Ncert Solutions Maths class 11th

Circumcentre (D) $\equiv \left(5,\frac{\alpha }{4}\right)$

${\left(5-\alpha \right)}^2+{\left(\frac{\alpha }{4}+2\right)}^2={\left(5-\alpha \right)}^2+{\left(\frac{\alpha }{4}-6\right)}^2...............\left(i\right)$

${\left(5-\frac{\alpha }{4}\right)}^2+{\left(\frac{\alpha }{4}+2\right)}^2....................\left(ii\right)$

(i) $\frac{\alpha }{4}+2=\pm \left(\frac{\alpha }{4}-6\right)$

(ii) 9 + 16 = 9 + 16

$\oplus \to x$

$\left(-\right)\to \frac{\alpha }{2}=4\Rightarrow \alpha =8$

ar $\left(ABC\right)=24$

2S = 24

R = 5, r = $\frac{\Delta }{s}=\frac{24}{12}=2$

2cos $\left(\frac{x^2+x}{6}\right)=4^x+4^{-x}$

$-2\le LHS\le 2LHS=2\&RHS=2\Rightarrow x=0only\to thenLHS=2also$

RHS $\ge$ 2

$m_1{m}_2=-1,$ for square a,b,c,d let

$A\left(10\left(cos\alpha -sin\alpha \right),10\left(sin\alpha +cos\alpha \right)\right)$

Diagonal : (cosα - sinα)x + (sinα + cosα)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{\left|10{\left(cos\alpha -sin\alpha \right)}^2+10{\left(sin\alpha +cos\alpha \right)}^2-10\right|}{\sqrt[]{2}}=\frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}}=\frac{a}{\sqrt[]{2}}\Rightarrow a=10$

Also, a² + 11a + 3 $\left(m_1^2+m_2^2\right)=220$

210 + 3 $\left(c{m}_1^2+m_2^2\right)=220$

$m_1^2+m_2^2=\frac{10}{3}$

Also, m₁ m₂ = -1

m² + $\frac{1}{m^2}=\frac{10}{3}$

or $-\sqrt[]{3},\frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3},\frac{-1}{\sqrt[]{3}}$

$m^4-\frac{10}{3}{m}^2+1=0\Rightarrow m^2=\frac{\frac{10}{3}\pm \sqrt[]{\frac{100}{9}-4}}{2}-\frac{\frac{10}{3}\pm \frac{8}{3}}{2}=3,\frac{1}{3}$

m = $\pm \sqrt[]{3},\pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$\left(sin\alpha +cos\alpha \right)x-\left(cos\alpha -sin\alpha \right)y$

=10 cos2α - 10cos2α = 0

Slope of AC = $\frac{sin\alpha +cos\alpha }{cos\alpha -sin\alpha }=\frac{tan\alpha +1}{1-tan\alpha }=tan\left(\alpha +\frac{\pi }{4}\right)\alpha =30°$

FIGURE

? = $72\left(\frac{1}{16}+\frac{9}{16}\right)+100-30+13=\frac{720}{16}+83=128$

${\displaystyle \sum _{r=1}^{20}\left(r^2+1\right)r!}$  

t_r = (r² + 1)r!

= r²r! + r!

= r(r + 1 – 1)r! + r!

= r(r + 1)! – (r – 1)r!

= V_r – V_r-1

  ${\displaystyle \sum _{r=1}^{20}\left(V_r-V_{r-1}\right)}$              

= V₁ – V₀

$+V_2-V_1$

$+V_3-V_2$

$+V_{20}-V_{19}$

$+V_{20}-V_{19}$

$=V_{20}-V_0=20\left(21!\right)-0$

$\left(22-2\right)\left(21!\right)=22!-2\left(21!\right)$        

$\left|z-\frac{1}{z}\right|=2$

${\left|z\right|}_{max}=?$

$\left|z-\frac{1}{2}\right|\ge \left|\left|z\right|-\frac{1}{\left|z\right|}\right|$

$2\ge \left|r-\frac{1}{r}\right|$

0 $\le r^2+2r-1\&r^2-2r-1\le 0$

$r=\frac{-2\pm \sqrt[]{8}}{2}$ $r=\frac{2\pm \sqrt[]{8}}{2}$

$=-1\pm \sqrt[]{2}$ $=1\pm \sqrt[]{2}$

r $\ge \sqrt[]{2}-1\&0\le r\le 1+\sqrt[]{2}$

$\sqrt[]{2}-1\le r\le \sqrt[]{2}+1$

$e=\frac{5}{4}$

$b^2=a^2\left(e^2-1\right)$

$\Rightarrow b=\frac{3a}{4}$

$p\left(\frac{8}{\sqrt[]{5}},\frac{12}{5}\right)$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\Rightarrow \frac{64}{5a^2}-\frac{144}{25b^2}=1$

$\Rightarrow \frac{\sqrt[]{5}x}{3}+\frac{5}{8}y=\frac{8}{3}+\frac{3}{2}=\frac{25}{6}$

$a\in \left\{1,2,3,....,9\right\}$  

  $b\in \left\{0,1,2,...,9\right\}$              

9 * 9 = 81

81 + 81 + 81 = 243

  $\frac{x^2}{4}+\frac{y^2}{2}=1$              

lim PQ is y = x + 1

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

$\Rightarrow {\left(x_1-x_2\right)}^2={\left(-\frac{4}{3}\right)}^2-4\left(-\frac{2}{3}\right)=\frac{16}{9}+\frac{8}{3}=\frac{40}{9}$               

y = x + 1

$y_1-y_2=x_1-x_2$

$P{Q}^2=\frac{80}{9}\Rightarrow \frac{PQ}{2}=r=\frac{\sqrt[]{80}}{6}$

$r=\frac{4\sqrt[]{5}}{6}\Rightarrow {\left(3r\right)}^2={\left(2\sqrt[]{5}\right)}^2=20$               

y = x – x²

p (t, t - t²)

$T_{\left(p\right)}:y+t-t^2$

$=x+t-2x.t$

$\Rightarrow y=\left(1-2t\right)x+t^2$

$\Rightarrow 1-2t=k,t^2=4,t=\pm 2$

$\Rightarrow y=5x+4$                

$\left|h\right|=rand\frac{\left|h\right|+k_1}{\sqrt[]{2}}=r$

$\frac{{\left(h+k\right)}^2}{2}=h^2\Rightarrow h^2=k^2+2hk$

$x^2-y^2=2xy$