Ncert Solutions Maths class 12th

Data contradiction.

Common tangents

$T_1:y=mx\pm \sqrt[]{4m^2+9}$      

$T_2:y=mx\pm \sqrt[]{42m^2-13}$    

So, 4m² + 9 = 42 m² – 13 Þ m =   $\pm 2$

  $\Rightarrow c=\pm 5$              

So tangent  $y=\pm 2x\pm 5$

y = 2x + 5 does not pass through 4^th quadrant compare this tangent with T = 0 to get pt. of intersection

$\frac{x{x}_2}{42}-\frac{y{y}_2}{143}=1\Rightarrow x_2=\frac{-84}{5}$

Find a, α

$\left({\alpha }^2+{\alpha }^2-3\right)+{\left({\alpha }^2-{\alpha }^2-1\right)}^5=0$                

$\alpha =\sqrt[]{2}$      

zx differentiate the curve

$2x+2y.y^1+5{\left(x^2-y^2-1\right)}^4\left(2x-2y\right)=0$ ……. (i)

differentiate equation (i)

$\left(\alpha ,\alpha \right)\Rightarrow y\text{'}\text{'}=\frac{-23}{4\sqrt[]{2}}$          

$V=\frac{1}{3}\pi r^{2}h$

$V=\frac{\pi }{8}{h}^{3}ta{n}^2\alpha$

$\frac{dv}{dt}=\pi h^{2}ta{n}^2\alpha .\frac{dh}{dt}$

CSA = $\pi r\mathcal{l}$ $$

$\frac{d\left(CSA\right)}{dt}=\frac{15}{16}\pi 2h.\frac{dh}{dt}$

$\frac{15}{8}*\frac{2}{3}=5m^2/hrs$

take z = x + iy

$z^2+\overline{z}=0$

$\Rightarrow x^2-y^2+x+i2xy-yi=0$

$\Rightarrow x^2-y^2+x=0andy\left(2x-1\right)=0$

if y = 0 x = 0, 1

$ifx=\frac{1}{2}\Rightarrow y=\pm \frac{\sqrt[]{3}}{2}$

$\Sigma \left(Re\left(z\right)+lm\left(z\right)\right)=\left(0-1+\frac{1}{2}+\frac{1}{2}\right)+\left(0+0+\frac{\sqrt[]{3}}{2}-\frac{\sqrt[]{3}}{2}\right)=0$

$\left|f\left(x\right)\right|\le 800\Rightarrow 2n^2-n-1\le 800$

$\Rightarrow 2n^2-n-801\le 0$

${\displaystyle \sum _{x\in S}^{}f\left(x\right)={\displaystyle \sum _{}^{}\left(2x^2-x-1\right)}}$

$=2\left(19^2+18^2+........1^2+0^2+1^2+.....+20^2\right)$

= 10620

y⁵ – 9xy + 2x = 0

differentiate 5y⁴ – 9x $\frac{dy}{dx}$ 9y + 2 = 0

$\frac{dy}{dx}=\frac{9y-2}{5y^4-9x}$

For horizontal tangent $\frac{dy}{dx}=0\Rightarrow y=\frac{2}{9}$ which does not satisfy the equation so no horizontal

For vertical tangent $5y^4-9x=0$

m = 0, N = 2

Plane through 4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3

is $x\left(4a+2\lambda \right)+y\left(-1-5\lambda \right)+z\left(5-\lambda \right)=7a+3\lambda$

This plane contains 4, -1, 0

9a + 1 + 10 = 0…… (i)

Plane contains the line $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$

$4a+11\lambda +7=0$ ……. (ii)

From (i) & (ii) a = 1,  $\lambda$ =1

Equation of plane $\pi \equiv x+2y+3z-2=0$

$\Rightarrow$$7$$P$$+$$3$$-$$2$$P$$+$$4$$-$$1$$2$$P$$+$$9$$-$$2$$=$$0$$\Rightarrow$$P$$=2$$\mathrm{}$

$?x=sin\left(2ta{n}^{-1}\alpha \right)=\frac{2\alpha }{1+{\alpha }^2}$ ……. (i)

and

$y=sin\left(\frac{1}{2}ta{n}^{-1}\frac{4}{3}\right)=sin\left(si{n}^{-1}\frac{1}{\sqrt[]{5}}\right)=\frac{1}{\sqrt[]{5}}$

Now,

$y^2=1-x$

$\therefore \alpha =2,\frac{1}{2}\therefore {\displaystyle \sum _{a\in S}^{}16{\alpha }^3=16*2^3+16*\frac{1}{2^3}=130}$

$4x^3-3x{y}^2+6x^2-5xy-8y^2+9x+14=0$ differentiating both sides we get

$12x^2-3y^2-6xyy\text{'}+12x-5y-5xy=16yy\text{'}+9=0$

At the point (2, 3)

48 – 27 + 36y' – 24 – 15 + 10y' – 48y' + 9 = 0

$\therefore Area=\frac{1}{2}*Base*Height$

$A=\frac{1}{2}*\left(\frac{-4}{3}+\frac{31}{2}\right)\left(3\right)=\frac{1}{2}\left(\frac{85}{0}\right).3=\frac{85}{4}=8A=170$