Maths NCERT Exemplar Solutions Class 12th Chapter Ten

 $\overrightarrow{a}\cdot \widehat{c}=\frac{\alpha +6+2}{\sqrt[]{1}+4+4}$

$\frac{10}{3}=\frac{8+\alpha }{3}\Rightarrow \alpha =2$

$\overrightarrow{b}*\overrightarrow{c}=\widehat{i}\left(2\beta -8\right)+\widehat{j}\left(10\right)+\widehat{k}\left(6+\beta \right)$

$=-6\widehat{i}+10\widehat{j}+7\widehat{k}$

So = 1

$If{\widehat{n}}_1$ is a vector normal to the plane determined by $\widehat{i}and\widehat{i}+\widehat{j}then{\widehat{n}}_1=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right|=\widehat{k}$

$If{\widehat{n}}_2$ is a vector normal to the pane determined by $\widehat{i}-\widehat{j},$ and $\widehat{i}+\widehat{k}$ then ${\widehat{n}}_2$ = $\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=\widehat{i}-\widehat{j}+\widehat{k}$

Vector $\widehat{a}$ is parallel to ${\widehat{n}}_1*{\widehat{n}}_2$ i.e. $\widehat{a}$ is parallel to $\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right|=\widehat{i}-\widehat{j}$

Given $\overrightarrow{b}=\widehat{i}-2\widehat{j}+2\widehat{k}$

consine of acute angle between $\widehat{a}and\widehat{b}$ = $\left|\frac{\widehat{a}.\widehat{b}}{\left|\widehat{a}\right|.\left|\widehat{b}\right|}\right|=\frac{1}{\sqrt[]{2}}$

obtuse angle between $\widehat{a}and\widehat{b}=\frac{3\pi }{4}$

For angle to the acute $\overrightarrow{U}.\overrightarrow{V}>0$

$\Rightarrow a{\left(lo{g}_{e}b\right)}^2-12+6a\left(lo{g}_{e}b\right)>0$

$\forall b>1$

Let log_e b = t > 0 as b > 1

Y = at² + 6at – 12 & y > 0 $\forall$ t > 0

$\Rightarrow a\in ?$

 $x=2\sqrt[]{2}cost\sqrt[]{sin2t}$

$\frac{dx}{dt}=\frac{2\sqrt[]{2}cos3t}{\sqrt[]{sin2t}}$

$y\left(t\right)=2\sqrt[]{2}sint\sqrt[]{sin2t}$

$\frac{dy}{dx}=\frac{2\sqrt[]{2}sin3t}{\sqrt[]{sin2t}}$

$\therefore \frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}=\frac{1+1}{-3}=\frac{-2}{3}$

$\left(\frac{\mathrm{K}-2}{\mathrm{h}-1}\right)\left(\frac{1-2}{3-1}\right)=-1\Rightarrow \mathrm{K}=2\mathrm{h}$

$?[?\mathrm{A}\mathrm{B}\mathrm{C}]=5\sqrt[]{5}$

$\Rightarrow \frac{1}{2}\left(\sqrt[]{5}\right)\sqrt[]{(\mathrm{h}-1{)}^2+(\mathrm{K}-2{)}^2}=5\sqrt[]{5}$

$\mathrm{t}\mathrm{a}\mathrm{n}?\theta =\frac{10}{\mathrm{x}}=\frac{\mathrm{h}}{{\mathrm{x}}_2}\Rightarrow {\mathrm{x}}_2=\frac{\mathrm{h}\mathrm{x}}{10}$

$\mathrm{t}\mathrm{a}\mathrm{n}??=\frac{15}{\mathrm{x}}=\frac{\mathrm{h}}{\mathrm{x}}\Rightarrow {\mathrm{x}}_1=\frac{\mathrm{h}\mathrm{x}}{15}$

Now,  $x_1+x_2=x=\frac{hx}{15}+\frac{hx}{10}$

$\Rightarrow 1=\frac{\mathrm{h}}{10}+\frac{\mathrm{h}}{15}\Rightarrow \mathrm{h}=6$

$f\left(x\right)=\stackrel{?}{a}\cdot (\stackrel{?}{b}*\stackrel{?}{c})=\left|\begin{array}{ccc}x& -2& 3\\ -2& x& -1\\ 7& -2& x\end{array}\right|$

$=x^3-27x+26$

$f^{\mathrm{\text{'}}}\left(x\right)=3x^2-27=0\Rightarrow x=\pm 3$ and $f^{\mathrm{\text{'}}\mathrm{\text{'}}}(-3)<0$

$\Rightarrow$ local maxima at $\mathrm{x}={\mathrm{x}}_0=-3$

Thus, $\stackrel{?}{a}=-3\hat{i}-2\hat{j}+3\hat{k},\stackrel{?}{b}=2\hat{i}-3\hat{j}-\hat{k}$ , and $\stackrel{?}{c}=7\hat{i}-2\hat{j}-3\hat{k}$

$\Rightarrow \stackrel{?}{a}\cdot \stackrel{?}{b}+\stackrel{?}{b}\cdot \stackrel{?}{c}+\stackrel{?}{c}\cdot \stackrel{?}{a}=9-5-26=-22$

$y+2x=\sqrt[]{11}+7\sqrt[]{7}$ ……… (i)

$2y+x=2\sqrt[]{11}+6\sqrt[]{7}$ ……… (ii)

$\Rightarrow x+y=\sqrt[]{11}+\frac{13}{3}\sqrt[]{7}$ ……… (iii)

Centre of the circle given by solving (i) & (ii)

$as\left(\frac{8\sqrt[]{7}}{3},\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)$

Again $\sqrt[]{11}y-3x=\frac{5\sqrt[]{77}}{3}$ is tangent to the circle.

$\therefore r=\left|\frac{\sqrt[]{11}\left(\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)-8\sqrt[]{7}-\frac{5\sqrt[]{77}}{3}}{\sqrt[]{11+9}}\right|=\left|\frac{11-8\sqrt[]{7}}{\sqrt[]{20}}\right|$

$\therefore {\left(5h-8k\right)}^2+5r^2=816$

$1<\left|z-3+2i\right|<4$

$z=a+iba,b\in I$              

⇒ 1 < (a 3)² + (b + 2)² < 16

in equivalent to 1 < ² + β² < 16  $\alpha =a-3\in l$               

$\left(\pm 2,0\right)\left(\pm 3,0\right),\left(0,\pm 2\right)\left(0,\pm 3\right)$             $\beta =b+2\in l$

Total 40 such points are possible

${\left(2x^3+\frac{3}{x^k}\right)}^{12}$

gen term =

${=}^{12}{C}_r{2}^{12-r}.3^r.x^{36-3r-rk}$

For constant term

36 – 3r – rk = 0

$\Rightarrow k=\frac{36-3r}{r}$

for r = 1, 2, 4

¹²C_r2^12-r>2⁸

Possible values of k = 3, 1