Ncert Solutions Maths class 11th

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side

  $?a=\sqrt[]{{\left(3?1\right)}^2+{\left(6?2\right)}^2}=\sqrt[]{20}\text{?}\text{?}and$

$\frac{b}{2}=\frac{4}{\sqrt[]{5}}?b=\frac{8}{\sqrt[]{5}}$

Area

$ab=2\sqrt[]{5}.\frac{8}{\sqrt[]{5}}=16$

  $T_r=\frac{r}{{\left(2r^2\right)}^2+1}$

$=\frac{r}{{\left(2r^2+1\right)}^2-{\left(2r\right)}^2}$

$=\frac{1}{4}\frac{4r}{\left(2r^2+2r+1\right)\left(2r^2-2r+1\right)}$

$S_{10}=\frac{1}{4}{\displaystyle \sum _{r=1}^{10}\left(\frac{1}{\left(2r^2-2r+1\right)}-\frac{1}{\left(2r^2+2r+1\right)}\right)}$

$\Rightarrow S_{10}=\frac{1}{4}.\frac{220}{221}=\frac{55}{221}=\frac{m}{n}$

$\therefore m+m=276$

  $T_{r+1}{=}^{60}{C}_r{\left(x^{\frac{1}{2}}\right)}^{60-r}{\left(x^{-\frac{1}{3}}\right)}^r{\left(5^{\frac{-1}{4}}\right)}^{60-r}{\left(5^{\frac{1}{2}}\right)}^r$

for

$x^{10}\frac{60-r}{2}-\frac{r}{3}=10$    

$\Rightarrow 180-3r-2r=60$             

->r = 24

k = 3 + exponent of 5 in 

= $3+\left(\left[\frac{60}{5}\right]+\left[\frac{60}{5^2}\right]-\left[\frac{24}{5}\right]-\left[\frac{24}{5^2}\right]-\left[\frac{36}{5}\right]-\left[\frac{36}{5^2}\right]\right)$  

= 3 + (12 + 2 – 4 – 0 – 7 – 1)

= 3 + 2 = 5

Let e^x = t then equation reduces to

$t^2-11t-\frac{45}{t}+\frac{81}{2}=0$                

$\Rightarrow 2t^3-22t^2+81t-45=0$                     ……. (i)

If roots of

$e^{2x}-11e^x-45e^{-x}+\frac{81}{2}=0$            

$\Rightarrow {\alpha }_1+{\alpha }_2+{\alpha }_3=\mathcal{l}n45\Rightarrow p=45$                

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{{\left(-4\sqrt[]{\frac{2}{5}}\right)}^2}{a^2}+\frac{32}{b^2}=1$                

$\Rightarrow \frac{32}{5a^2}+\frac{9}{b^2}=1$ ……. (i)

From (i)

$\frac{6}{b^2}+\frac{9}{b^2}=1\Rightarrow b^2=15\&a^2=16$

$a^2+b^2=15+16=31$

  $\begin{array}{l}2x+y=4\\ 2x+6y=14\end{array}\}y=2,x=3$              

B (1, 2)

Let C (k, 4 – 2k)

Now AB² = AC²

->5k² – 24k + 19 = 0

$\alpha =\frac{6+1+\frac{10}{5}}{3}=\frac{18}{5}$    

Now 15 (a + b)

$15\left(\frac{17}{5}\right)=51$                

  $f\left(x\right)=x^4-4x+1=0$              

$f\text{'}\left(x\right)=4x^3-4$

$=4\left(x-1\right)\left(x^2+1+x\right)$              

$\Rightarrow$ Two solution

  $\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$

$=bxdy+cydy+ady=axdx-bydx+adx$                

$=\frac{c{y}^2}{2}+ay-\frac{a{x}^2}{2}-ax+bxy=k$              

$a{x}^2+a{y}^2+2ax-2ay=k$            

$\Rightarrow x^2+y^2+2x-2y=\lambda$              

Short distance of (11,6)

= $\sqrt[]{12^2+5^2}-5$

= 13 – 5

= 8

  $x={\displaystyle \sum _{n=0}^{\infty }{a}^n}=\frac{1}{1-a}y={\displaystyle \sum _{n=0}^{\infty }{b}^n=\frac{1}{1-b}{\displaystyle \sum _{n=0}^{\infty }{c}^n=\frac{1}{1-c}}}$               

Now,

a, b, c -> AP

1 – a, 1 – b, 1 – c -> AP

$\frac{1}{1-a},\frac{1}{1-b},\frac{1}{1-c}\to HP$

x, y, z -> HP