Maths

 $\overrightarrow{a}*\left(\overrightarrow{a}*\overrightarrow{b}\right)=\left(\overrightarrow{a}.\overrightarrow{b}\right)\overrightarrow{a}-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}=0-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}\left(as\overrightarrow{a}.\overrightarrow{b}=0given\right)$

$\overrightarrow{a}*\left(\overrightarrow{a}*\left(\overrightarrow{a}*\overrightarrow{b}\right)\right)=-{\left|\overrightarrow{a}\right|}^2\overrightarrow{a}*\overrightarrow{b}$

$\overrightarrow{a}*\left(\overrightarrow{a}*\left(\overrightarrow{a}*\left(\overrightarrow{a}*\overrightarrow{b}\right)\right)\right)=-{\left|\overrightarrow{a}\right|}^2\overrightarrow{a}*\left(\overrightarrow{a}*\overrightarrow{b}\right)=-{\left|\overrightarrow{a}\right|}^2\left(-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}\right)={\left|\overrightarrow{a}\right|}^4\overrightarrow{b}$

$T_{r+1}{=}^{10}{C}_r{\left(t{x}^{\frac{1}{5}}\right)}^{10-r}{\left(\frac{{\left(1-x\right)}^{\frac{1}{10}}}{t}\right)}^r$

According to question, 10 – 2r = 0 ⇒ r = 5

$\therefore T_6{=}^{10}{C}_5*{\left(1-x\right)}^{\frac{1}{2}}$

T₆ is maximum, when f(x) = $x{\left(1-x\right)}^{\frac{1}{2}}$ is maximum.

$f\text{'}\left(x\right)={\left(1-x\right)}^{\frac{1}{2}}-\frac{x}{2\sqrt[]{1-x}}=\frac{2\left(1-x\right)-x}{2\sqrt[]{1-x}}$

For maximum, $f\text{'}\left(x\right)=0\Rightarrow x=\frac{2}{3}$

$\therefore T_6{=}^{10}{C}_5\frac{2}{3}.{\left(\frac{1}{3}\right)}^{\frac{1}{2}}=\frac{2\left(10!\right)}{3\sqrt[]{3}{\left(5!\right)}^2}$

Let n be the number of times.

$p=\frac{1}{2},q=\frac{1}{2}$

According to question,

${\Rightarrow }^n{C}_7{=}^n{C}_9\Rightarrow n-7=9\Rightarrow n=16$

$p\left(x=2\right){=}^{16}{C}_2{\left(\frac{1}{2}\right)}^2.{\left(\frac{1}{2}\right)}^{14}=\frac{15}{2^{13}}$

Put x = $\frac{1}{3}$

S = 1 + 2x + 7x² + 12x³ + 17x⁴ + 22x⁵ + .

$\underset{\_}{xS=x+2x^2+7x^3+12x^4+17x^5+.........}$

Subtracting,

$\left(1-x\right)S=1+x+5x^2+5x^3+5x^4+5x^5+......$

= $1-4x+5x+5x^2+5x^3+.........$

$=\frac{1-x-4x+4x^2+5x}{1-x}=\frac{4x^2+1}{1-x}$

$S=\frac{4x^2+1}{{\left(1-x\right)}^2}putx=\frac{1}{3}wegets=\frac{\frac{4}{9}+1}{\frac{4}{9}}=\frac{13}{4}$

 $\frac{dv}{dt}\propto V\Rightarrow \frac{dv}{dt}=\lambda V\Rightarrow {\displaystyle \underset{1000}{\overset{1200}{\int }}\frac{dv}{v}}=\lambda {\displaystyle \underset{0}{\overset{2}{\int }}dt}\Rightarrow \lambda =\frac{1}{2}\mathcal{l}n\frac{6}{5}$

${\displaystyle \underset{1000}{\overset{2000}{\int }}\frac{dv}{v}}=\frac{1}{2}\mathcal{l}n\left(\frac{6}{5}\right){\displaystyle \underset{0}{\overset{T}{\int }}dt}\Rightarrow T=\frac{2\mathcal{l}n2}{\mathcal{l}n\left(\frac{6}{5}\right)}\Rightarrow k=2ln2$

$A=\left[\begin{array}{cc}\hfill p\hfill & \hfill q\hfill \\ \hfill r\hfill & \hfill s\hfill \end{array}\right]$ is symmetric. So, q = r.

$AA=A^2=\left[\begin{array}{cc}\hfill p\hfill & \hfill q\hfill \\ \hfill r\hfill & \hfill s\hfill \end{array}\right]\left[\begin{array}{cc}\hfill p\hfill & \hfill q\hfill \\ \hfill r\hfill & \hfill s\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill p^2+qr\hfill & \hfill pq+qs\hfill \\ \hfill rp+rs\hfill & \hfill rq+s^2\hfill \end{array}\right]$

Sum of diagonal elements =

$p^2+qr+rq+s^2=1\Rightarrow p^2+2r^2+s^2=1,\left(asq=r\right),p=0,r=0ands=\pm 1orr=0,s=0andp=\pm 1$

Total number of matrices = 4.

All the points A (1, 5, 35), B (7, 5,5),  $C(1,\lambda ,7),D(2\lambda ,1,2)$ are coplanar. Hence

$\overrightarrow{AB}*\overrightarrow{AC}.\overrightarrow{AD}=\left|\begin{array}{ccc}\hfill 6\hfill & \hfill 0\hfill & \hfill 30\hfill \\ \hfill 0\hfill & \hfill \lambda -5\hfill & \hfill 28\hfill \\ \hfill 2\lambda -1\hfill & \hfill -4\hfill & \hfill 33\hfill \end{array}\right|=0$

$\Rightarrow 5{\lambda }^2-44\lambda +95=0$

$\Rightarrow sumofroots=\frac{44}{5}$

$\left|f\left(x\right)-f\left(y\right)\right|\le \left|{\left(x-y\right)}^2\right|$

$\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le x-y$

Taking the limit y x on both sides

$\underset{y\to x}{Lt}\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le \underset{y\to x}{Lt}\left(x-y\right)$

$\left|f\text{'}\left(x\right)\right|\le 0$

Hence, modulus cannot be zero. Hence f' (x) = 0. Integrating, we get f (x) = c

at x = 0, f (0) = c = 1

$\therefore f\left(x\right)=1>0,\forall x\in R$

Hence, option (A) is correct option.

x – y = 0 . (i)

x + 2y = 3. (ii)

2x + y = 6 . (iii)

Solving (i) & (ii), we get (1, 1)

Solving (ii) & (iii), we get (3, 0)

Solving (iii) & (i), we get (2, 2)     

$AB=\sqrt[]{5,}BC=\sqrt[]{5},CA=\sqrt[]{2}$

$\therefore AB=BC$ Isosceles triangle

Hence, option (B) is correct answer.