Ncert Solutions Maths class 11th

sin x = 1 – sin² x

=> sin x = $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$  

draw y = sin x

y = $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.

First common term to both AP's is 9

t₇₈ of $\left(3,6,9,......\right)=78*3=234$  

t₅₉ of $\left(5,9,13,........\right)=5+\left(5-1\right)4=237$  

n^th common term $\le 234$  

9 + (n – 1) 12 $\le$  234

n < $\frac{237}{12}\Rightarrow n=19$  

Now sum of 19 terms with a = 9, d = 12

  $=\frac{19}{2}\left(2.9+\left(19-1\right)\cdot 12\right)=2223$

Data contradiction.

$\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=\left(\overrightarrow{a}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\overrightarrow{c}$

pva $\left(\sim rvp\right)$ $$

$\sim \left(p\vee q\right)\vee \left(\sim rvp\right)$

its negation as asked in question

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

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

= $\left(p\wedge r\wedge \sim p\right)\left[asp\wedge \sim pisfalse\right]$

Let base = b

$tan60°=\frac{h}{b}$

$tan30°=\frac{h-20}{b}$

$?{\mathcal{l}}_{1}and{\mathcal{l}}_2$ are perpendicular, so

$3*1+\left(-2\right)\left(\frac{\alpha }{2}\right)+0*2=0$

⇒a = 3

Now angle between  ${\mathcal{l}}_{2}and{\mathcal{l}}_3$ ,

$cos\theta =\frac{1\left(-3\right)+\frac{\alpha }{2}\left(-2\right)+2\left(4\right)}{\sqrt[]{1+\frac{{\alpha }^2}{4}+4}.\sqrt[]{9+4+16}}$

$\Rightarrow cos\theta =\frac{\frac{2}{29}}{2}\Rightarrow \theta =co{s}^{-1}\left(\frac{4}{29}\right)=se{c}^{-1}\left(\frac{29}{4}\right)$

Let P (at², 2 at) where

$a=\frac{3}{2}$

T : yt = x + at² so point Q is

$\left(-a,at-\frac{a}{t}\right)$

N : y = -tx + 2at + at³ passes through (5, -8)

$-8=-5t+3t+\frac{3}{2}{t}^3$

$3t^3-4t+16=0$

⇒ t = -2

  So ordinate of point Q is $-\frac{9}{4}$  

Mid point of BC is $\frac{1}{2}\left(5\widehat{i}+\left(\alpha -2\right)\widehat{j}+9\widehat{k}\right)$ $\frac{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$

$\overline{AB}=\widehat{i}+\left(\alpha -4\right)\widehat{j}+\widehat{k}$

$\overline{AC}=\widehat{i}+\left(-2-\alpha \right)\widehat{j}+\widehat{k}$

For = 1, $\overline{AB}$ $\stackrel{}{}$ and $\overline{AC}$ $\stackrel{}{}$ will be collinear. So for non collinearity

= 2

Equation of perpendicular bisector of AB is

$y-\frac{3}{2}=-\frac{1}{5}\left(x-\frac{5}{2}\right)\Rightarrow x+5y=10$

Solving it with equation of given circle

${\left(x-5\right)}^2+{\left(\frac{10-x}{5}-1\right)}^2=\frac{13}{2}$

$\Rightarrow x-5=\pm \frac{5}{2}\Rightarrow x=\frac{5}{2}or\frac{15}{2}$

But $x\ne \frac{5}{2}$

because AB is not the diameter.

So, centre will be

$\left(\frac{15}{2},\frac{1}{2}\right)$

Now,

$r^2={\left(\frac{15}{2}-2\right)}^2+{\left(\frac{1}{2}+1\right)}^2=\frac{65}{2}$

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

here AB = $\sqrt[]{2}$ $\text{exception 25:}$ , BC = 2, AC = $\sqrt[]{2}$ $\text{exception 25:}$

area = $\frac{1}{2}*\sqrt[]{2}*\sqrt[]{2}=1$