Class 12th

  $A\left(a\right)=2{\displaystyle \underset{0}{\overset{\sqrt[]{1-a}}{\int }}\left(\left(1-x^2\right)-a\right)dx=\frac{4}{3}{\left(1-a\right)}^{3/2}}$  

  $\therefore A\left(0\right)=\frac{4}{3}$            

and    ^{$A\left(\frac{1}{2}\right)=\frac{4}{3}{\left(\frac{1}{2}\right)}^{\frac{3}{2}}\Rightarrow \frac{A\left(0\right)}{A\left(\frac{1}{2}\right)}=2\sqrt[]{2}$}

As observer is at O So height of water observed by observer

$\Rightarrow \frac{H}{{\mu }_{\omega }}=\frac{H}{\left(4/3\right)}=\frac{3H}{4}$

given diagram (17.5 H) is height of observer

$So\frac{3H}{4}=17.5-H$

$\frac{7H}{4}=17.5$

7H = 70

H = 10

$\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}\right|=\sum {\left|\overrightarrow{a}\right|}^2+2\sum \overrightarrow{a}.\overrightarrow{b}$  

$=4+2\overrightarrow{d}.\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)$  (a, b, c are mutually ^ r)

Let   $\overrightarrow{d}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}+v\overrightarrow{c}$

Also   ${\lambda }^2+{\mu }^2+v^2=1=3co{s}^2\theta orcos\theta =\frac{1}{\sqrt[]{3}}$

$\therefore {\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}\right|}^2=4\pm 2.\frac{3}{\sqrt[]{3}}$        

$=4\pm 2\frac{3}{\sqrt[]{3}}$       

$y(ydx+xdy)=x^5\left(\frac{xdy-ydx}{x^2}\right)$

$\Rightarrow d\left(xy\right)=x^5{y}^{-1}\left(\frac{xdy-ydx}{x^2}\right)\Rightarrow d\left(xy\right)=(xy{)}^{\alpha }{\left(\frac{y}{x}\right)}^{\beta }d\left(\frac{y}{x}\right)$

$\alpha -\beta =5\mathrm{}\alpha +\beta =-1$

$2\alpha =4\mathrm{}\alpha =2\mathrm{}\beta =-3$

$\Rightarrow \mathrm{}\left(xy{)}^{-2}d\right(xy)={\left(\frac{y}{x}\right)}^{-3}d\left(\frac{y}{x}\right)\Rightarrow -(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}+c$

$\Rightarrow \mathrm{}-(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}+c\mathrm{}$ Passing $\left(\mathrm{1,1}\right)$

$\Rightarrow \mathrm{}-1=-\frac{1}{2}+c\Rightarrow c=-\frac{1}{2};-(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}-\frac{1}{2}$

$Inx\in \left(-2,2\right)$  

 |x|- 1| is not differentiable at x = -1, 0, 1

|cospx| is not differentiable at x =  $\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}$ -  

$\int {\mathrm{c}\mathrm{o}\mathrm{t}}^{-1}?\left(\frac{1}{1+x}\right)dx$

$=\int {\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)dx$

$=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\int \frac{x}{1+(1+x{)}^2}dx;=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\int \frac{x}{x^2+2x+2}dx$

$=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\int \frac{\frac{1}{2}(2x+2)-1}{x^2+2x+2}dx=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\frac{1}{2}\mathrm{l}\mathrm{o}\mathrm{g}?\left(x^2+2x+2\right)+\int \frac{dx}{(x+1{)}^2+1}$

$=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\frac{1}{2}\mathrm{l}\mathrm{o}\mathrm{g}?\left(x^2+2x+2\right)+{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(x+1)+c$

$A^2{\left|A\right|}^2-{\left|B\right|}^{3.2}B$

$=25A^2-\left|B\right|.B$

$\begin{array}{l}=25A^2+\left|B\right|A^2\\ =0\end{array}$

Put x³/² = t
√xdx = (2/3)dt
⇒ (2/3) ∫ dt/√ (1-t²) = (2/3)sin? ¹t + c
= (2/3)sin? ¹x³/² + c
⇒ g (x) = sin? ¹x
g (0) = 0

A: runner succeeded exactly 3 times out of 5.
B: runner succeeds on the first trial.
P (B/A) = P (B ∩ A) / P (A) = (p (? C? p² (1−p)²) / (? C? p³ (1-p)²) = 3/5 = 0.6

The equation of the line through P (x, y) making an angle with the x-axis which is supplementary to the angle made by the tangent at P (x, y) is

$Y-y=-\frac{dy}{dx}\left(X-x\right)$                    …. (1)

At the point where it meets the x-axis

Y = 0, X = x    $+\frac{y}{\frac{dy}{dx}}\Rightarrow OA=x+\frac{y}{\frac{dy}{dx}}$    …. (2)

The line through P (x, y) and perpendicular to (1) is

$Y-y=\frac{dx}{dy}\left(X-x\right)$           

At the point where it meets the y-axis

$X=0,Y=y=\frac{x}{\frac{dy}{dx}}\Rightarrow OB=y-\frac{x}{\frac{dy}{dx}}$           .

$\boxed{x^2+2xy-y^2=2}$