Ncert Solutions Maths class 12th

$\underset{n\to \infty }{lim}{\left(1+\frac{1+\frac{1}{2}+.......+\frac{1}{n}}{n^2}\right)}^n$ limit is in the form of $1^{\infty }$  

$\mathcal{l}=exp\left(\underset{n\to \infty }{lim}\frac{1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}}{n^2}\right)$             

$0\le 1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}\le 1+\frac{1}{\sqrt[]{2}}+\frac{1}{\sqrt[]{3}}+....+\frac{1}{\sqrt[]{n}}\le 2\sqrt[]{n}-1$        

Taking limit   $\left(n\to \infty \right)$

$\mathcal{l}$ = exp (0) (from sandwich)

  $\mathcal{l}=1$          

Second Method :

$1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}\ge ln\left(n+1\right).......\left(i\right)$

$1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}\le 1+{\displaystyle {\int }_1^2\frac{1}{2}dx\le 1+lnn..........\left(ii\right)}$

From (i) & (ii)

$ln\left(n+1\right)\le 1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}\le 1+Inn,\forall n\in N,n\ge 2$           

As $\underset{n\to \infty }{lim}\frac{ln\left(n+1\right)}{n}=0$  

and $\underset{n\to \infty }{lim}\frac{1+ln\left(n\right)}{n}=0$  

$\therefore$ from sandwich theorem

$\underset{n\to \infty }{lim}\frac{1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}}{n}=0$  

$\therefore e^0=1$   

$\overrightarrow{O}P=\left(2,-1,1\right)$

Normal vector to the plane

$=\overrightarrow{A}B*A\overrightarrow{C}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left(3,-1,1\right)=\overrightarrow{n}$

Projection of $\overrightarrow{O}Pon\overrightarrow{n}is\left|\frac{\overrightarrow{O}P.\overrightarrow{n}}{\left|\overrightarrow{n}\right|}\right|$

PN = $\frac{6+1+1}{\sqrt[]{11}}=\frac{8}{\text{exception 25:}}$

$\frac{\text{exception 25:}}{}$Projection of OP on plane = ON =  $\sqrt[]{O{P}^2-P{N}^2}=\sqrt[]{6-\frac{64}{11}}=\sqrt[]{\frac{2}{11}}$

  $\frac{x^2}{a}+\frac{y^2}{b}=1,\frac{x^2}{c}-\frac{y^2}{\left(-d\right)}=1$

Ellipse and Hyperbola are orthogonal so these will be confocal.

$\sqrt[]{a-b}=\sqrt[]{c+\left(-d\right)}$            

a – b = c – d

$l={\displaystyle \int \frac{8x^3+20x^2}{x^4+5x^3-7x^2}}dx=4{\displaystyle \int \frac{2x+5}{x^2+5x-7}dx=4ln\left|x^2+5x-7\right|+c}$

$f\left(x\right)=x^3-a{x}^2+bx-4$

f (1) = f (2)

->1 – a + b – 4 = 8 – 4a + 2b – 4

->3a – b = 7 . (i)

8a = 16 + 3b . (ii)

(i) and (ii) -> (a, b) = (5, 8)

As per questions

$\frac{dy}{dx}=\frac{x^2-4x+y+8}{x-2}$           

$\frac{dy}{dx}=\frac{{\left(x-2\right)}^2+\left(y+4\right)}{\left(x-2\right)}$           

$\frac{dy}{dx}=\left(x-2\right)+\frac{y+4}{x-2}$                       .(i)

Let $\frac{y+4}{x-2}=t$  

(y + 4) = t(x – 2)

Putting in equation (i)

$\left(x-2\right)\frac{dt}{dx}+t=\left(x-2\right)+t$        

    $\frac{dt}{dx}=1$        

dt = dx

Integrating on both the sides t = x + c

$\frac{y+4}{x-2}=x+c$  

Passing through origin C = -2

  $\therefore$         equation of curve $\frac{y+4}{x-2}=x-2$

$l_n={\displaystyle \underset{\pi /4}{\overset{\pi /2}{\int }}co{t}^{n}xdx=}{\displaystyle \underset{\pi /4}{\overset{\pi /2}{\int }}co{t}^{n-2}x\left(cose{c}^{2}x-1\right)dx}$

$={\left[\frac{-co{t}^{n-1}x}{n-1}\right]}_{\pi /4}^{\pi /2}-l_{n-2}=\frac{1}{n-1}-l_{n-2}$

$l_n+l_{n-2}=\frac{1}{n-1}$

$\begin{array}{l}n=4\Rightarrow l_4+l_2=\frac{1}{3}\\ n=5\Rightarrow l_5+l_3=\frac{1}{4}\\ n=6\Rightarrow l_6+l_4=\frac{1}{5}\end{array}\}\Rightarrow \frac{1}{l_2+l_4},\frac{1}{l_3+l_5},\frac{1}{l_4+l_6}$ are in A.P.

$f:\mathbf{R}\to \mathbf{R}$

$$$(\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{x}\mathrm{c}\mathrm{o}\mathrm{s}?\mathrm{y})\left(f\right(2\mathrm{x}+2\mathrm{y})-f(2\mathrm{x}-2\mathrm{y}\left)\right)=(\mathrm{c}\mathrm{o}\mathrm{s}?\mathrm{x}$ Put $\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{y}\left)\right(f(2\mathrm{x}+2\mathrm{y})+f(2\mathrm{x}-2\mathrm{y}))$

$$$\mathrm{x},\mathrm{y}\in \mathbf{R}$ Put $f^{\text{'}}\left(0\right)=\frac{1}{2}$

$24f^{\text{'}\text{'}}\left(\frac{5\pi }{3}\right)$

$\mathrm{}(\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{x}\mathrm{c}\mathrm{o}\mathrm{s}?\mathrm{y})\left(\mathrm{f}\right(2\mathrm{x}+2\mathrm{y})-\mathrm{f}(2\mathrm{x}-2\mathrm{y}\left)\right)=(\mathrm{c}\mathrm{o}\mathrm{s}?\mathrm{x}\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{y})$

$\left(f\right(2x+2y)+f(2x-2y\left)\right)$

$\mathrm{f}(2\mathrm{x}+2\mathrm{y})(\mathrm{s}\mathrm{i}\mathrm{n}?(\mathrm{x}-\mathrm{y}\left)\right)=\mathrm{f}(2\mathrm{x}-2\mathrm{y})\mathrm{s}\mathrm{i}\mathrm{n}?(\mathrm{x}+\mathrm{y})$

: $?2$ adj $(3\mathrm{A}$  adj(2A))|
${=2}^3.?3\mathrm{A}$ adj(2A)| ${\left.\right|}^2$

${=2}^3\cdot {\left(3^3\right)}^2\cdot |\mathrm{A}{|}^2\cdot |\mathrm{a}\mathrm{d}\mathrm{j}\left(2\mathrm{A}\right){|}^2$  intersect the line $=2^3\cdot 3^6\cdot |\mathrm{A}{|}^2\cdot {\left(|2\mathrm{A}{|}^2\right)}^2$ at the point

$=2^3\cdot 3^6\cdot |\mathrm{A}{}^{}$

We can convert 50! In terms of prime factor: $\mathrm{}{2}^{\alpha }\cdot 3^{\beta }\cdot 5^{\gamma }$ & using the greatest integer function.

n $=\left. [\frac{50}{3}\right]+\left. [\frac{50}{3^2}\right]+\left. [\frac{50}{3^3}\right]+\left. [\frac{50}{3^4}\right]$
$=16+5+1+0$ The maximum value of n is 22.