Class 12th

Given f(x) = ${\displaystyle {\int }_e^x{e}^{t}f\left(t\right)dt+e^x..........\left(i\right)}$  

using Leibniz rule then

f'(x) = e^xf(x) + e^x

  _{$\Rightarrow \frac{dy}{dx}=e^{x}y+e^{x}wherey=f\left(x\right)then\frac{dy}{dx}=f\text{'}\left(x\right)$}                

P = -e^x, Q = e^x

Solution be y. (I.F.) =   ${\displaystyle \int Q\left(I.F.\right)dx+c}$

I. f. =   $e^{{\displaystyle \int -e^{x}dx}=e^{-e^x}}$

$\Rightarrow y.\left(e^{-e^x}\right)={\displaystyle \int e^x.e^{-e^x}dx+c}$          

  $y.e^{-e^x}=-{\displaystyle \int dt+c=-t+c=-e^{-e^x}+c..........\left(ii\right)}$          

Put x = 0 , in (i) f (0) = 1

$From\left(ii\right),\frac{1}{e}=-\frac{1}{e}+cgivenc=\frac{2}{e}$        

Hence f(x) = 2. $e^{\left(e^x-1\right)}-1$  

$\frac{dq}{dt}=t=20t+8t^2$

$\Rightarrow {\displaystyle \underset{0}{\overset{q}{\int }}dq={\displaystyle \underset{0}{\overset{15}{\int }}\left(20t+8t^2\right)dt}}$

$\Rightarrow q=20*\frac{15^2}{2}+8.\frac{15^3}{3}=11250C$

$\frac{dy}{dx}=\frac{1}{1+sin2x}$

$dy=\frac{se{c}^{2}xdx}{{\left(1+tanx\right)}^2}$

$\Rightarrow y=-\frac{1}{1+tanx}+c$

when   $x=\frac{\pi }{4},y=\frac{1}{2}$ gives c = 1

so $x+\frac{\pi }{4}=\frac{5\pi }{6}or\frac{13\pi }{6}\Rightarrow x=\frac{7\pi }{12}or\frac{23\pi }{12}$

sum of all solutions =

$\pi +\frac{7\pi }{12}+\frac{23\pi }{12}=\frac{42\pi }{12}$

Hence k = 42

Each element of ordered pair (i, j) is either present in A or in B.

              So, A + B = Sum of all elements of all ordered pairs {i, j} for $1\le i\le 10$ and $1\le j\le 10$  

              = 20 (1 + 2 + 3 + … + 10) = 1100

$l=\frac{48}{{\pi }^4}{\displaystyle {\int }_0^{\pi }\left[{\left(\frac{\pi }{2}-x\right)}^3-\frac{3{\pi }^2}{4}\left(\frac{\pi }{2}-x\right)+\frac{{\pi }^3}{4}\right]\frac{sinxdx}{1+co{s}^{2}x}}$

Using ${\displaystyle {\int }_a^{b}f\left(x\right)dx={\displaystyle {\int }_a^{b}f\left(a+b-x\right)dx}}$

we get $l=\frac{48}{{\pi }^4}{\displaystyle {\int }_0^{\pi }\left[-{\left(\frac{\pi }{2}-x\right)}^3+\frac{3{\pi }^2}{4}\left(\frac{\pi }{2}-x\right)+\frac{{\pi }^3}{4}\right]\frac{sinxdx}{1+co{s}^{2}x}}$

Adding these two equations, we get

$\Rightarrow l=\frac{12}{\pi }{\left[-ta{n}^{-1}\left(cosx\right)\right]}_0^{\pi }=\frac{12}{\pi }.\frac{\pi }{2}=6$

Sum of all elements of $A\cap B=2$   [Sum of natural number upto 100 which are neither divisible by 3 nor by 5]

$=2\left[\frac{100*101}{2}-3\left(\frac{33*34}{2}\right)-5\left(\frac{20*21}{2}\right)+15\left(\frac{6*7}{2}\right)\right]$

= 10100 – 3366 – 2100 + 630

              = 5264

$\left(sin10°.sin50°.sin70°\right).\left(sin10°.sin20°.sin40°\right)$

$=\left(\frac{1}{4}sin30°\right).\left[\frac{1}{2}sin10°\left(cos20°-cos60°\right)\right]$

$=\frac{1}{32}\left[sin30°-sin10°-sin10°\right]$

$\frac{1}{64}-\frac{1}{16}sin10°$

Clearly $\alpha =\frac{1}{64}$  

              Hence 16 + a^-1 = 80

$\left(4+x^2\right)dy-2x\left(x^2+3y+4\right)dx=0$

$\Rightarrow \frac{dy}{dx}=\left(\frac{6x}{x^2+4}\right)y+2x$

$e^{-3\mathcal{l}n\left(x^2+4\right)}=\frac{1}{{\left(x^2+4\right)}^3}$

so $\frac{y}{{\left(x^2+4\right)}^3}={\displaystyle \int \frac{2x}{{\left(x^2+4\right)}^3}dx+c}$

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

When x = 0, y = 0 gives $c=\frac{1}{32}$

So, for x = 2, y = 12

$\therefore {\displaystyle \underset{-6}{\overset{0}{\int }}f\left(x\right)dx=2*\frac{1}{2}\left(2+5\right)*3=21}$

  ${E^o}_{Cell}={E^o}_{A{g}^{+}/Ag}-{E^o}_{Z{n}^{+2}/Zn}=0.8+0.76=1.56V$

Anode :  $Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$    

Cathode :  2Ag⁺(aq) + 2e^- -> 2Ag(s)

Zn(s) + 2Ag⁺(aq) -> Zn²⁺ (aq) + 2Ag(s)

  $E_{cell}={E^o}_{Cell}-\frac{0.0591}{n}log\frac{\left[Z{n}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}=1.56-\frac{0.0591}{2}log\left(\frac{0.1}{10^{-4}}\right)$          

x = 147