Maths

Sum of all given numbers = 31

Difference between odd and even positions must be 0,11 or 22 but 0 and 22 are not possible.

$\therefore$ Hence 11 is possible.

This is possible only when either 1, 2, 3, 4 if filled in odd places in order and remaining in other order.

Hence 2, 3, 5 or 7, 2, 1 or 4, 5, 1 at even places.

$\therefore$ Total possible ways = (4! * 3!) * 4 = 576

 $S=\{\left(\begin{array}{cc}\hfill 1\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill b\hfill \end{array}\right),a,b\in 1,2,3,.....100$

$\therefore A=\left(\begin{array}{cc}\hfill -1\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill b\hfill \end{array}\right)$ then even power of A as A = $\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right).$

If b = 1 & $a\in \left\{1,2,3.....100\right\}$ and n (n + 1) is always even

$\therefore T_1,T_2,T_3,........,T_n$ are all 1 for b = 1 and each value of a.

$\therefore {\displaystyle \underset{n=1}{\overset{1000}{\cap }}{T}_n=100}$

Circle $\left|z-3\right|\le 1\Rightarrow {\left(x-3\right)}^2+y^2\le 1$

and line $z\left(4+3i\right)+\overline{z}\left(4-3i\right)\le 24$

$\Rightarrow 4x-3y\le 12\therefore slope=tan\theta =\frac{4}{3}$

$?f_{\lambda }\left(x\right)=4\lambda x^3-36\lambda x^2+36x+48$

$f_{\lambda }\left(x\right)=12\left(\lambda x^2-6\lambda x+3\right)$

For increasing $f_{\lambda }\left(x\right)\ge 0$

$f_{\lambda }^{*}\left(x\right)=\frac{4}{3}{x}^3-12x^2+36x+48\therefore f_{\lambda }^{*}\left(1\right)+f_{\lambda }^{*}\left(-1\right)=73\frac{1}{2}-1\frac{1}{2}=72$

 $?-\frac{dx}{dy}=\frac{x^2}{xy-x^2{y}^2-1}$

$\therefore \frac{dy}{dx}=\frac{xy-x^2{y}^2-1}{x^2}$

$Letxy=v\Rightarrow x\frac{dy}{dx}+y=\frac{dv}{dx}$

Put x = 1, y = 1 $\Rightarrow ta{n}^{-1}=c\Rightarrow c=\frac{\pi }{4}$

$\therefore ta{n}^{-1}\left(xy\right)=\mathcal{l}nx=\frac{\pi }{4}$

$e\left(y\left(e\right)\right)=tan\left(1+\frac{\pi }{4}\right)=\frac{tan1+1}{1-tan1}$

$?$ given statement is

$\left(A\wedge C\right)\to B$ then its negation is $\left\{\left(A\wedge C\right)\to B\right\}$

$Let{x}^3=\theta \Rightarrow \frac{\theta }{2}\in \left(\frac{\pi }{4},\frac{3\pi }{4}\right)$

$\therefore y=ta{n}^{-1}\left(sec\theta -tan\theta \right)$

$ta{n}^{-1}\left(\frac{1-sin\theta }{cos\theta }\right)$

$\Rightarrow \frac{dy}{dx}=\frac{-3x^2}{2}\Rightarrow \frac{d^{2}y}{d{x}^2}=-3x$

$\therefore x^2\frac{d^{2}y}{d{x}^2}-6y+\frac{3\pi }{2}=0\Rightarrow x^2{y}^{11}-6y+\frac{3\pi }{2}=0$

$\left|\overrightarrow{a}+\overrightarrow{b}+2\left(\overrightarrow{a}*\overrightarrow{b}\right)\right|=2,\theta \in \left(0,\pi \right)$

squaring on both sides, we get = $\frac{2\pi }{3}$

where θ is angle between $\widehat{a}and\widehat{b}.$

$2\left|\widehat{a}*\widehat{b}\right|=\sqrt[]{3}=\left|\widehat{a}-\widehat{b}\right|,S_1$ is correct.

and projection of $\widehat{a}on\widehat{a}+\widehat{b}=\left|\frac{\widehat{a}.\left(\widehat{a}+\widehat{b}\right)}{\left|\widehat{a}+\widehat{b}\right|}\right|=\frac{1}{2}$

So (S₂) is correct.

Let P (x, y, z) be any point on plane P₁ then

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

$\therefore cos\theta =\frac{\left|6-2+3\right|}{14}\Rightarrow \theta =\frac{\pi }{3}$

S. D. = $\frac{1}{\sqrt[]{3}}$

$b_1*b_2=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill \lambda \hfill \\ \hfill 1\hfill & \hfill 4\hfill & \hfill 5\hfill \end{array}\right|=\widehat{i}\left(15-4\lambda \right)+\widehat{j}\left(\lambda -10\right)+\widehat{k}\left(5\right)$

$\frac{\left|\left(-\widehat{i}-2\widehat{j}-2\widehat{k}\right).\left\{\left(15-4\lambda \right)\widehat{i}+\left(\lambda -10\right)\widehat{j}+5\widehat{k}\right\}\right|}{\sqrt[]{{\left(15-4\lambda \right)}^2+{\left(\lambda -10\right)}^2+25}}=\frac{1}{\sqrt[]{3}}$

$\Rightarrow \lambda =16$