82^\circ$ \begin... )\sin \left( \dfrac{\alpha -\beta }{2} \right).$$ Put $\alpha =\dfrac{P+Q}{2}$ and $\beta =\dfrac{P-Q}
- Question 8, Exercise 1.2 @math-11-kpk:sol:unit01
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... how that $\overline{z}=-z$ if and only if $z$ is pure imaginary. ====Solution==== Suppose that $z=a+... it in (1), we get $$z=0+bi=bi,$$ that is, $z$ is pure imaginary. Conversly, suppose that $z$ is pure imaginary, then its real part will be zero, that is
- Question 1, Exercise 1.3 @math-11-kpk:sol:unit01
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... &=\dfrac{11-11i}{11}\\ \implies w&=1-i\end{align} Put value of $w$ in (i). \begin{align}z&-4(1-i)=3i\\... =6i & -2\\ \end{array}$$ \[-w=6i-2\] \[w=2-6i\] Put value of $w$ in (i).\\ \begin{align}z&+(2-6i)=3i... hline 8z & 0 & =8\\ \end{array}\\ \implies z=1$$ Put value of $z$ in (i).\\ \begin{align} &3(1)+(2+i)
- Question 3 & 4, Exercise 1.3 @math-11-kpk:sol:unit01
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... tion==== Given: $$z^2+2z_1+2=0\quad \ldots (i)$$ Put the value of $z_1=-1+i$ in (i) \begin{align}L.H.... es $z_1=-1+i$ satisfied the given equation.\\ Now put $z_2=-1-i$ in (i) \begin{align} L.H.S&=(-1-i)^2+... ===Solution==== Given: $$z^2-2z+5=0 \ldots (i)$$ Put $z=1+2i$ in equaiton (i), we have \begin{align}L
- Question 2 & 3 Exercise 4.4 @math-11-kpk:sol:unit04
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... htarrow r^2&=9 \text { or } r= \pm 3 .\end{align} Putting this in (i), then\\ $$a_1(9)=27 \quad \text... \ \Rightarrow r&=-\dfrac{1}{\sqrt{2}}.\end{align} Putting in (i), then\\ \begin{align}\Rightarrow a_1(... that\\ \begin{align}a_n&=a_1 r^{n-1}, \\ \text { putting } n&=7, \text { and } a_1, r \text { then we
- Question 13 & 14 Exercise 4.5 @math-11-kpk:sol:unit04
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... ts and is given by $S_{\infty}=\dfrac{a_1}{1-r}$, putting $a_1, \quad r$, we get $$S_{\infty}=\dfrac{1}{1-\dfrac{x}{3}}=\dfrac{3}{3-x}$$ putting in (i), we have \begin{align}1+y&=\dfrac{3}{... htarrow S_7&=10,0.9921875\\ &=9.921875\end{align} Putting (ii) in (i), we get\\ \begin{align}S&=10+2(9
- Question 2 Exercise 7.2 @math-11-kpk:sol:unit07
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... r !}(2)^{7-r } a^r $$ For $4^{\text {th }}$ term, putting $r=3$ \begin{align} & T_{3+1}=\dfrac{7 !}{(7... r}(-\dfrac{3}{y})^r$$ For $8^{\text {th }}$ term, putting $r=7$ \begin{align}T_{7+1}&=\dfrac{10 !}{(10... t}=x^0$ $$\Rightarrow 21-3 r=0 \Rightarrow r=7 $$ Putting $r=7$ in the above, we get \begin{align}T_{7
- Question 3 Exercise 7.2 @math-11-kpk:sol:unit07
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... ow 18-3 r=0 \Rightarrow 3 r=18 \Rightarrow r=6 $$ Putting $r=6$ in the above $T_{r+1}$ \begin{align}T_... $ $$\Rightarrow 10-5 r=0 \quad \Rightarrow r=2 $$ Putting $r=2$ in the above, we get \begin{align} T_{... t}=x^0$ $$\Rightarrow 21-3 r=0 \Rightarrow r=7 $$ Putting $r=7$ in the above, we get \begin{align}T_{7
- Question 4 Exercise 7.2 @math-11-kpk:sol:unit07
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... row 40-r&=23\\ \Rightarrow r&=40-23=17\end{align} Putting $r=17$. in $T_{r+1}$ we get \begin{align}T_{... x^r}&=\dfrac{1}{x^4}\\ \Rightarrow 4=r\end{align} Putting $r=4$ in $T_{r+ 1}$ we get \begin{align}T_{4... ^3\\ \Rightarrow 9 -r=6 \text{or} r=3\end{align} Putting $r=3$ in the above, we get \begin{align}T_{3
- Question 2, Exercise 1.3 @math-11-kpk:sol:unit01
- unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB... orem, $(z-a)$ is a factor of $P(z)$ iff $P(a)=0$. Put $z=-2$ \begin{align} P(-2)&=(-2)^3+6(-2)+20\\ &... eorem $(z-a)$ is a factor of $()$ iff $P(a)=0$.\\ Put $z=2$\\ \begin{align}P(-2)&=(-2)^3+6(-2)+20\\ &