Maths NCERT Exemplar Solutions Class 12th Chapter Eight

${\left(3x^3-2x^2+\frac{5}{x^5}\right)}^{10}$

General term $10!\frac{{\left(3x^3\right)}^{\alpha }\cdot {\left(-2^x\right)}^{\beta }\cdot {\left(5x^{-5}\right)}^{\gamma }}{\alpha !\beta !\gamma !}$

$=10!\frac{3^{\alpha }\cdot {\left(-2\right)}^{\beta }\cdot 5^{\gamma }\cdot x^{3\alpha +2\beta -5\gamma }}{\alpha !\beta !\gamma !}$

Now for constant term 3 + 2 $-5\gamma =0$ ………….(i)

$\alpha +\beta +\gamma =10$ …………(ii)

From equation (i) & (ii)

$3\alpha +2\left(10-\alpha -\gamma \right)=5\gamma$

$\alpha +20=7\gamma$

= 1, $\gamma$ = 3, = 6

Constant term $10!\cdot \frac{3^1\cdot {\left(-2\right)}^6\cdot 5^3}{3!6!}=2^9\cdot 3^2\cdot 5^4\cdot 7^1$

Area of $\Delta ABC$

= $\frac{1}{2}AB\cdot BC$

$=\frac{1}{2}\cdot \sqrt[]{2}\cdot 1$

$=\frac{1}{\sqrt[]{2}}$

now required area

$={\displaystyle \underset{0}{\overset{4}{\int }}\left(2\sqrt[]{2}-\sqrt[]{2}x\right)dx-\frac{1}{\sqrt[]{2}}}$

$=\frac{32\sqrt[]{2}}{3}=8\sqrt[]{2}-\frac{1}{\sqrt[]{2}}$

$=\frac{13\sqrt[]{2}}{6}$

 ${\displaystyle \sum _{k=1}^{10}\frac{k}{k^4+k^2+1}}$

$=\frac{1}{2}{\displaystyle \sum _{k=1}^{10}\left[\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right]}$

$=\frac{1}{2}\left[1-\frac{1}{111}\right]=\frac{110}{222}=\frac{55}{111}=\frac{m}{n}$

$\therefore m+n=166$

 $Area=2{\displaystyle \underset{0}{\overset{\sqrt[]{2}}{\int }}\left(1-\left|x^2-1\right|\right)dx=2\left[{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-\left(1-x^2\right)\right)dx+{\displaystyle \underset{1}{\overset{\sqrt[]{2}}{\int }}\left(2-x^2\right)dx}}\right]}=\frac{8}{3}\left(\sqrt[]{2}-1\right)$

Kindly consider the following figure

Required area is

= ${\displaystyle \underset{e-e^2}{\overset{0}{\int }}\mathcal{l}n\left(x+e^2\right)-1dx+{\displaystyle \underset{0}{\overset{\mathcal{l}n2}{\int }}2e^{-x}-1dx}}$

$=1+e-\mathcal{l}n2$

${\sum }_{r=0}^{20}?{}^{50-r}{\mathrm{C}}_6={}^{50}{\mathrm{C}}_6+{}^{49}{\mathrm{C}}_6+{}^{48}{\mathrm{C}}_6+\dots .+{}^{30}{\mathrm{C}}_6$

$\begin{array}{rr}& ={}^{50}{\mathrm{C}}_6+{}^{49}{\mathrm{C}}_6+{}^{48}{\mathrm{C}}_6+\dots .+\left({}^{30}{\mathrm{C}}_6+{}^{30}{\mathrm{C}}_7\right)-{}^{30}{\mathrm{C}}_7\\ & ={}^{50}{\mathrm{C}}_6+{}^{49}{\mathrm{C}}_6+{}^{48}{\mathrm{C}}_6+\dots .+\left({}^{31}{\mathrm{C}}_6+{}^{31}{\mathrm{C}}_7\right)-{}^{30}{\mathrm{C}}_7\\ & ={}^{50}{\mathrm{C}}_6+{}^{50}{\mathrm{C}}_7-{}^{30}{\mathrm{C}}_7\\ & ={}^{51}{\mathrm{C}}_7-{}^{30}{\mathrm{C}}_7\end{array}$

 $\overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k}$

$\begin{array}{l}\overrightarrow{b}=3\widehat{i}+3\widehat{j}+\widehat{k}\\ \overrightarrow{c}=c_1\widehat{i}+c_2\widehat{j}+c_3\widehat{k}\end{array}$

$Coplnanar\Rightarrow \left|\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right|=0$

$\begin{array}{l}\Rightarrow -8c_1+7c_2+12c_3=0........(i)\\ \overrightarrow{a}.\overrightarrow{c}=5\Rightarrow 2c_1+c_2+3c_3=5........(ii)\\ \overrightarrow{b}.\overrightarrow{c}=0\Rightarrow 3c_1+3c_2+c_3=0........(iii)\end{array}$

Solving (i), (ii), (iii)

$C_1=\frac{10}{122},c_2=\frac{-85}{122},c_3=\frac{225}{122}$

$\therefore 122\left(c_1+c_2+c_3\right)=150$

Let   $\frac{A_2}{A_1}=\frac{A_3}{A_2}=...=r$

$\therefore A_1{A}_3{A}_5{A}_7=\frac{1}{1296}$              

$\Rightarrow A_1{r}^3=\frac{1}{6}.......\left(i\right)$               

Again, A₂ + A₄ = $\frac{7}{36}$

$A_{1}r=\frac{7}{36}-\frac{1}{6}=\frac{1}{36}........\left(ii\right)$

$\left(i\right)\&\left(ii\right)r=\sqrt[]{6}\&A_1=\frac{1}{36\sqrt[]{6}}$

$\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$