Class 11th

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

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

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

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

$x=-1,1$

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.

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

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

Now at x = π,

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

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

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

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

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

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

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

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

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

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

${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$

$\Rightarrow \frac{n}{a}{\left(\frac{x}{a}\right)}^{n-1}+\frac{n}{b}{\left(\frac{y}{b}\right)}^{n-1}\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{b}{a}{\left(\frac{bx}{ay}\right)}^{n-1}$

$\frac{dy}{d{x}_{\left(a,b\right)}}=-\frac{b}{a}$

So line always touches the given curve.

$2021\equiv -2$ mod (7)

$\Rightarrow {\left(2021\right)}^{2023}\equiv {\left(-2\right)}^{2023}mod\left(7\right)$ …… (i)

Now,  ${\left(-2\right)}^3\equiv -1mod\left(7\right)$

$\Rightarrow {\left(-2\right)}^{2023}\equiv \left(-2\right)mod\left(7\right)\equiv 5mod\left(7\right)$ ……. (ii)

(i) & (ii)

$\Rightarrow {\left(2021\right)}^{2023}\equiv 5mod\left(7\right)$

$\therefore$ Remainder = 5

System of equation can be written as

$\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -8\hfill & \hfill 9\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill a\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}b\\ 3\\ -1\end{array}\right)$

$\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 15\hfill & \hfill -24\hfill & \hfill 27\hfill \\ \hfill 6\hfill & \hfill 3\hfill & \hfill 3a\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}b\\ 9\\ -3\end{array}\right)$

$R_3-2R_1,R_2-5R_1$

for no solution

3a + 9 = 0 but $\frac{3}{2}-\frac{9b}{2}\ne 0$

$\Rightarrow a=-3\Rightarrow b\ne \frac{1}{3}$

$\left|adj\left(24A\right)\right|=\left|adj\left(3adj\left(2A\right)\right)\right|$

$\Rightarrow {\left|24A\right|}^2={\left|3adj\left(2A\right)\right|}^2$

$24^6{\left|A\right|}^2=3^6.{\left(2^3\right)}^4{\left|A\right|}^4$

${\left|A\right|}^2=\frac{24^6}{3^6.2^{12}}=\frac{2^{18}.3^6}{3^6.2^{12}}=2^6$

pH of acidic buffer is given by,

pH = pKa + log₁₀ $\frac{\left[Salt\right]}{\left[acid\right]}$ $\frac{}{}$

or, $p=pka+lo{g}_{10}\frac{\left[C{H}_{3}C{H}_{2}CO{O}^{-}\right]}{\left[C{H}_{3}C{H}_{2}COOH\right]}$ ${}_{}\frac{{}_{}{}_{}{}^{}}{{}_{}{}_{}}$

or, $\left[H^{+}\right]=k_a\frac{\left[C{H}_{3}C{H}_{2}COOH\right]}{\left[C{H}_{3}C{H}_{2}CO{O}^{-}\right]}$ ${}^{}{}_{}\frac{{}_{}{}_{}}{{}_{}{}_{}{}^{}}$

$\frac{\left[C{H}_{3}C{H}_{2}CO{O}^{-}\right]}{\left[C{H}_{3}C{H}_{2}COOH\right]}=\frac{k_a}{\left[H^{+}\right]}=\frac{1.3*10^{-5}}{10^{-4}}=0.13$