Maths

For $x^2+\alpha x=\beta >0\forall x\in R$  to hold, we should have   ${\alpha }^2-4\beta <0$

If $\alpha =1,\beta$  can be 1, 2, 3, 4, 5, 6 i.e., 6 choices

If a  = 4, b can be 5 or 6 i.e., 2 choices

hence total favourable outcomes

= 6 + 5 + 4 + 2 + 0 + 0 = 17

Required probability =  $\frac{17}{36}$

If n is number of trials, p is probability of success and q is probability of unsuccess then,

Mean = np and variance = npq.

Here np + npq = 24         …. (i)

np. npq = 128                   …. (ii)

and q = 1 – p                    …… (iii)

From eq. (i), (ii) and (iii) :

$=\left(32+\frac{32*31}{2}\right).\frac{1}{2^{32}}$      

$=\frac{33}{2^{28}}$

  $?\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$ …….(i)

then

$\overrightarrow{a}+\overrightarrow{c}=-\overrightarrow{b}$

then

$\left(\overrightarrow{a}+\overrightarrow{c}\right)*\overrightarrow{b}=-\overrightarrow{b}*\overline{b}$

$\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$  $\therefore \overrightarrow{a}*\overrightarrow{b}+\overrightarrow{c}*\overrightarrow{b}=\overrightarrow{0}$ …….(ii)

Now (S1) : $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\text{exception 25:}$

$\left|\overrightarrow{a}*\overrightarrow{b}+\overrightarrow{c}*\overrightarrow{b}\right|-\left|\overrightarrow{c}\right|=6\left(2\sqrt[]{2}-1\right)$

$\therefore cos\left(\angle ACB\right)=\sqrt[]{\frac{2}{3}}$

$\therefore \angle ACB=co{s}^{-1}\sqrt[]{\frac{2}{3}}$

$$ $\therefore S\left(2\right)$  is correct

Let a, b, c be direction ratios of plane containing lines

$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$

and

$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$

Equation of plane P is : 1 (x – 3) 1 (y + 4) + 2 (z – 7) = 0

$\Rightarrow x-y+2z-21=0$

Distance from point (2, 5, 11) is

$d=\frac{\left|2+5+22-2\right|}{\sqrt[]{6}}$

$\therefore d^2=\frac{32}{3}$

Radius of circle S touching x-axis and centre  $\left(\alpha ,\beta \right)$ $$ is |. According to given conditions

${\alpha }^2+{\left(\beta -1\right)}^2={\left(\left|\beta \right|+1\right)}^2$

${\alpha }^2=4\beta as\beta >0$

$$ $\therefore$  Required locus is L : x2 = 4y

The area of shaded region = $2{\displaystyle {\int }_0^{4}2\sqrt[]{y}dy}$ ${\displaystyle {}_{}^{}\text{exception 25:}}$

= $4.{\left[\frac{\frac{y^{\frac{3}{2}}}{3}}{2}\right]}_0^4$ ${\frac{\frac{{}^{\frac{}{}}}{}}{}}_{}^{}$

= $\frac{64}{3}$ $\frac{}{}$ square units.

Let inclination of required line is,

So the coordinates of point B can be assumed as

$\left(4-\frac{\sqrt[]{29}}{3}cos\theta ,3-\frac{\sqrt[]{29}}{3}sin\theta \right)$

which satisfies x – y – 2 = 0

$sin\theta -cos\theta =\frac{3}{\sqrt[]{29}}$

By squaring

$sin2\theta =\frac{20}{29}=\frac{2tan\theta }{1+ta{n}^2\theta }$

Point B : $\left(\frac{10}{3},\frac{4}{3}\right)$ $\frac{}{}\frac{}{}$

which also satisfies x + 2y = 6

$\left(x-y^2\right)dx+y\left(5x+y^2\right)dy=0$                

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

Let y² = t

$\frac{1}{2}.\frac{dt}{dx}=\frac{t-x}{5x=t}$  

Now substitute, t = yx

$\frac{dt}{dx}=v+x\frac{dv}{dx}$

  $\left|\frac{{\left(v+1\right)}^4}{{\left(v+2\right)}^3}\right|=\frac{C}{x}$

$\left|{\left(y^2+x\right)}^4\right|=C\left|{\left(y^2+2x\right)}^3\right|$    

$\frac{dy}{dx}=\frac{2e^{2x}-6e^{-x}+9}{2+9e^{-2x}}=e^{2x}-\frac{6e^{-x}}{2+9e^{-2x}}$

${\displaystyle \int dy={\displaystyle \int e^{2x}dx-3{\displaystyle \int \frac{e^{-z}}{\underset{put{e}^{-x}=t}{\underset{?}{1+\left(\frac{3e^{-x}}{\sqrt[]{2}}\right)}}}}dx}}$

= $y=\frac{e^{2x}}{2}=\sqrt[]{2}ta{n}^{-1}\left(\frac{3e^{-x}}{\sqrt[]{2}}\right)+C$

It is given that the curve passes through

$\left(0,\frac{1}{2}+\frac{\pi }{2\sqrt[]{2}}\right)$

$\frac{3}{\sqrt[]{2}}{e}^{\alpha }-\frac{3}{\sqrt[]{2}}=e^{\alpha }+\frac{9}{2}$

$e^{\alpha }=\frac{\frac{9}{2}+\frac{3}{\sqrt[]{2}}}{\frac{3}{\sqrt[]{2}}-1}=\frac{3}{\sqrt[]{2}}\left(\frac{3+\sqrt[]{2}}{3-\sqrt[]{2}}\right)$

$f\left(x\right)=\{\begin{array}{l}x-\left[x\right]if\left[x\right]isodd\\ 1+\left[x\right]-x,if\left[x\right]iseven\end{array}$

Graph of f (x)

So

$=2{\pi }^2{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-x\right)cos\pi xdx}$

=4

$A=\left\{\left(x,y\right):x^2\le y\le min\{x+2,4-3x\right\}$

So, area of the required region

$A={\displaystyle \underset{-1}{\overset{\frac{1}{2}1}{\int }}\left(x+2-x^2\right)dx+{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}\left(4-3x-x^2\right)}dx}$

$={\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}^{\frac{1}{2}}+{\left[4x-\frac{3x^2}{2}-\frac{x^3}{3}\right]}^1$

= $\frac{17}{6}$