Ncert Solutions Maths class 12th

Given P (X = 3) = 5P (X = 4) and n = 7

${\Rightarrow }^7{C}_3{p}^3{q}^4=5^7{C}_4{p}^4{q}^3$

-> q = 5p and also p + q = 1

$\Rightarrow p=\frac{1}{6}andq=\frac{5}{6}$  

Mean = $\frac{7}{6}$  and variance =   $\frac{35}{36}$

Mean + Variance

$=\frac{7}{6}+\frac{35}{36}+\frac{77}{36}$

Total case = ¹⁸C₅

Favourable cases

(Select x₁) (Select x₃)       (Select x₅)

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

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

Now,

$\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}$      

$\Rightarrow \left(\widehat{i}+\widehat{j}-\widehat{k}\right)\left(2\widehat{i}-3\widehat{j}+2\widehat{k}\right)=0$            

 = 2 – 3 – 2 = 0

->-3 = 0 (Not possible)

->No possible value of

$\overrightarrow{b}$  is possible.

 l + m – n = 0 Þ n = l + m

$3l^2+m^2+cnl=0$  

$3l^2+m^2+cl\left(l+m\right)=0$    

$=\left(3+c\right){\left(\frac{l}{m}\right)}^2+c\left(\frac{l}{m}\right)+1=0$    

$?$    Lines are parallel

D = 0

$c=4$     (as c > 0)

  $l={\displaystyle \underset{-2}{\overset{2}{\int }}\frac{\left|x^3+x\right|}{e^{x\left|x\right|}+1}dx}$ ……. (i)

$l={\displaystyle \underset{-2}{\overset{2}{\int }}\frac{\left|x^3+x\right|}{e^{-x\left|x\right|}+1}dx}$ …. (ii)

$=\left(\frac{16}{4}+\frac{4}{2}\right)$ - 0

= 4 + 2 = 6

$co{s}^{-1}\left(\frac{y}{2}\right)=lo{g}_e{\left(\frac{x}{5}\right)}^5,\left|y\right|<2$  

 Differentiating on both side

$-\frac{1}{\sqrt[]{1-{\left(\frac{y}{2}\right)}^2}}\times \frac{y\text{'}}{2}=\frac{5}{\frac{x}{5}}\times \frac{1}{5}$  

$\frac{-xy\text{'}}{2}=5\sqrt[]{1-{\left(\frac{y}{2}\right)}^2}$

Square on both side

$\frac{x^{2}y{\text{'}}^2}{4}=25\left(\frac{4-y^2}{4}\right)$

Diff on both side

$xy\text{'}+y\text{'}\text{'}{x}^2+25y=0$  

  $CD=\sqrt[]{{\left(10+x^2\right)}^2-{\left(10-x^2\right)}^2}=2\sqrt[]{10}\left|x\right|$

$=\frac{1}{2}\times CD\times AB=\frac{1}{2}\times 2\sqrt[]{10}\left|x\right|\left(20-2x^2\right)$

$\Rightarrow 10-x^2=2x$              

 3x² = 10

x = k

3k² = 10

  $\underset{x\to 7}{lim}\frac{18-\left[1-x\right]}{\left[x-3a\right]}$

exist  & $a\in I.$  

$=\underset{x\to 7}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}$      

exist

RHL = $\underset{x\to 7^{+}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{25}{7-3a}\left[a\ne \frac{7}{3}\right]$  

$LHL=\underset{x\to 7^{-}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{24}{6-3a}\left[a\ne 2\right]$

LHL = RHL

$\Rightarrow \frac{25}{7-3a}=\frac{8}{2-a}$

$\therefore a=-6$  

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

$\overrightarrow{a}\times \overrightarrow{b}=2\widehat{i}-\widehat{k}$

$\overrightarrow{a}.\overrightarrow{b}=3$

${\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2.{\left|\overrightarrow{b}\right|}^2$$$

$=\frac{\overrightarrow{b}.\overrightarrow{a}-{\left|\overrightarrow{b}\right|}^2}{\left|\overrightarrow{a}-\overrightarrow{b}\right|}=\frac{3-\frac{7}{3}}{\sqrt[]{\frac{7}{3}}}=\frac{2}{\sqrt[]{21}}$

 $A\left(-7\widehat{i}+6\widehat{j}\right)C\left(7\widehat{i}+2\widehat{j}+6\widehat{k}\right)$

$\overrightarrow{b}=-6\widehat{i}+7\widehat{j}+\widehat{k}$ $\overrightarrow{d}=-2\widehat{i}+\widehat{j}+\widehat{k}$

$\overrightarrow{b}\times \overrightarrow{d}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill -6\hfill & \hfill 7\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right|=3\widehat{i}+2\widehat{j}+4\widehat{k}$

the shortest distance between the lines

$=$$\left|\frac{42-8+24}{\text{exception 25:}}\right|$$=$$\frac{58}{\text{exception 25:}}$$=$$2$