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- Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- lign}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Rightarrow \quad \sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alph... gn}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} ... -\sqrt{1+\frac{9}{16}}=-\sqrt{\frac{25}{16}\,}\\ \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now ... }{\sec \alpha }=\frac{1}{-\tfrac{5}{4}}$$ $$\Rightarrow \quad \cos\alpha=-\dfrac{4}{5}$$ Now \begin{align
- Question 5, Exercise 10.1 @math-11-kpk:sol:unit10
- lign}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Rightarrow \quad \sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alph... gn}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} ... -\sqrt{1+\frac{9}{16}}=-\sqrt{\frac{25}{16}\,}\\ \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now ... }{\sec \alpha }=\frac{1}{-\tfrac{5}{4}}$$ $$\Rightarrow \quad \cos\alpha=-\dfrac{4}{5}$$ Now \begin{align
- Question 1 Exercise 6.3 @math-11-kpk:sol:unit06
- on==== We are given: \begin{align}&^n C_2=36\\ & \Rightarrow \dfrac{n !}{(n-2) ! 2 !}=36 \\ & \Rightarrow \dfrac{n(n-1)(n-2) !}{(n-2) ! \cdot 2}=36 \\ & \Rightarrow n(n-1)=72 \\ & \Rightarrow n^2-n-72=0 \\ & \Rightarrow n^2-9 n+8 n-72=0\\ & \Rightarrow n(n-9)+8(n-9)=0 \\ &
- Question 2 and 3 Exercise 3.3 @math-11-kpk:sol:unit03
- t{j}-5 \hat{k})+(2 \hat{i}+\hat{j}-7 \hat{k}) \\ \Rightarrow &=4 \hat{i}+3 \hat{j}-12 \hat{k}\\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{(4)^2+(3)^2+(-12)^2} \\ \Rightarrow &=\sqrt{16+9+144} \\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{169}=13\end{align} Now let say $\hat{c}$ be the unit
- Question 2 Exercise 3.4 @math-11-kpk:sol:unit03
- =(12-12) \hat{i}-(-6+6) \hat{j}+(4-4) \hat{k} \\ \Rightarrow \vec{a} \times \vec{b}&=0 . \\ & \Rightarrow \vec{a} \| \vec{b} .\end{align} Second Way \begin{align}\vec{a}... \hat{k}) \cdot(2 \hat{i}-4 \hat{j}+6 \hat{k}) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=-1(2)+2(-4)-3(6) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=-28 .\end{align} Als
- Question 1 and 2 Exercise 6.2 @math-11-kpk:sol:unit06
- We are given: \begin{align}^n P_5&=56(^n P_3) \\ \Rightarrow \dfrac{n !}{(n-5) !}&=56 \dfrac{n !}{(n-3) !} \\ \Rightarrow \dfrac{1}{(n-5) !}&=\dfrac{56}{(n-3) !} \\ \Rightarrow \dfrac{1}{(n-5) !}&=\dfrac{56}{(n-3)(n-4)(n-5) !} \\ \Rightarrow(n-3)(n-4)&=56 \\ \Rightarrow n^2-7 n+12-56&=0 \\
- Question 2 Review Exercise 6 @math-11-kpk:sol:unit06
- begin{align} { }^{2 n} C_r&={ }^{2 n} C_{r+2} \\ \Rightarrow \dfrac{(2 n) !}{(2 n-r) ! r !}&=\dfrac{(2 n) !}{(... ing both sides by $(2 n)$ ! we get \begin{align} \Rightarrow \dfrac{1}{(2 n-r) ! r !}&=\dfrac{1}{(2 n-r-2) !(r+2) !} \\ \Rightarrow \dfrac{1}{(2 n-r)(2 n-r-1)(2 n-r-2) ! r !}&=\dfrac{1}{(2 n-r-2) !(r+2) !} \\ \Rightarrow \dfrac{1}{(2 n-r)(2 n-r-1) r !}& =\dfrac{1}{(r+2)
- Question 9 & 10 Exercise 4.5 @math-11-kpk:sol:unit04
- c{a_1(r^6-1)}{r-1}&=9 \dfrac{a_1(r^3-1)}{r-1} \\ \Rightarrow r^6-1-9(r^3-1) \\ \Rightarrow r^6-1&=9 r^3-9 \\ \Rightarrow r^6&=9 r^3-8 \\ \Rightarrow r^6-9 r^3+8&=0, \\ \Rightarrow r^6-r^3-8 r^3+8&=0 \\ \Rightarrow r^3(r^3-1)-8(r^3
- Question 3, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- \cos $ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\sin }^{2}}u}\\ &=\s... qrt{1-\dfrac{9}{25}}\,\,=\sqrt{\dfrac{16}{25}}\\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Also $\... $\cos $ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{\sin }^{2}}v}\\ &=\s... qrt{1-\dfrac{16}{25}}\,\,=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begi
- Question 2 Exercise 6.3 @math-11-kpk:sol:unit06
- gn} & { }^n C_4=\dfrac{n !}{(n-4) ! 4 !}=35 \\ & \Rightarrow \dfrac{n(n-1)(n-2)(n-3)(n-4) !}{(n-4) !}=35 \times 24=840 \\ & \Rightarrow n(n-1)(n-2)(n-3)=840 \\ & \Rightarrow n(n-3)(n-1)(n-2)=840 \\ & \Rightarrow(n^2-3 n)(n^2-3 n+2)=840 \end{align} Let $y=n^2-3 n$ then the above last
- Question 3, Exercise 10.1 @math-11-kpk:sol:unit10
- \cos $ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\sin }^{2}}u}\\ &=\s... qrt{1-\dfrac{9}{25}}\,\,=\sqrt{\dfrac{16}{25}}\\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Also $\... $\cos $ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{\sin }^{2}}v}\\ &=\s... qrt{1-\dfrac{16}{25}}\,\,=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begi
- Question 1 Exercise 4.5 @math-11-kpk:sol:unit04
- ^{n-1} \text { or }(2)^{n-1}=\dfrac{3.2^9}{3} \\ \Rightarrow(2)^{n-1}&=2^9 \\ \Rightarrow n-1&=9 \text { or } n=10 \\ \text {. Now }\quad S_n&=\dfrac{a_1(r^n-1)}{r-1},\e... \begin{align}S_{10}&=\dfrac{3[2^{10}-1]}{2-1} \\ \Rightarrow \quad S_{10}&=3(2^{10}-1)\end{align}\\ is the req... \ (\dfrac{1}{2})^{n-1}&=\dfrac{1}{8 \times 16}\\ \Rightarrow(\dfrac{1}{2})^{n-1}&=(\dfrac{1}{2})^7\\ \Rightarr
- Question 2 Exercise 4.5 @math-11-kpk:sol:unit04
- n-1}$, therefore\\ \begin{align}64&=(-2)^{n-1}\\ \Rightarrow(-2)^{n-1}&=(-2)^6 \\ \Rightarrow n-1&=6 \\ \Rightarrow n&=7\\ S_7&=\dfrac{a_1[r^{\prime \prime}-1]}{r-1}\\ \text{then}\\ S_7&=\dfrac{1[(-2)^7-1]}{-2-1}\\ \Rightarrow S_7&=\dfrac{-128-1}{-3}\\ s_7&=\dfrac{129}{3}\end
- Question 5 & 6 Exercise 4.3 @math-11-kpk:sol:unit04
- , 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{... lign}(a-3 d)^2+(a-d)^2+(a+d)^2+(a+2 d)^2&=120 \\ \Rightarrow a^2-6 a d+9 d^2+a^2-2 a d+d^2+a^2+2 a d+d^2+a^2+6 a d+9 d^2&=120 \\ \Rightarrow 4 a^2+20 d^2&=120 \\ \Rightarrow 20 d^2&=120-4 \c
- Question 7 & 8 Exercise 4.5 @math-11-kpk:sol:unit04
- rac{1}{2}[1-(\dfrac{1}{2})^n]}{1-\dfrac{1}{2}}\\ \Rightarrow S_n&=\dfrac{\dfrac{1}{2}[1-\dfrac{1}{2^n}]}{\dfrac{1}{2}} \\ \Rightarrow S_n&=1-\dfrac{1}{2^n}\end{align}\\ is the require... gin{align}\dfrac{a}{r} \cdot a \cdot a r&=1728\\ \Rightarrow a^3&=1728\\ \Rightarrow \quad a&=12,\\ \text{putting}\text{in} (1)\\ \dfrac{12}{r}+12+12 r&=38\\ \Rightarrow