Maths NCERT Exemplar Solutions Class 12th Chapter Seven

$dist.=\frac{{\overrightarrow{A}}_1{A}_2.\left({\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right)}{\left|{\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right|}$

$=\frac{\left(-4-\alpha ,-2,-3\right).\left(2,2,1\right)}{3}$

$\left|-5-\frac{2\alpha }{3}\right|=9\Rightarrow -5-\frac{2\alpha }{3}=\pm 9$

$\Rightarrow \frac{2\alpha }{3}=4,-14,$

but a > 0

a = 6

$\underset{x\to 0}{lim}{\left(2-cosx.\sqrt[]{cos2x}\right)}^{\frac{x+2}{x^2}}$ $\left(1^{\infty }\right)$

= $\underset{x\to 0}{lim}{e}^{2\left(\frac{sinx\sqrt[]{cos2x}-cosx.\frac{1\left(-2sin2x\right)}{2\sqrt[]{cos2x}}}{2x}\right)}$

$=\underset{x\to 0}{lim}{e}^{2\left(\frac{1}{2}+1\right)}=e^3=e^a$

a = 3

For the plane P,

$\overrightarrow{n}=\frac{1}{2}\left(-2\widehat{j}\right)*\left(-\widehat{i}+\widehat{j}-3\widehat{k}\right)=\widehat{j}*\widehat{i}+3\widehat{j}*\widehat{k}=-\widehat{k}+3\widehat{i}$

$\overrightarrow{a}.\overrightarrow{n}=0\Rightarrow 3\alpha -\gamma =0.......\left(i\right)$

$\overrightarrow{a}.\left(1,2,3\right)=0\Rightarrow \alpha +2\beta +3\gamma =0..........\left(ii\right)$

$\overrightarrow{a}.\left(1,1,2\right)=2\Rightarrow \alpha +\beta +2\gamma =2...........\left(iii\right)$

(i), (ii) & (iii) Þ a = 1, β = -5, $\gamma =$ 3

$T_{r+1}{=}^{120}{C}_r{4}^{\frac{120-r}{4}}.5^{r/6}$

For to be rational

r should be multiple of 6

Number of such terms = 21

$Area=\frac{1}{2}\left(\sqrt[]{5}-1\right)\frac{9-\sqrt[]{5}}{4}-{\displaystyle \underset{1}{\overset{\sqrt[]{5}}{\int }}\frac{1}{2}\sqrt[]{5-x^2}dx}$

$=\frac{1}{8}\left(-14+10\sqrt[]{5}\right)-\frac{1}{2}{\displaystyle \underset{co{s}^{-1}\frac{1}{\sqrt[]{5}}}{\overset{0}{\int }}\left(-si{n}^2\theta \right)d\theta }$

$A=\frac{-14}{8}+\frac{5}{4}\sqrt[]{5}-\left(\frac{5}{4}co{s}^{-1}\frac{1}{\sqrt[]{5}}-\frac{1}{2}\right)$

$=\frac{5}{4}\sqrt[]{5}-\frac{5}{4}-\frac{5}{4}co{s}^{-1}\frac{1}{\sqrt[]{5}}$

$\alpha =\frac{5}{4},\beta =\frac{-5}{4},\gamma =\frac{-5}{4}$

$\left|\alpha +\beta +\gamma \right|=\frac{5}{4}$

$\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=\mathcal{l}$

$\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$

${\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}^2=3{\mathcal{l}}^2$

$\Rightarrow {\mathcal{l}}^2=\sqrt[]{3}{\mathcal{l}}^{2}cos\theta \Rightarrow cos\theta =\frac{1}{\sqrt[]{3}}$

36 cos² 2q = 36 ${\left(\frac{2}{3}-1\right)}^2=4$

$R_1\to R_1-R_2,R_2\to R_2-R_3$

$\Delta =\left|\begin{array}{ccc}\hfill a-c+1\hfill & \hfill b-c\hfill & \hfill a-b\hfill \\ \hfill b-d-1\hfill & \hfill c-d\hfill & \hfill b-c\hfill \\ \hfill x-b+d\hfill & \hfill x+d\hfill & \hfill x+c\hfill \end{array}\right|$

$C_1\to C_1-C_2\&C_2\to C_2-C_3$

$\Delta =\left|\begin{array}{ccc}\hfill a+1-b\hfill & \hfill 2b-c-a\hfill & \hfill a-b\hfill \\ \hfill b-1-c\hfill & \hfill 2c-b-d\hfill & \hfill b-c\hfill \\ \hfill -b\hfill & \hfill d-c\hfill & \hfill x+c\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill -\lambda +1\hfill & \hfill 0\hfill & \hfill -\lambda \hfill \\ \hfill -\lambda -1\hfill & \hfill 0\hfill & \hfill -\lambda \hfill \\ \hfill -b\hfill & \hfill \lambda \hfill & \hfill x+c\hfill \end{array}\right|,R_1\to R{}_1-R_2$

$=\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -\lambda -1\hfill & \hfill 0\hfill & \hfill -\lambda \hfill \\ \hfill -b\hfill & \hfill \lambda \hfill & \hfill x+c\hfill \end{array}\right|=2{\lambda }^2=2\Rightarrow {\lambda }^2=1$

0 = -16 m + c . (i)

$\frac{\left|m\left(-10\right)+C\right|}{\sqrt[]{m^2+1}}=2.............\left(ii\right)$

(i) & (ii) Þ 36 m² = 4 (m² + 1)

$m=\frac{1}{2\sqrt[]{2}},C=\frac{8}{\sqrt[]{2}}$

$4\sqrt[]{2}\left(m+c\right)=34$

0 = 16 m + c . (i)

 $\frac{\left|m\left(?10\right)+C\right|}{\sqrt[]{m^2+1}}=2.............\left(ii\right)$

(i) & (ii) 36 m2 = 4 (m2 + 1)

$m=\frac{1}{2\sqrt[]{2}},C=\frac{8}{\sqrt[]{2}}$

$4\sqrt[]{2}\left(m+c\right)=34$