Maths

RM = $\left|\frac{3+7-5}{\sqrt[]{2}}\right|=\frac{5}{\sqrt[]{2}}$

$\mathcal{l}sin60°=\frac{5}{\sqrt[]{2}}\Rightarrow \mathcal{l}=5\sqrt[]{\frac{2}{3}}$

$\therefore Areaof\Delta PQR=\frac{\text{exception 25:}}{4}{\mathcal{l}}^2=$=25/2√3

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

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

$\frac{dz}{dx}=e^x.2\left(x-1\right)+e^x{\left(x-1\right)}^2=0$

$e^x\left(x-1\right)\left(2+x-1\right)=0$

$x=-1,1$

x = 1 local maxima. Then maximum value is

$z\left(-1\right)=\frac{4}{e}-e$

$\left|x^2-9\right|=3$

$\Rightarrow x=\pm 2\sqrt[]{3},\pm \sqrt[]{6}$

Required area = A

$\frac{A}{2}={\displaystyle \underset{0}{\overset{\sqrt[]{6}}{\int }}\left(9-x^2-3\right)dx+{\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{9+y}-\sqrt[]{9-y}\right)dy}}$

$A=16\sqrt[]{6}+32\sqrt[]{3}-72=8\left[2\sqrt[]{6}+4\sqrt[]{3}-9\right]$

Note : No option in the question paper is correct.

Put x = cos 2θ

dx = -2 sin 2θ . dθ

$={\displaystyle \int \frac{1}{cos2\theta }tan\theta \left(-4sin\theta .cos\theta \right)d\theta }$

$g\left(\frac{1}{2}\right)=\mathcal{l}n\left|2-\sqrt[]{3}\right|+\frac{\pi }{3}$

= $\mathcal{l}n\left|\frac{\sqrt[]{3}-1}{\sqrt[]{3}+1}\right|+\frac{\pi }{3}$  

Let p₁ : y² = 8x

p₂ : y² = 16 (3 – x) = -16 (x – 3)

finding their intersection points.

y² = 8x & y² = -16 (x – 3)

8x = -16x + 48

$=2.{\displaystyle \underset{0}{\overset{4}{\int }}\left(3-\frac{y^2}{16}-\frac{y^2}{8}\right)dy}$

$=2{\left(3y-\frac{y^3}{3*16}-\frac{y^3}{3*8}\right)}_0^4$ = 16

$?s_1+s_2=k$                

76x² + 3πr² = k

$\therefore 152x\frac{dx}{dr}+6\pi r=0$

Now

$V=40x^3+\frac{2}{3}\pi r^3$

$\Rightarrow \left(\frac{x}{r}\right)=\frac{152}{3}\frac{1}{120}=\frac{19}{45}$

  $f\left(x\right)=min\left\{1,1+xsinx\right\},0\le x\le 2\pi$

$f\left(x\right)=\{\begin{array}{l}1,0\le x<\pi \\ 1+xsinx,\pi \le x\le 2\pi \end{array}$

Now at x = π,

$\therefore f\left(x\right)$ is not differentiable at x = π

$\therefore$ (m, n) = (1, 0)

  $\underset{x\to 0}{lim}\frac{cos\left(sinx\right)-cosx}{x^4}=\underset{x\to 0}{lim}\frac{2sin\left(x+sinx\right).sin\left(\frac{x-sinx}{2}\right)}{x^4}$ $\underset{}{}\frac{}{{}^{}}\underset{}{}\frac{\frac{}{}}{{}^{}}$

$=\underset{x\to 0}{lim}2.\left(\frac{\left(\frac{x+sinx}{2}\right)\left(\frac{x-sinx}{2}\right)}{x^4}\right)$

= $\underset{x\to 0}{lim}\frac{1}{2}.\left(2-\frac{x^2}{3!}+\frac{x^4}{5!}........\right)\left(\frac{1}{3!}-\frac{x^2}{5!}-1\right)$ $\underset{}{}\frac{}{}\frac{{}^{}}{}\frac{{}^{}}{}\frac{}{}\frac{{}^{}}{}$

= $\frac{1}{6}$ $\frac{}{}$

$A={\displaystyle {\sum }_{n=1}^{\infty }\frac{{\left(-1\right)}^n}{{\left(3+{\left(-1\right)}^n\right)}^n}}$

$B={\displaystyle {\sum }_{n=1}^{\infty }\frac{{\left(-1\right)}^n}{{\left(3+{\left(-1\right)}^n\right)}^n}}$

$A=\frac{11}{15},B=\frac{-9}{15}$

$\therefore \frac{A}{B}=\frac{-11}{9}$

Given system of equations

αx + y + z = 5

x + 2y + 3z = 4, has infinite solution

x + 3y + 5z =β

$\begin{array}{c}\hfill \hfill \end{array}$ $\therefore \Delta =\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}\right|$  = 0

α (1) – 1 (2) + 1 (1) = 0

α= 1 and

${\Delta }_3=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill \beta \hfill \end{array}\right|=0$

β= 3

$\therefore \left(\alpha ,\beta \right)=\left(1,3\right)$