Ncert Solutions Maths class 12th

  $E:\frac{x^2}{4}+\frac{y^2}{2}=1$

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

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

GOF is differentiable at x = 0

So R.H.D = L.H.D.

$\frac{d}{dx}\left(4e^x+k_2\right)=\frac{d}{dx}\left({\left(-\left|x+3\right|\right)}^2-k_1\left|x+3\right|\right)$                

⇒ 4 = 6 – k₁ Þ k₁ = 2

Now g (f (-4) + g (f (4)

= 2 (2e⁴ – 1)

The direction of ratios of the lines,   $\frac{x-3}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\&\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ , are $-3,2k,2and3k,1,-5$ respectively.

It is known that two lines with direction ratios,   $a_1, b_1, c_1 and a_2, b_2,c_2$ , are perpendicular, if  $a_1{a}_2 + b_1{b}_2 + c_1{c}_2 =0$

$\begin{array}{l}\therefore -3(3k)+2k*1+2(-5)=0\\ \Rightarrow -9k+2k-10=0\\ \Rightarrow 7k=-10\\ \Rightarrow k=\frac{-10}{7}\end{array}$

Therefore, for k= -10/7, the given lines are perpendicular to each other.

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