Ncert Solutions Maths class 11th

$y^2=2x-3$ ……. (i)

Equation of chord of contact

PQ : r = O

(y * 1) = (x + 0) – 3

y = x – 3              ……… (ii)

From (i) and (ii)

y = 1 or 3

$MPQ=\frac{4}{4}=1$

  $MQR=\frac{2}{6}=\frac{1}{3}$              

$M_{PQ}*M_{PR}=-1\Rightarrow PQ\perp PR$      

Orthocentre = P (2, -1)

S  $\equiv sint,C\equiv cost$ $$

Let orthocenter be (h, k)

Since it if an equilateral triangle hence orthocenter coincides with centroid.

$\therefore a+s+c=3h,b+s-c=3k$

$\frac{a}{3}=1,\frac{b}{3}=\frac{1}{3}\Rightarrow a=3,b=1$

$\therefore a^2-b^2-8$

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

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

_{$+V_2-V_1$}

_{$+V_3-V_2$}

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

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

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

(A) $\left(\wedge ,\vee \right):\left(P\vee Q\right)\wedge \left(P\wedge \sim Q\right)$ $$ not a tautology

(B) $\left(\vee ,\vee \right):\left(P\vee Q\right)\vee \left(P\vee \sim Q\right)=\left(P\vee T\right)$ $$ in a tautology

$$  $\left(p\vee q\right)\vee \left(p\vee \sim q\right)=$ T using venn diagrams

y = 2x - $\left|3x^2-5x+2\right|,0\le x\le 1$  

$=\{\begin{array}{cc}\hfill -3x^2+7x-2,\hfill & \hfill 3x^2-3x+2,\hfill \end{array}\begin{array}{cc}\hfill 0\le x\le \frac{2}{3}\hfill & \hfill \frac{2}{3}<x\le 1\hfill \end{array}$               

 3x² – 5x + 2 = 0

$x=\frac{+5\pm \sqrt[]{25-24}}{6}$               

= $\frac{5+1}{6}=1,\frac{2}{3}$

3x² – 7x + 3 = 0

x = $\frac{7\pm \sqrt[]{49-39}}{6}=\frac{7\pm \sqrt[]{13}}{6}$

$-3x^2+7x-2$

$\frac{-7\pm \sqrt[]{49-24}}{-6}=\frac{7+5}{6}=2,\frac{1}{3}$  

3x² + 7x – 2 = -1

->3x² – 7x + 1 = 0

x =   $\frac{7\pm \sqrt[]{49-12}}{6}=\frac{7\pm \sqrt[]{37}}{6}$

I =   ${\displaystyle \underset{0}{\overset{6}{\int }}\left(\begin{array}{cc}\hfill \left(-2\right)\hfill & \hfill +1\hfill \end{array}\right)dx+{\displaystyle \underset{\frac{7-\sqrt[]{37}}{6}}{\overset{\frac{1}{3}}{\int }}\left(\begin{array}{cc}\hfill \left(-1\right)\hfill & \hfill +1\hfill \end{array}\right)dx}+{\displaystyle \underset{-\frac{1}{3}}{\overset{\frac{7-\sqrt[]{13}}{6}}{\int }}\left(\begin{array}{cc}\hfill 0\hfill & \hfill +1\hfill \end{array}\right)dx+{\displaystyle \underset{\frac{7-\sqrt[]{13}}{6}}{\overset{\frac{2}{3}}{\int }}\left(1+1\right)dx}}}+{\displaystyle \underset{\frac{2}{3}}{\overset{1}{\int }}\left(1+1\right)dx}$

$=\frac{\sqrt[]{37}+\sqrt[]{13}-4}{6}$

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

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

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

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

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

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

L₁ : 3x – 4y + 12 = 0

L₂ : 8x + 6y + 11 = 0

$\left(\alpha ,\beta \right)$ lies on that angle which contain origin

$\therefore$ Equation of angle bisector of that angle which contain origin is

$\frac{3x-4y+12}{5}=\frac{8x+6y+11}{10}$

$\Rightarrow 2x+14y-13=0$

$\left(\alpha ,\beta \right)$ lies on it

$\Rightarrow 2\alpha +14\beta -13=0$ …… (i)

$\Rightarrow 3\alpha -4\beta +7=0$ ……. (ii)

Solving (i) & (ii)

$\alpha =-\frac{23}{25}\&\beta =\frac{53}{50}$

$\therefore \alpha +\beta =\frac{7}{50}$

$\Rightarrow 100\left(\alpha +\beta \right)=14$

$l=\underset{n\to x}{lim}\frac{1}{2^n}\left(\frac{1}{\sqrt[]{1-\frac{1}{2^n}}}+\frac{1}{\sqrt[]{1-\frac{2}{2^n}}}+\frac{1}{\sqrt[]{1-\frac{3}{2^n}}}+.....+\frac{1}{\sqrt[]{1-\frac{2^n-1}{2^n}}}\right)$

Let 2n = t and if n $\infty$ then t $\infty$

$l=\underset{n\to x}{lim}\frac{1}{t}\left({\displaystyle \sum _{r=1}^{t=1}\frac{1}{\sqrt[]{1+\frac{r}{t}}}}\right)$

$={\left[2x^{\frac{1}{2}}\right]}_0^1=2$

$\underset{x\to \frac{x}{4}}{lim}\frac{8\sqrt[]{2}-{\left(cosx+sinx\right)}^7}{\sqrt[]{2}-\sqrt[]{2}sin2x}\left(\frac{0}{0}form\right)$

$=\underset{x\to \frac{x}{4}}{lim}\frac{-7{\left(cosx+sinx\right)}^6\left(-sinx+cosx\right)}{-2\sqrt[]{2}cos2x}$ $\underset{x\to \frac{\pi }{4}}{Lt}\frac{{\left(cosx+sinx\right)}^5\left(co{s}^{2}x-si{n}^{2}x\right)}{2\sqrt[]{2}cos2x}=\frac{7{\left(\sqrt[]{2}\right)}^5}{2\sqrt[]{2}}=14$

${\displaystyle \sum _{n=1}^{21}\frac{3}{\left(4n-1\right)\left(4n+3\right)}}$

$=\frac{3}{4}{\displaystyle \sum _{n=1}^{21}\frac{\left(4n+3\right)-\left(4n-1\right)}{\left(4n-1\right)\left(4n+3\right)}}$

$=\frac{3}{4}\frac{4n+3-2}{3\left(4n+3\right)}=\frac{n}{4n+3}$

for n = 21

$S_{21}=\frac{21}{84+3}=\frac{21}{87}=\frac{7}{29}$