Ncert Solutions Maths class 12th

 Normal vector to the given plane be

$2\widehat{i}-\widehat{j}+3\widehat{k}so$                   

Equation of line QS :

$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\lambda$

So let P $\left(2\lambda +1,-\lambda +3,\lambda +4\right)$  

Now P lies on given plane so

$4\lambda +2+\lambda -3+8\lambda +4+3=0$  

So, S (-3, 5, 2)

also given R lies on given plane so

6 – 5 + $\gamma$ + 3 = 0 so   $\gamma$ = -4

So, R (3,

$y=lo{g}_{10}x+lo{g}_{10}{x}^{1/3}+lo{g}_{10}{x}^{1/9}+........upto\infty terms$

$=lo{g}_{10}x\left(1+\frac{1}{3}+\frac{1}{9}+......\right)$   

y = log₁₀ x × $\frac{1}{1-\frac{1}{3}}=\frac{3}{2}lo{g}_{10}x$  

Now, $\frac{2+4+6+....+2y}{3+6+9+....+3y}=\frac{4}{lo{g}_{10}x}$

$\Rightarrow \frac{2\times y\frac{\left(y+1\right)}{2}}{3y\frac{\left(y+1\right)}{2}}=\frac{4}{lo{g}_{10}x}solo{g}_{10}x=6$

$\Rightarrow y=\frac{3}{2}\times 6=9$

So, (x, y) = (10⁶, 9)

$y=\frac{x^2}{2}+\frac{2}{3}{x}^3+\frac{3}{4}{x}^4+.....$

$=\left(x^2+x^{}+x^4+.....\right)+\left(-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}........\right)$

$y=\frac{x^2}{1-x}+log\left(1-x\right)+x=\frac{x}{1-x}+log\left(1-x\right)$

$atx=\frac{1}{2},y=1+ln\frac{1}{2}=1-ln2$

$e^{y+1}=e^{1-ln2+1}=e^{2-ln2}=\frac{e^2}{2}$

Required equation of plane will be (x – y – z – 1) + $\lambda$ (2x + y – 3z + 4) = 0

Given ${\perp }^r$ distance of (i) from origin = $\sqrt[]{\frac{2}{21}}$  

$\frac{\left|4\lambda -1\right|}{\sqrt[]{{\left(2\lambda +1\right)}^2+{\left(\lambda -1\right)}^2+{\left(3\lambda +1\right)}^2}}=\sqrt[]{\frac{2}{21}}$   

$\Rightarrow \lambda =\frac{1}{2}or\frac{15}{154}$

So plane be 4x – y – 5z + 2 = 0 for $\lambda =\frac{1}{2}$

Breadth = b – 2x

& height = x

Let volume V = $\left(a-2x\right)\left(b-2x\right)x$  

For minimum volume $\frac{dv}{dx}=0$  

$\left(a-2x\right)\left(b-2x\right)-2\left(a-2x\right)x-2\left(b-2x\right)x=0$              

Since x =  $\left\{\left(a+b\right)+\sqrt[]{a^2+b^2-ab}\right\}/6$ not possible because maxima occurs

Required probability = probability of both getting 0 head or 1 head or 2 head or 3 head

 = $\frac{5}{16}$

$l=\underset{n\to \infty }{lim}{\left(u_n\right)}^{\frac{-4}{n^2}}=\underset{n\to \infty }{lim}{\left\{\left(1+\frac{1}{n^2}\right){\left(1+\frac{2^2}{n^2}\right)}^2{\left(1+\frac{3^2}{n^2}\right)}^3......{\left(1+\frac{n^2}{n^2}\right)}^n\right\}}^{\frac{-4}{n^2}},$ Taking log on both the sides

$\underset{n\to \infty }{lim}\frac{-4}{n^2}{\displaystyle \sum _{r=1}^{n}rlog\left(1+\frac{r^2}{n^2}\right)}=\underset{n\to \infty }{lim}\frac{-4}{n}{\displaystyle \sum _{r=1}^n\frac{r}{n}log\left(1+{\left(\frac{r}{n}\right)}^2\right)}$

$\therefore logl=-4{\displaystyle \underset{0}{\overset{1}{\int }}xlog\left(1+x^2\right)dx}$

$logl=-2\left[log4-1\right]=-2log\frac{4}{e}=log\frac{e^2}{16}$       

$l=\frac{e^2}{16}$

Required probability of obtaining a (sum = 7) = $\frac{13}{96}\left(given\right)$  

    $i.e.2\left(\left(\frac{1}{6}+x\right)\left(\frac{1}{6}-x\right)+\frac{1}{6}\times \frac{1}{6}+\frac{1}{6}\times \frac{1}{6}\right)=\frac{13}{96}$

$x=\frac{1}{8}$           

Equation of line PQ :

$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=k$

So, let Q (2k + 1, 3k – 2, -6k + 3) Q lies on given plane so

2k + 1 – 3k + 2 – 6k + 3 = 5

$-7k=-1ork=\frac{1}{7}$

So, $Q\left(\frac{9}{7},\frac{-11}{7},\frac{15}{7}\right)$

Now required distance = PQ = $\sqrt[]{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=1$

Let square is made with piece of length x metre & hexagon with piece of length y metre

x + y = 20 .(i)

$a=\frac{x}{4}\&b=\frac{y}{6}$       

Now let A = area of square + area of hexagon

$A=\frac{x^2}{16}+6\times \frac{\sqrt[]{3}}{4}\times \frac{y^2}{36}=\frac{x^2}{16}+\frac{\sqrt[]{3}{y}^2}{24}=\frac{x^2}{16}+\frac{\sqrt[]{3}}{24}{\left(20-x\right)}^{2}from\left(i\right)$ ,

for minimum area

$x=\frac{80\sqrt[]{3}}{6+4\sqrt[]{3}}=\frac{80\sqrt[]{3}}{2\sqrt[]{3}\left(\sqrt[]{3}+2\right)}=\frac{40}{2+\sqrt[]{3}}$

$x=40\left(2-\sqrt[]{3}\right)$

=> side of hexagon = $\frac{y}{6}=\frac{20\sqrt[]{3}\left(2-\sqrt[]{3}\right)}{6}=\frac{20\sqrt[]{3}}{6\left(2+\sqrt[]{3}\right)}=\frac{10\sqrt[]{3}}{3\left(2+\sqrt[]{3}\right)}=\frac{10}{2\sqrt[]{3}+3}$