Class 12th

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

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

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

Given system of equations

αx + y + z = 5

x + 2y + 3z = 4, has infinite solution

x + 3y + 5z =β

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

α (1) – 1 (2) + 1 (1) = 0

α= 1 and

${\Delta }_3=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill \beta \hfill \end{array}\right|=0$

β= 3

$\therefore \left(\alpha ,\beta \right)=\left(1,3\right)$

f : R defined as

$f\left(x\right)=x-1andg\left(x\right):R-\left\{1,-1\right\}\to R$

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

$\therefore fog\left(x\right)=\frac{x^2}{x^2-1}-1=\frac{1}{x^2-1}$

domain of fog (x) R – {-1, 1} and range   $(-\infty ,-1]\cup (0,\infty )$ $\therefore$ $$

$$ fog (x) is neither one-one nor onto

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

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

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

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)

$\underset{x\to \frac{1}{\sqrt[]{2}}}{lim}\frac{sin\left(co{s}^{-1}x\right)-x}{1-tan\left(co{s}^{-1}x\right)}$

let $co{s}^{-1}x=\frac{\pi }{4}+\theta$

= $\underset{\theta \to \infty }{lim}\frac{\text{exception 25:}sin\theta }{-2tan\theta }\left(1-tan\theta \right)=-$$\frac{1}{}$

 $f\left(x\right)=\frac{x-1}{x+1}\Rightarrow f\left(f\left(x\right)\right)=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=-\frac{1}{x}$

$f^3\left(x\right)=-\frac{x+1}{x-1}\Rightarrow f^4\left(x\right)=\frac{\frac{x-1}{x+1}+1}{\frac{x-1}{x+1}-1}=x$

$So,f^6\left(6\right)+f^7\left(7\right)=f^2\left(6\right)+f^3\left(7\right)$

= $-\frac{1}{6}-\frac{7+1}{7-1}=-\frac{9}{6}=-\frac{3}{2}$