Ncert Solutions Maths class 11th

$\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1+\lambda \left(\overrightarrow{r}.\left(\widehat{i}-2\widehat{j}\right)+2\right)=0$

$\therefore point\left(1,0,2\right)=\widehat{i}+2\widehat{k}$

$\therefore \overrightarrow{r}\left(\frac{\widehat{i}}{3}+\frac{7\widehat{j}}{3}+\widehat{k}\right)-\frac{7}{3}=0$

$\overrightarrow{r}.\left(\widehat{i}+7\widehat{j}+3\widehat{k}\right)=7$

${=}^{n+1}{C}_2+2{\displaystyle \sum _{r=1}^{n-1}\frac{\left(r+1\right)!}{\left(r-1\right)!2!}}{=}^{n+1}{C}_2+{\displaystyle \sum _{r=1}^{n-1}\left(r+1\right)r}$

  =   $\frac{\left(n+1\right)!}{2!\left(n+1\right)!}+\frac{\left(n-1\right)n\left(2\left(n-1\right)+1\right)}{6}+\frac{\left(n-1\right)n}{2}=\frac{n\left(n+1\right)}{2}+\frac{n\left(n-1\right)\left(2n-1\right)}{6}+\frac{n\left(n-1\right)}{2}$

=   $\frac{n\left(6n+2n^2-3n+1\right)}{6}=\frac{\left(2n^2+3n+1\right)}{6}=\frac{n\left(2n+1\right)\left(n+1\right)}{6}$

$x+\sqrt[]{3}y=2\sqrt[]{3},andm=-\frac{1}{\sqrt[]{3}},C=2$          not possible

 (2) , $\left(1\right)\frac{x^2}{\frac{9}{2}}-\frac{y^2}{\frac{1}{2}}=1\Rightarrow c=\sqrt[]{a^2{m}^2-b^2}=\sqrt[]{\frac{9}{2}*\frac{1}{3}-\frac{1}{2}}=1,$ not possible

(3) $c=a\sqrt[]{1+m^2}=\sqrt[]{7}*2=2\sqrt[]{7},$ not possible

(4) $\frac{x^2}{9}+\frac{y^2}{1}=1,C=\sqrt[]{9m^2+1}=\sqrt[]{9*\frac{1}{3}+1}=2$  

 $x=y^4,xy=k$

$\frac{dy}{dx}=\frac{1}{4y^3},\frac{dy}{dx}=\frac{-k}{x^2}$

P (x₁, y₁)

where $x_1=y_1^4\&x_1{y}_1=k$

$\Rightarrow y_1=k^{1/5},x_1{y}_1=k$

$m_1{m}_2=-1$

$\Rightarrow \frac{-1}{4.k^{6/5}}=-1\Rightarrow k^{6/5}=\frac{1}{4}\Rightarrow k^6=\frac{1}{4^5}=\frac{1}{2024}$

${\left(4k\right)}^6=2^{12}.\frac{1}{2^{10}}=2^2=4$

A tangent to y² = 4x is x – ty + t² = 0

$\frac{3+t^2}{\sqrt[]{1+t^2}}=3$

(3 + t²)² = 9 (1 + t²)

$9+t^4+6t^2=9+9t^2$

Point of contact $\left(3,2\sqrt[]{3}\right)=\left(a,b\right)$

$x-\sqrt[]{3}y+3=0$

$\frac{\sqrt[]{3}x+y-3\sqrt[]{3}=0]}{4x-6=0}$

$x=\frac{3}{2},y=\frac{\sqrt[]{3}}{2}+\sqrt[]{3}$

& $\left(\frac{3}{2},\frac{\sqrt[]{3}}{2}+2\right)=\left(c,d\right),2\left(a+c\right)=2\left(3+\frac{3}{2}\right)=9$

 $\underset{x\to 0}{lim}\frac{ax-\left(e^{4x}-1\right)}{ax\left(e^{4x}-1\right)}=b,$ use of L' Hospital rule implies

$\underset{x\to 0}{lim}\frac{a-4e^{4x}}{a\left(e^{4x}-1\right)+ax\left(4e^{4x}\right)}$

$=\frac{a-4}{0}\Rightarrow a=4$

$\Rightarrow \underset{n\text{?}\to 0}{lim}\frac{4\left(-4.e^{4x}\right)}{4.4e^{4x}+16e^{4x}+16x.4e^{4x}}$

$=\frac{-16}{16+16}=-\frac{1}{2}=b$

a – 2b = 4 – (1) = 5

  $\frac{a+2+a}{3}=\frac{10}{3}$

2a + 2 = 0

2a = 8 -> a = 4       .(i)

and            $\frac{c+b+b}{3}=\frac{7}{3}$

2b + c = 7 .(ii)

Since         a, b, c are in A.P.

2b = a + c

From (i)   $\therefore$ 2b = 4 + c .(iii)

Solving (ii) and (iii)

4 + c + c = 7

2c = 3

$c=\frac{3}{2}$    

$\therefore 2b=4+\frac{3}{2}=\frac{11}{2}$     

$\therefore b=\frac{11}{4}$  

As per question

$\alpha +\beta =\frac{-b}{a}and\alpha \beta =\frac{1}{a}$       

$=\frac{121-192}{256}=\frac{-71}{256}$

K $=\frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}$  

K =    $\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$

$\therefore \frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}=\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$   

$\frac{x^2}{16}-\frac{y^2}{48}=1$

48 = 16 (e² – 1)

$e=\sqrt[]{4}=2$      

$f\left(x\right)=\frac{5^x}{5^x+5}$

$f\left(2-x\right)=\frac{5^{2-x}}{5^{2-x}+5}=\frac{25/5^x}{\frac{25}{5^x}+5}=\frac{25}{25+5^{x+1}}=\frac{5}{5+5^x}$

$f\left(x\right)+f\left(2-x\right)=1$

$S={\displaystyle \sum _{k=1}^{39}f\left(\frac{k}{20}\right).....\left(i\right)}$

$S={\displaystyle \sum _{k=1}^{39}f}\left(2-\frac{k}{20}\right)..........\left(ii\right)$

(i) + (ii) 2S = ${\displaystyle \sum _{k=1}^{39}\left(f\left(\frac{k}{20}\right)+f\left(2-\frac{k}{20}\right)\right)}$

$S=\frac{39}{2}$

Let a_n be the side of square A_n

$a_n=\sqrt[]{2}{a}_{n+1}$       

a₁ = 12

a_n = 12 $*{\left(\frac{1}{\sqrt[]{2}}\right)}^{n-1}$  

${\left(a_n\right)}^2<1$

$\frac{144}{2^{\left(n-1\right)}}<1$

$\Rightarrow 2^{\left(n-1\right)}>144$   

$\Rightarrow n-1\ge 8$

$\Rightarrow n\ge 9$