Maths

Given, mean = np = a. and variance = npq = $\frac{\alpha }{3}\Rightarrow q=\frac{1}{3}andp=\frac{2}{3}$   

  $P\left(X=1\right)=n{p}^1{q}^{n-1}=\frac{4}{243}\Rightarrow n{\left(\frac{2}{3}\right)}^1{\left(\frac{1}{3}\right)}^{n-1}=\frac{4}{243}\Rightarrow n=6$  

  $P\left(X=4or5\right){=}^6{C}_4{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^2{+}^6{C}_5{\left(\frac{2}{3}\right)}^5{\left(\frac{1}{3}\right)}^1=\frac{16}{27}$  

Given : $\overrightarrow{a}=\left(\alpha ,1,-1\right)and\overrightarrow{b}=\left(2,1,-\alpha \right)\overrightarrow{c}=\overrightarrow{a}*\overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \alpha \hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill -\alpha \hfill \end{array}\right|$  

$=\left(-\alpha +1\right)\widehat{i}+\left({\alpha }^2-2\right)\widehat{j}+\left(\alpha -2\right)\widehat{k}$ Projection of $\overrightarrow{c}$ on $\overrightarrow{d}=-\widehat{i}+2\widehat{j}-2\widehat{k}=\left|\overrightarrow{c}.\frac{\overrightarrow{d}}{\left|\overrightarrow{d}\right|}\right|=30\left\{Given\right\}$

$\Rightarrow \left|\frac{\alpha -1-4+2{\alpha }^2-2\alpha +4}{\sqrt[]{1+4+4}}\right|=30$  

On solving Þ $\alpha =\frac{-13}{2}$  (Rejected as a > 0) and a = 7

The line x + y – z = 0 = x – 2y + 3z – 5 is parallel to the vector

  $\overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right|=\left(1,4,-3\right)$ Equation of line through P(1, 2, 4) and parallel to $\overrightarrow{b}\frac{x-1}{1}=\frac{y-2}{-4}=\frac{z-4}{-3}$  

Let  $N\equiv \left(\lambda +1,-4\lambda +2,-3\lambda +4\right)\overline{QN}=\left(\lambda ,-4\lambda +4,-3\lambda -1\right)$  

$\overline{QN}$ is perpendicular to $\overrightarrow{b}\Rightarrow \left(\lambda ,-4\lambda +4,-3\lambda -1\right).\left(1,4,-3\right)=0\Rightarrow \lambda =\frac{1}{2}.$  

Hence  $\overline{QN}\left(\frac{1}{2},2,\frac{-5}{2}\right)and\overline{\left|QN\right|}=\sqrt[]{\frac{21}{2}}$  

Any tangent to y² = 24x at (a, b) is by = 12 (x + a) therefore   Slope = $\frac{12}{\beta }$  

and perpendicular to 2x + 2y = 5 Þ 12 = b and a = 6 Hence hyperbola is $\frac{x^2}{6^2}-\frac{y^2}{12^2}$  = 1 and normal is drawn at (10, 16)

therefore equation of normal  $\frac{36x}{10}+\frac{144y}{16}=36+144\Rightarrow \frac{x}{50}+\frac{y}{20}=1$  This does not pass through (15, 13) out of given option.

Let point P : (h, k)

Therefore according to question,  ${\left(h-1\right)}^2+{\left(k-2\right)}^2+{\left(h+2\right)}^2+{\left(k-1\right)}^2=14$

$\therefore$ locus of P(h, k) is $x^2+y^2+x-3y-2=0$  

Now intersection with x – axis are  $x^2+x-2=0\Rightarrow x=-2,1$  

Now intersection with y – axis are  $y^2-3y-2=0\Rightarrow y=\frac{3\pm \sqrt[]{17}}{2}$  

Therefore are of the quadrilateral ABCD is =  $\frac{1}{2}\left(\left|x_1\right|+\left|x_2\right|\right)\left(\left|y_1\right|+\left|y_2\right|\right)=\frac{1}{2}*3*\sqrt[]{17}=\frac{3\sqrt[]{17}}{2}$  

$\frac{dy}{dx}+2ytanx=sinx,\therefore I.F.e^{{\displaystyle \int 2tanxdx}}=se{c}^{2}x$  

= cos x – 2 cos² x = $-2{\left(cosx-\frac{1}{4}\right)}^2+\frac{1}{8}\therefore y_{max}=\frac{1}{8}$

$a=\underset{n\to \infty }{lim}{\displaystyle \sum _{k=1}^n\frac{2n}{n^2+k^2}=\underset{n\to \infty }{lim}\frac{1}{n}}{\displaystyle \sum _{k=1}^n\frac{2}{1+{\left(\frac{k}{n}\right)}^2}}$

 $\therefore a={\displaystyle {\int }_0^1\frac{2}{1+x^2}dx=2ta{n}^{-1}x}]{}_0^1=\frac{\pi }{2}$

⇒ $f\text{'}\left(\frac{a}{2}\right)=\sqrt[]{2}f\left(\frac{a}{2}\right)$

$f\left(x\right)=\{\begin{array}{cc}\hfill x^3-x^2+10x-7,\hfill & \hfill x\le 1\hfill \\ \hfill -2x+lo{g}_2\left(b^2-4\right),\hfill & \hfill x>1\hfill \end{array}$  

If f(x) has maximum value at x = 1 then

$f\left(1\right)\le f\left(1\right)\Rightarrow -2+lo{g}_2\left(b^2-4\right)\le 1-1+10-7$

$\Rightarrow lo{g}_2\left(b^2-4\right)\le 5\Rightarrow 0<b^2-4\le 32$

$\Rightarrow b^2-4>0\Rightarrow b\in \left(-\infty ,-2\right)\cup \left(2,\infty \right)$         …….(i)

$And{b}^2-4\le 32\Rightarrow b\in \left[-6,6\right]$                      …….(ii)

From (i) and (ii) we get  $b\in [-6,-2)\cup (2,6]$  

$f\left(x\right)=\{\begin{array}{cc}\hfill x+a,\hfill & \hfill x?0\hfill \\ \hfill \left|x?4\right|,\hfill & \hfill x>0\hfill \end{array}\text{?}\text{?}\text{?}\text{?}and\text{?}\text{?}\text{?}g\left(x\right)=\{\begin{array}{cc}\hfill x+1,\hfill & \hfill x<0\hfill \\ \hfill {\left(x?4\right)}^2+b,\hfill & \hfill x?0\hfill \end{array}$

$?$    f (x) and g (x) are continuous on R $?$  a = 4 and b = 1 – 16 = 15

then (gof) (2) + (fog) (-2) = g (2) + f (-1) = -11 + 3 = -8

$f\left(x\right)=\{\begin{array}{l}\frac{lo{g}_e\left(1-x+x^2\right)+lo{g}_e\left(1+x+x^2\right)}{secx-cosx},x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)-\left\{0\right\}\\ k,x=0\end{array}$ for continuity at x = 0

      $\underset{x\to 0}{lim}f\left(x\right)=k\therefore k=\underset{x\to 0}{lim}\frac{lo{g}_e\left(1+x^2+x^4\right)}{secx-cosx}\left(\frac{0}{0}form\right)=\underset{x\to 0}{lim}\frac{cosxlo{g}_e\left(1+x^2+x^4\right)}{si{n}^{2}x}=1$