Ncert Solutions Maths class 12th

The given two lines are coplanar

$\therefore \left|\begin{array}{ccc}\hfill 0\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill -3\hfill \\ \hfill 1-\alpha \hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=0\Rightarrow \alpha =\frac{5}{3}$

Now, $\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill -3\hfill \end{array}\right|=\widehat{i}\left(-9\right)-\widehat{j}\left(2\right)+\widehat{k}\left(-6\right)=\left(9,2,6\right)$

Equation of plane :

$=\left|\frac{\left(9.\frac{5}{3}+0+0-13\right)}{\sqrt[]{81+36+4}}\right|=\frac{2}{\sqrt[]{121}}=\frac{2}{11}$

  $P_1:2x-y-52=0,P_2:3x-y+4z-7=0\text{?}\text{?}P$

Equation of plane passing through the line of intersection between planes p₁ = 0 & p₂ = 0 is

    $P:P_1+\lambda P_2$            

$\therefore P:8x-y+32-14=0$      

It passes through the point (1, 0, 2)

Given hyperbola :

$\frac{x^2}{a^2}-\frac{y^2}{9}=1$

$$ $?$  it passes through

$\left(8,3\sqrt[]{3}\right)$

$?\frac{64}{a^2}-\frac{27}{9}=1\Rightarrow a^2=16$

Now, equation of normal to hyperbola

$\frac{16x}{8}+\frac{9y}{3\sqrt[]{3}}=16+9$

$\text{exception 25:}$ $\left(-1,9\sqrt[]{3}\right)$  satisfied

  $E:\frac{x^2}{4}+\frac{y^2}{2}=1$

any pt on it is P  $\left(2cos\theta ,\sqrt[]{2}sin\theta \right)$

$\therefore$ M (h, k) be mid point of P & A (4, 3)

$\Rightarrow {\left(h-2\right)}^2+{\left(\frac{2k-3}{\sqrt[]{2}}\right)}^2=1$        

Required locus (x – 2)² +   $\frac{{\left(y-\frac{3}{2}\right)}^2}{\frac{1}{2}}=1$

$\therefore e=\sqrt[]{1-\frac{1}{2}}=\frac{1}{\sqrt[]{2}}$

Equation of any tangent to   $\frac{x^2}{16}+\frac{y^2}{9}=1$

is y = mx + $\sqrt[]{16m^2+9}$  if this line is also tangent to x² + y² = 12

then  $\sqrt[]{12}$ =   $\left|\frac{\sqrt[]{16m^2+9}}{\sqrt[]{1+m^2}}\right|$

^{$\Rightarrow 12+12m^2=16m^2+9$}

$\Rightarrow 12m^2=9$        

RM = $\left|\frac{3+7-5}{\sqrt[]{2}}\right|=\frac{5}{\sqrt[]{2}}$

$\mathcal{l}sin60°=\frac{5}{\sqrt[]{2}}\Rightarrow \mathcal{l}=5\sqrt[]{\frac{2}{3}}$

$\therefore Areaof\Delta PQR=\frac{\text{exception 25:}}{4}{\mathcal{l}}^2=$=25/2√3

Put x = cos 2θ

dx = -2 sin 2θ . dθ

$={\displaystyle \int \frac{1}{cos2\theta }tan\theta \left(-4sin\theta .cos\theta \right)d\theta }$

$g\left(\frac{1}{2}\right)=\mathcal{l}n\left|2-\sqrt[]{3}\right|+\frac{\pi }{3}$

= $\mathcal{l}n\left|\frac{\sqrt[]{3}-1}{\sqrt[]{3}+1}\right|+\frac{\pi }{3}$  

Let p₁ : y² = 8x

p₂ : y² = 16 (3 – x) = -16 (x – 3)

finding their intersection points.

y² = 8x & y² = -16 (x – 3)

8x = -16x + 48

$=2.{\displaystyle \underset{0}{\overset{4}{\int }}\left(3-\frac{y^2}{16}-\frac{y^2}{8}\right)dy}$

$=2{\left(3y-\frac{y^3}{3*16}-\frac{y^3}{3*8}\right)}_0^4$ = 16

$?s_1+s_2=k$                

76x² + 3πr² = k

$\therefore 152x\frac{dx}{dr}+6\pi r=0$

Now

$V=40x^3+\frac{2}{3}\pi r^3$

$\Rightarrow \left(\frac{x}{r}\right)=\frac{152}{3}\frac{1}{120}=\frac{19}{45}$

 $f\left(x\right)=\left|x^2-3x-2\right|-x$

$=\left|\left(x-\frac{3-\sqrt[]{17}}{2}\right)\left(x-\frac{3+\sqrt[]{17}}{2}\right)\right|-x$

$\Rightarrow f\left(x\right)=\left[\begin{array}{cc}\hfill x^2-4x-2-1\le x\le \frac{3-\sqrt[]{17}}{2}\hfill & \hfill -x^2+2x+2\frac{3-\sqrt[]{17}}{2}<x\le 2\hfill \end{array}\right]$

absolute minimum $f\left(\frac{3-\sqrt[]{17}}{2}\right)=\frac{-3+\sqrt[]{17}}{2}$

absolute maximum = 3

$\therefore sum3+\frac{-3+\sqrt[]{17}}{2}=\frac{3+\sqrt[]{17}}{2}$