Ncert Solutions Maths class 11th

Ellipse : $\frac{x^2}{16}+\frac{y^2}{7}=1$

Eccentricity = $\sqrt[]{1-\frac{7}{10}}=\frac{3}{4}$

Foci $\equiv \left(\pm ae,0\right)$ $\equiv \left(\pm 3,0\right)$

Hyperbola : $\frac{x^2}{\left(\frac{144}{25}\right)}-\frac{y^2}{\left(\frac{\alpha }{25}\right)}=1$

Eccentricity = $\sqrt[]{1+\frac{\alpha }{144}}=\frac{1}{12}\sqrt[]{144+\alpha }$

$=2.\frac{\frac{81}{25}}{\frac{12}{5}}=\frac{27}{10}$

$y^2-2x-2y=1$

$\Rightarrow {\left(y-1\right)}^2=2\left(x+1\right)$ …. (i)

Equation of tangent at A is 2x – y – 5 = 0 ………. (ii)

D is mid point of AB solving (ii) with y = 1 P (3, 1)

$\therefore PD=4,AD=2$

$Areaof\Delta APD=\frac{1}{2}\left(PD\right)\left(AD\right)=4$

$\therefore Areaof\Delta APB=8sq.units$

$\left(A\right)\sim \left(p\wedge \left(\sim r\right)\right)\vee q=\left(\sim p\right)\vee \left(r\right)\vee q$

(B) $$ $q\vee \left(-r\vee p\right)$

(C)  $\left(\sim p\right)\vee \left(q\vee r\right)$

(D) $$ $\left(\sim p\vee q\right)\vee r$

Using Venn diagram we get B as the correct option.

Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through the points (7, 0) & (0,  $-2\sqrt[]{6}$ )

$\therefore a^2=49\&b^2=24$

$\therefore e=\sqrt[]{1-\frac{b^2}{a^2}}=\sqrt[]{1-\frac{24}{49}}=\frac{5}{7}$

A(2, 3, 9)

B(5, 2, 1)

C(1, $\lambda$ , 8)

D( $\lambda$ ,2, 3)

$\Delta =\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -8\hfill \\ \hfill -4\hfill & \hfill \lambda -2\hfill & \hfill 7\hfill \\ \hfill \lambda -2\hfill & \hfill -1\hfill & \hfill -6\hfill \end{array}\right|$           

    $\overrightarrow{AB}=\left(3,-1,-8\right)$

$\overrightarrow{AC}=\left(-4,\lambda -2,7\right)$            

$\overrightarrow{AD}=\left(\lambda -2,-1,-6\right)$

$\Delta =3\left[-6\lambda +12+7\right]-1\left(7\lambda -14-24\right)-8\left(4-{\left(\lambda -2\right)}^2\right)$

$=57-18\lambda \Rightarrow +38-32+8\left({\lambda }^2-4\lambda +4\right)$            

$=95-57\lambda +8{\lambda }^2$

$\Delta =0\Rightarrow {\lambda }_1{\lambda }_2=\frac{95}{8}$                       

x + 2y + z = 14

${\perp }^{r}linePQ:\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{1}=t$              

Q (1 + t, 2 + 2t, 3 + t)

x + 2y + z = 14 -> 1 + t + 4 + 4t + 3 + t = 14 Þ 6t = 6

t = 1

-> Q (2, 4, 4)

PQ = $\sqrt[]{1+4+1}=\sqrt[]{6}$  

$tan60°=\frac{PQ}{QR}\Rightarrow QR=\frac{PQ}{\sqrt[]{3}}=\sqrt[]{2}$

ar (PQR) = $\frac{1}{2}\sqrt[]{6}\cdot \sqrt[]{2}=\sqrt[]{3}$  

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 =   $\frac{\Delta }{s}=\frac{24}{12}=2$

$a_{n+2}{a}_{n+1}-a_{n+1}{a}_n=2$

Series will satisfy

$a_1{a}_2,a_2{a}_3,a_3{a}_4,......a_4{a}_5$

$\begin{array}{l}1.22.22.32.4\\ \frac{a_n+\frac{1}{a_{n+1}}}{a_{n+2}}=\frac{a_{n+2}-\frac{1}{a_{n+1}}}{a_{n+2}}\end{array}$

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

Now, proof = $\frac{30}{11}\frac{\left(2r+1\right)}{2\left(r+1\right)}$ $\frac{}{}\frac{}{}$

r = 1

$=\frac{\left(1.3.5......61\right)}{2^{30}\left(2.3......31\right)}$

$\frac{\overline{)61}}{2^{60}.\overline{)31}.\overline{)30}}=\alpha =-60$

$x^2+y^2-x+2y=\frac{11}{4}$

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

or $\Delta PQRPR=QRsin2\ge \frac{1}{3}$

$=4.6sin\frac{\pi }{8}$

$as\Delta PQR=\frac{1}{2}PR*PQ$

$=4sin\frac{\pi }{4}=\frac{4}{\sqrt[]{2}}=2\sqrt[]{2}$