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- Question 6 Exercise 4.1 @math-11-kpk:sol:unit04
- xt{ where } r=0,1,2,3,\ldots.$$ For $r=0$ \begin{align}&P_{0+1}=\dfrac{5-0}{0+1} P_0\\ \implies &P_1=5.\end{align} For $r=1$ \begin{align}&P_{1+1}=\dfrac{5-1}{1+1} P_1\\ \implies &P_2=2\cdot 5=10.\end{align} For $r=2$ \begin{align}&P_{2+1}=\dfrac{5-2}{2+1}
- Question 5, Exercise 10.1 @math-11-kpk:sol:unit10
- nt, therefor it lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Righ... sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alpha}.\end{align} Since terminal arm of $\alpha$ is in the 3rd quadrant, value of $\sec$ is –ive \begin{align}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Ri... \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now $$\cos\alpha =\frac{1}{\sec \alpha }=\frac{
- Question 13, Exercise 10.1 @math-11-kpk:sol:unit10
- nd $3=r\sin\varphi$. Squaring and adding \begin{align} &r^2\cos^2 \varphi+r^2\sin^2\varphi = 4^2+3^2\\ ... = 16+9\\ \implies &r^2 = 25\\ \implies &r=5 \end{align} Also \begin{align}r\cos\varphi = 4 \implies 5\cos\varphi = 4 \implies \cos\varphi =\dfrac{4}{5}\end{align} and \begin{align}r\sin\varphi = 3 \implies 5\sin
- Question 7, Exercise 10.2 @math-11-kpk:sol:unit10
- dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\sin }^{4}}\theta \... dentity)}\\ &=\dfrac{1}{\sec 2\theta }=R.H.S.\end{align} =====Question 7(ii)===== Prove the identity $\t... \dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+co\operatorname{t}... (By \,using\, double\, angle\, identity)\end{align} =====Question 7(iii)===== Prove the identity $\
- Question 3, Exercise 10.1 @math-11-kpk:sol:unit10
- t and $\cos $ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\sin }^{2... \\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Also $\cos v=\sqrt{1-{{\sin }^{2}}v}$ As $v$ lie... nt and $\cos $ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{\sin }^{2... 25}}\\ \Rightarrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begin{align}\cos \left( u+v \right)&=\cos
- Question 1, Exercise 1.3 @math-11-kpk:sol:unit01
- linear equation with complex coefficient. \begin{align}&z-4w=3i\\ &2z+3w=11-5i\end{align} ====Solution==== Given that \begin{align}z-4w&=3i …(i)\\ 2z+3w&=11-5i …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{align}2z-8w&=6i …(i
- Question 9 Exercise 6.3 @math-11-kpk:sol:unit06
- actly contain four men and four women are: \begin{align}{ }^7 C_4 \cdot{ }^6 C_4&=\dfrac{7 !}{(7-4) ! 4 !} \cdot \dfrac{6 !}{(6-4)}\\\ &= 525\end{align} =====Question 9(ii)===== An $8$-persons committ... s case the total number of committees are: \begin{align}{ }^7 C_2 \cdot{ }^6 C_6&=\dfrac{7 !}{(7-2) ! 2 !} \cdot \dfrac{6 !}{(6-6) ! 6 !}\\ &=21\end{align} If committee contains $3$ men then it will cont
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- prove that $$(AB)C=A(BC)$$ First, we take \begin{align} AB&=\begin{bmatrix}x & y & z\end{bmatrix}\begin{... {{y}^{2}}+2fyz+c{{z}^{2}} \right] \ldots (1) \end{align} Now we take \begin{align}BC&=\begin{bmatrix}a & h & g\\h & b & f\\g & f & c \end{bmatrix}\begin{bmatrix... ix}ax+hy+gz\\hx+by+fz\\gx+fy+cz \end{bmatrix}\end{align} \begin{align} R.H.S &= A(BC)\\ &=\begin{bmatrix}
- Question 13 Exercise 6.2 @math-11-kpk:sol:unit06
- are $L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutations are} &=\left(... ac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\ &=37,800 \end{align} Begin with $\mathrm{E}$ If we have to pick the ... $ are $L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{Number of permulations are} &=\left(\begin{... rac{9 !}{3 ! \cdot 2 ! \cdot 2 !}\\ &=15,120 \end{align} =====Question 13(ii)===== Find the number of pe
- Question 1, Exercise 10.1 @math-11-kpk:sol:unit10
- n {{22}^{\circ }}$ ==== Solution ==== As \begin{align} \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta, \end{align} Therefore \begin{align} \sin {{37}^{\circ }}\cos {{22}^{\circ }}+\cos {{37}^{\circ }}\sin {{22}^{\cir... t( 37+22 \right) \\ & =\sin {{59}^{\circ }}. \end{align} ===== Question 1(ii)===== Write as a trigonom
- Question 7 Exercise 6.4 @math-11-kpk:sol:unit06
- The sample space rolling a pair of dice is \begin{align}S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin... 3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \t... imes 6=36$$ doublet of even numbers. Let \begin{align}A&=\{(2,2),(4,4),(6,6)\}\\ n(A)&=3\end{align} Hence the possibility of getting doublet of an even number
- Question 3 and 4 Exercise 4.1 @math-11-kpk:sol:unit04
- ce to pick the pattern of the sequence as: \begin{align} &(-1)^2 \cdot 2 \cdot 1, (-1)^3 \cdot 2 \cdot 2,... \cdot 3, (-1)^{4+1} \cdot 2 \cdot 4, \ldots \end{align} Hence the general term of the sequence is $(-1)^... ce to pick the pattern of the sequence as: \begin{align}(-1)^2,(-1)^3,(-1)^4,(-1)^5, \ldots, (-1)^{n+1}, \ldots \end{align} Hence the general term of the sequence is $(-1)^
- Question 5 & 6 Exercise 4.3 @math-11-kpk:sol:unit04
- $Condition-1$\\ Their sum is $20$ , thus\\ \begin{align}a-3 d+a-d+a+d+a+3 d&=20 \\ \Rightarrow 4 a&=20\\ \Rightarrow a&=5 .\end{align} $Condition-2$\\ The sum of their square is $120$, therefore\\ \begin{align}(a-3 d)^2+(a-d)^2+(a+d)^2+(a+2 d)^2&=120 \\ \Righ... 0 \\ \Rightarrow d^2&=1 \text { or } d= \pm 1\end{align} When $a=5$ and $d=1$ then the numbers are\\ \beg
- Question 1 Exercise 5.2 @math-11-kpk:sol:unit05
- +3.2^3+4.2^4+\ldots$. ====Solution==== Let \begin{align} & S_n=1.2+2.2^2+3 \cdot 2^3+4 \cdot 2^4+\ldots +... 2.2^3+3.2^4+4.2^5+\ldots +n \cdot 2^n....(ii)\end{align} Suburacting the (ii) from (i), we get \begin{align} (1-2) S_n&=1 \cdot 2+(2-1) 2^2+(3-2) 2^2+(4-3) 2^3+... cdot 2^{n+1}-2^{n+1} \\ S_n& =2+(n-1) 2^{n+1}\end{align} =====Question 1(ii)===== Sum up to $n$ terms th
- Question 1 Exercise 5.3 @math-11-kpk:sol:unit05
- $A+B=0 \text{and} A=1$$ Putting $A=1$,then \begin{align}1+B&=0\\ B&=-1\end{align} hence \begin{align} & \dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1} \\ & \Rightarrow T_n=\dfrac{1}{n}-\dfrac{1}{n+1}\end{align} Taking summation of the both sides of the above