Maths

Area of $\Delta ABC$

= $\frac{1}{2}AB\cdot BC$

$=\frac{1}{2}\cdot \sqrt[]{2}\cdot 1$

$=\frac{1}{\sqrt[]{2}}$

now required area

$={\displaystyle \underset{0}{\overset{4}{\int }}\left(2\sqrt[]{2}-\sqrt[]{2}x\right)dx-\frac{1}{\sqrt[]{2}}}$

$=\frac{32\sqrt[]{2}}{3}=8\sqrt[]{2}-\frac{1}{\sqrt[]{2}}$

$=\frac{13\sqrt[]{2}}{6}$

 $\overrightarrow{a}\cdot \widehat{c}=\frac{\alpha +6+2}{\sqrt[]{1}+4+4}$

$\frac{10}{3}=\frac{8+\alpha }{3}\Rightarrow \alpha =2$

$\overrightarrow{b}*\overrightarrow{c}=\widehat{i}\left(2\beta -8\right)+\widehat{j}\left(10\right)+\widehat{k}\left(6+\beta \right)$

$=-6\widehat{i}+10\widehat{j}+7\widehat{k}$

So = 1

Draw g(t) = t³ – 3t

g'(t) = 3(t² – 1)

g(1) is maximum in (-2, 2)

So, maximum (t³ – 3t) = $\{\begin{array}{cc}\hfill t^3-3t\hfill & \hfill ;-2<t<-1\hfill \\ \hfill 2\hfill & \hfill ;-1<t<2\hfill \end{array}$

$I={\displaystyle \underset{-2}{\overset{2}{\int }}f}\left(x\right)dx$

= ${\displaystyle \underset{-2}{\overset{-1}{\int }}\left(t^3-3t\right)dt}+{\displaystyle \underset{-1}{\overset{2}{\int }}2dt}$

I = $\frac{27}{4}$

again rewrite the f(x)

$f\left(x\right)=\left\{\begin{array}{ccc}\hfill \begin{array}{cc}\hfill x^3-3x\hfill & \hfill 2\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{cc}\hfill x\le -1\hfill & \hfill -1<x\le 2\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\hfill x^2+2x-6\hfill & \hfill 9\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{cc}\hfill 2<x<3\hfill & \hfill 3\le x<4\hfill \end{array}\hfill \\ \hfill \begin{array}{ccc}\hfill 10\hfill & \hfill 11\hfill & \hfill 2x+1\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{ccc}\hfill 4\le x<5\hfill & \hfill x=5\hfill & \hfill x>5\hfill \end{array}\hfill \end{array}\right\}$

$f\text{'}\left(x\right)=\left\{\begin{array}{cccccc}\hfill 3x^2-3\hfill & \hfill ;\hfill & \hfill x<-1\hfill & \hfill 0\hfill & \hfill ;\hfill & \hfill -1<x<2\hfill \\ \hfill 2x+2\hfill & \hfill ;\hfill & \hfill 2<x<3\hfill & \hfill 0\hfill & \hfill ;\hfill & \hfill 3<x<4\hfill \\ \hfill 0\hfill & \hfill ;\hfill & \hfill 4<x<5\hfill & \hfill 2\hfill & \hfill ;\hfill & \hfill x>5\hfill \end{array}\right\}$

So f(x) is not differentiable at x = 2, 3, 4, 5

so m = 4

Image of pt (2,4,7) in the plane 3x – y + 4z = 2 is

$\frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=\frac{-2\left(6-4+28-2\right)}{3^2+{\left(-1\right)}^2+4^2}$

Let $\frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=\frac{-28}{13}=\lambda$

$x=3\lambda +2y=-\lambda +4z=4\lambda +7$

Now according to the question

(a, b, c) = $\left(3\lambda +2,-\lambda +4,4\lambda +7\right)$

Now $2a+b+2c=6\lambda +4-\lambda +4+8\lambda +14$

= $13\lambda +22$

= 28 + 22 [Use $\lambda =\frac{-28}{13}$ ]

= 6

 $\frac{dy}{dx}=\frac{y}{x}+\frac{\sqrt[]{y^2+16x^2}}{x}$

Put y = $V\cdot x$

differentiable worst = x

$\frac{dy}{dx}=V+x\cdot \frac{dV}{dx}$

V + $x\frac{dV}{dx}=V+\sqrt[]{V^2+16}$

$x\frac{dV}{dx}=\sqrt[]{V^2+16}$

apply variable separable method

${\displaystyle \int \frac{dV}{\sqrt[]{V^2+16}}={\displaystyle \int \frac{dx}{x}+\mathcal{l}nC}}$

$\Rightarrow \mathcal{l}n\left|V+\sqrt[]{V^2+16}\right|=\mathcal{l}nCx$

$\frac{y}{x}+\frac{\sqrt[]{y^2+16x^2}}{x}=Cx$

Given y(1) = 3 C = 8

Now at x = 2

$\frac{y}{2}+\frac{\sqrt[]{y^2+16.4}}{2}=8.2$

y2 + 64 = (32 – y)2

y = 15

Set of first 10 prime numbers

= {2,3,5,7,11,13,17,19,23,29,31}

So sample space = 104.

Favourable cases

So required probability

$=\frac{10+4{*}^{10}\text{?}{C}_2}{10^4}=\frac{10+\frac{4*10.9}{2}}{10^4}=\frac{19}{1000}$

Let PT perpendicular to QR

$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}=\lambda \Rightarrow T\left(2\lambda -1,3\lambda -2,2\lambda +1\right)$ therefore

$2\left(2\lambda -5\right)+3\left(3\lambda -4\right)+2\left(2\lambda -6\right)=0\Rightarrow \lambda =2$

$T\left(3,4,5\right)\therefore PT=\sqrt[]{1+4+4}=3\therefore QT=\sqrt[]{26-9}=\sqrt[]{17}$

$\therefore \Delta PQR=\frac{1}{2}*2\sqrt[]{17}*3=3\sqrt[]{17}$

Therefore square of $ar\left(\Delta PQR\right)$ = 153.

 $f\text{'}\left(x\right)=\frac{4x^2?1}{x}$ so f (x) is decreasing in $\left(0,\frac{1}{2}\right)and\text{?}\text{?}\left(\frac{1}{2},?\right)$ $?a=\frac{1}{2}$

Tangent at y² = 2x is y = mx + $\frac{1}{2m}$ it is passing through (4, 3) therefore we get m = $\frac{1}{2}or\text{?}\text{?}\frac{1}{4}$

So tangent may be $y=\frac{1}{2}x+1\text{?}\text{?}or\text{?}\text{?}y=\frac{1}{4}x+2\text{?}\text{?}\text{?}but\text{?}\text{?}y=\frac{1}{2}x+1$ passes through (-2, 0) so rejected.

Equation of normal $\frac{x}{9}+\frac{y}{36}=1$

Slope of AH = $\frac{a+2}{1}$ slope of BC = $-\frac{1}{p}\therefore p=a+2$ $\therefore C\left(\frac{18p-30}{p+1},\frac{15p-33}{p+1}\right)$

slope of HC = $\frac{16p-p^2-31}{16p-32}$

slope of BC * slope of HC = -1 p = 3 or 5

hence p = 3 is only possible value.

 ${\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}={\left(\frac{f\left(x\right)}{x}\right)}^3\Rightarrow x^3{\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}=f^3\left(x\right),$ differentiating w.r.to x

$x^{3}f\left(x\right)+3x^2\frac{f^3\left(x\right)}{x^3}=3f^2\left(x\right)f\text{'}\left(x\right)\Rightarrow 3y^2\frac{dy}{dx}=x^{3}y=\frac{3y^3}{x}\Rightarrow 3xy\frac{dy}{dx}=x^4+3y^2$

After solving we get $y^2=\frac{x^4}{3}+c{x}^2$ also curve passes through (3, 3) c = -2

$\therefore y^2=\frac{x^4}{3}-2x^2$ which passes through $\left(\alpha ,6\sqrt[]{10}\right)$ $\therefore \frac{{\alpha }^4-6{\alpha }^2}{3}=360\Rightarrow \alpha =6$