Maths

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

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

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

$\underset{h\to 0}{Lt}\frac{2*2\left[\frac{\sqrt[]{3}}{2}sin\left(\frac{\pi }{6}+h\right)-\frac{1}{2}cos\left(\frac{\pi }{6}+h\right)\right]}{2\sqrt[]{3}h\left[\frac{\sqrt[]{3}}{2}cosh-\frac{1}{2}sinh\right]}$

$=\underset{h\to 0}{Lt}4\frac{sin\left(\frac{\pi }{6}+h-\frac{\pi }{6}\right)}{2\sqrt[]{3}hsin\left(\frac{\pi }{3}-h\right)}=\frac{2}{\sqrt[]{3}}*1*\frac{1}{\frac{\sqrt[]{3}}{2}}=\frac{4}{3}$

 ${\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx={\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx}}}}$

${\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x-\left[x\right]}dx+{\displaystyle \underset{1}{\overset{2}{\int }}{e}^{x-\left[x\right]}dx}}+{\displaystyle \underset{2}{\overset{3}{\int }}{e}^{x-\left[x\right]}dx+.....+{\displaystyle \underset{99}{\overset{100}{\int }}{e}^{x-\left[x\right]}dx}}$

$=e-1+\frac{1}{e}\left(e^2-e\right)+\frac{1}{e^2}\left(e^3-e^2\right)+......+\frac{1}{e^{99}}\left(e^{100}-e^{99}\right)$

$=e-1+\left(e-1\right)+\left(e-1\right)+......+\left(e-1\right),100times$

$=100\left(e-1\right)$

2nd method

${\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx={\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{\left\{x\right\}}dx=}}{\displaystyle \sum _{n=1}^{100}\left({\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}\right)}}}={\displaystyle \sum _{n=1}^{100}\left(e-1\right)}=100\left(e-1\right)$

The points on the curve are (0, 0), (2, 2) and $\left(3,\frac{21}{2}\right)$

$\therefore \frac{dy}{dx}=2x^3-15x^2+36x-19$

$at\left(0,0\right),\frac{dy}{dx}=-19,at\left(2,2\right),\frac{dy}{dx}=9at\left(3,\frac{21}{2}\right),\frac{dy}{dx}=8$

Hence maximum slope at (2, 2) is 9.