Maths

Area of shaded region

$2{\displaystyle \underset{0}{\overset{\sqrt[]{3}}{\int }}\left(2x^2+9-5x^2\right)dx}$

$=2{\displaystyle \underset{0}{\overset{\sqrt[]{3}}{\int }}\left(9-3x^2\right)dx}$

${\displaystyle \underset{0}{\overset{\sqrt[]{3}}{\int }}\left(3-x^2\right)dx=6{\left[3x-\frac{x^3}{3}\right]}_0^{\sqrt[]{3}}=6\left[\frac{9\sqrt[]{3}-3\sqrt[]{3}}{3}\right]}=12\sqrt[]{3}$

 $f\left(x\right)=\left\{\begin{array}{cc}\hfill min\left\{\left|x\right|,2-x^2\right\},\hfill & \hfill -2\le x\le 2\hfill \\ \hfill \left[\left|x\right|\right],\hfill & \hfill 2\le \left|x\right|\le 3\hfill \end{array}\right\}$

Number of points where f is not differentiable = 5

$I=3{\displaystyle \underset{-2}{\overset{2}{\int }}\left|x^2-x-2\right|dx}$

$=3\left({\displaystyle \underset{-2}{\overset{-1}{\int }}\left(x^2-x-2\right)dx-{\displaystyle \underset{-1}{\overset{2}{\int }}\left(x^2-x-2\right)dx}}\right)$

$=3\left[\left(\frac{7}{6}+\frac{2}{3}\right)-\left\{-\frac{10}{3}-\frac{7}{6}\right\}\right]$

$=3\left(\frac{11}{6}+\frac{27}{6}\right)=\frac{38}{2}=19$

Total possibilities = 2⁵ * 2⁵

Farounable case = ⁵C₂ * 3³ = 10 * 3³

$\therefore requiredprobability=\frac{10*3^3}{2^5*2^5}=\frac{5*27}{2^9}=\frac{135}{2^9}$

$2x{y}^2-y)dx+xdy=0$

$\Rightarrow \frac{dy}{dx}+2y^2-\frac{y}{x}=0$

$Put\frac{1}{y}=z$

Then $-\frac{1}{y^2}\frac{dy}{dx}=\frac{dz}{dx}$

$\frac{dz}{dx}+\frac{1}{x}z=2$

Point (2, 1) c = 2 – 4 = 2 y = $\frac{x}{x^2-2}$

$\left|y\left(1\right)\right|=1$

let f : R -> R

$f\text{'}\left(x\right)=\{\begin{array}{l}-55,x<-5\\ 6x^2-6x-120,-5<x<4\\ 6x^2-6x-36,x>4\end{array}$            

f' (x) increasing in x $\in \left(-5,-4\right)\cup \left(4,\infty \right)$  

A tangent to y² = 4x is x – ty + t² = 0

$\frac{3+t^2}{\sqrt[]{1+t^2}}=3$

(3 + t²)² = 9 (1 + t²)

$9+t^4+6t^2=9+9t^2$

Point of contact $\left(3,2\sqrt[]{3}\right)=\left(a,b\right)$

$x-\sqrt[]{3}y+3=0$

$\frac{\sqrt[]{3}x+y-3\sqrt[]{3}=0]}{4x-6=0}$

$x=\frac{3}{2},y=\frac{\sqrt[]{3}}{2}+\sqrt[]{3}$

& $\left(\frac{3}{2},\frac{\sqrt[]{3}}{2}+2\right)=\left(c,d\right),2\left(a+c\right)=2\left(3+\frac{3}{2}\right)=9$

 $\underset{x\to 0}{lim}\frac{ax-\left(e^{4x}-1\right)}{ax\left(e^{4x}-1\right)}=b,$ use of L' Hospital rule implies

$\underset{x\to 0}{lim}\frac{a-4e^{4x}}{a\left(e^{4x}-1\right)+ax\left(4e^{4x}\right)}$

$=\frac{a-4}{0}\Rightarrow a=4$

$\Rightarrow \underset{n\text{?}\to 0}{lim}\frac{4\left(-4.e^{4x}\right)}{4.4e^{4x}+16e^{4x}+16x.4e^{4x}}$

$=\frac{-16}{16+16}=-\frac{1}{2}=b$

a – 2b = 4 – (1) = 5

${\mathcal{l}}_1:\frac{x-3}{1}=\frac{y+1}{2}=\frac{z-4}{2}=t$

${\mathcal{l}}_2:\frac{x-3}{2}=\frac{y-3}{2}=\frac{z-2}{1}=s$

$\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=\left(-2\widehat{i}+3\widehat{j}-2\widehat{k}\right)$

$\mathcal{l}:\overrightarrow{r}=\overrightarrow{0}+\lambda \left(-2\widehat{i}+3\widehat{j}-2\widehat{k}\right)$

$\mathcal{l}\&{\mathcal{l}}_1\to$

Point of intersection $\mathcal{l}\&{\mathcal{l}}_1$ is P (2, 3, 2)

Point Q on ${\mathcal{l}}_2$ is (3 + 2s, 3 + 2s, 2 + s).

  $\frac{a+2+a}{3}=\frac{10}{3}$

2a + 2 = 0

2a = 8 -> a = 4       .(i)

and            $\frac{c+b+b}{3}=\frac{7}{3}$

2b + c = 7 .(ii)

Since         a, b, c are in A.P.

2b = a + c

From (i)   $\therefore$ 2b = 4 + c .(iii)

Solving (ii) and (iii)

4 + c + c = 7

2c = 3

$c=\frac{3}{2}$    

$\therefore 2b=4+\frac{3}{2}=\frac{11}{2}$     

$\therefore b=\frac{11}{4}$  

As per question

$\alpha +\beta =\frac{-b}{a}and\alpha \beta =\frac{1}{a}$       

$=\frac{121-192}{256}=\frac{-71}{256}$