Ncert Solutions Maths class 12th

 f (3x)- f (x) = x

Replace  $x\to \frac{x}{3}\Rightarrow f\left(x\right)-f\left(\frac{x}{3}\right)=\frac{x}{3}$  

Again replace  $x\to \frac{x}{3}\Rightarrow f\left(\frac{x}{3}\right)-f\left(\frac{x}{3^2}\right)-f\left(\frac{x}{3^2}\right)=\frac{x}{3^2}$

$\Rightarrow f\left(3x\right)-f\left(0\right)=\frac{3x}{2}puttingx=\frac{8}{3}\Rightarrow f\left(8\right)-f\left(0\right)=4\therefore f\left(0\right)=3$  

Also putting x =  $\frac{14}{3}$ in f (3x) – 3 = $\frac{3x}{2}\Rightarrow$ F (14) – 3 = 7 Þ f (14) = 10

$f\left(A,B\right)=\frac{1-sinB+sinA-sinAsinB+cosAcosB}{cosA\left(1-sinB\right)}$

$\begin{array}{l}\Rightarrow f\left(A,B\right)\left(cosA-cosAsinB-cosB+sinAcosB\right)\\ =2sinA-2sinB\end{array}$

$\Rightarrow f\left(A,B\right)=\frac{2\left(sinA-sinB\right)}{cosA-cosB+sin\left(A-B\right)}=2g\left(A,B\right)$

$x_1=\underset{x\to \infty }{lim}\frac{2^{\frac{x^n}{e^x}}-3^{\frac{x^n}{e^x}}}{\frac{x^n}{e^x}},put\frac{x^n}{e^x}=t$

$x_2=\underset{x\to \infty }{lim}\frac{co{t}^{-1}\left(\sqrt[]{x+1}-\sqrt[]{x}\right)}{se{c}^{-1}\left({\left(\frac{2x+1}{x-1}\right)}^x\right)}$

$x_2=\frac{2}{\pi }\underset{x\to \infty }{lim}ta{n}^{-1}\left(\sqrt[]{x+1}+\sqrt[]{x}\right)$

$x_2=\frac{2}{\pi }.\frac{\pi }{2}=1$

$\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{1}{4},\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{1}{12}$

divide both $\frac{P\left(B\right)}{P\left(A\right)}=\frac{1}{3}$

$P\left(A\right)=\frac{1}{11},P\left(B\right)=\frac{1}{33}$

Line L is   $\frac{x-1}{1}=\frac{y-2}{2\sqrt[]{2}}=\frac{z}{0}$

this line L makes an angle of 45° with the plane  $\sqrt[]{2}x+y-z=1$

$\therefore$ Required distance PQ is perpendicular distance of plane from P is i.e.,

$PQ=\frac{\left|\sqrt[]{2}+7-\sqrt[]{2}-1\right|}{\sqrt[]{2+1+1}}=3$

Plane containing both line $x=1$

Image of point $(\mathrm{2,1},3)$  in the plane $\frac{\alpha -2}{1}=\frac{\beta -1}{0}=\frac{\gamma -3}{0}=-2\left(1\right)$

$\alpha =0,\beta =1,\gamma =3;\alpha +\beta +\gamma =4$

  $x-cy-bz=0$

$cx-y+az=0$

$bx+ay-z=0$

Equation of any plane passing through the line of intersection of planes (1) and (2)

$x-cy-bz+\lambda (cx-y+az)=0$

$(1+\lambda c)x-(c+\lambda )y+(-b+a\lambda )z=0$

If (3) &  (4) is same plane.

$\frac{1+c\lambda }{b}=\frac{-(c+\lambda )}{a}=\frac{-b+a\lambda }{-1}$

(i)

(ii) (iii)

By (i)  & (ii) $\lambda =-\frac{(a+bc)}{ac+b}$

By (ii) & (iii

Mean $=np$ , variance $=npq$

$np-npq=1;p+q=1$

$n^2{p}^2-n^2{p}^2{q}^2=11$

We get   $q=\frac{5}{6},p=\frac{1}{6},n=36\mathrm{}$  Probability of ' 3 ' success $={}^{36}{C}_3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^{33}$

$\bar{a}+\bar{b}+\bar{c}+\bar{d}=0$

Magnitude of all vectors will be same as well as angle between these vectors will also be same. $\bar{a}+\bar{b}+\bar{c}=-\bar{d}$

$\begin{array}{c}|\bar{a}{|}^2+|\bar{b}{|}^2+|\bar{c}{|}^2+2\bar{a}\cdot \bar{b}+2\bar{b}\cdot \bar{c}+2\bar{c}\cdot \bar{a}=|d{|}^2\\ \mathrm{}1+1+1+2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha +2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha +2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha =1\\ 6\mathrm{c}\mathrm{o}\mathrm{s}?\alpha =-2\\ \mathrm{c}\mathrm{o}\mathrm{s}?\alpha =-\frac{1}{3}\\ \alpha ={\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?\left(\frac{-1}{3}\right)=\pi -{\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?\frac{1}{3}\end{array}$

P(W) = $\frac{1}{3};$  P(B) =   $\frac{2}{3}$

$\Rightarrow p=\frac{1}{3};q=\frac{2}{3}$           and r = 4 or 5 and n = 5

Use   $P\left(r\right){=}^n{C}_r{p}^r{q}^{n-r}$

 P(4) + P(5)

${=}^5{C}_4{\left(\frac{1}{3}\right)}^4{\left(\frac{2}{3}\right)}^1{+}^5{C}_5{\left(\frac{1}{3}\right)}^5$            

$=\frac{10}{3^5}+\frac{1}{3^5}=\frac{11}{3^5}=\frac{11}{243}$