Class 12th

  $f\left(x\right)=2x-1g:R-\left\{1\right\}\to R$       

  $g\left(x\right)=\frac{x-1/2}{x-1}$         

  $f\left\{g\left(x\right)\right\}=2\left(\frac{x-1/2}{x-1}\right)-1=\frac{2x-1-x+1}{x-1}=\frac{x}{x-1},x\ne 1$          

$y=\frac{x}{x-1}\Rightarrow x=\frac{y}{y-1}\therefore y\ne 1$           

One – one but not onto

$\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=\frac{2\left(a+b+c\right)}{24}=\frac{c}{5}=\frac{a}{4}=\frac{b}{3}$

$r=\frac{\Delta }{S}=\frac{6k^2}{6k}=k$

$R=\frac{5k}{2}\Rightarrow \frac{R}{r}=\frac{5}{2}$

$\underset{x\to 0}{lim}\frac{{\displaystyle \underset{0}{\overset{x^2}{\int }}\left(sin\sqrt[]{t}\right)dt}}{x^3}$ $\left(\frac{0}{0}\right)$   aplying L' Hospital rule

$=\underset{x\to 0}{lim}\frac{sin\left(\sqrt[]{x^2}\right).2x}{3x^2}$

$=\frac{2}{3},forx>0$

$l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{e^{cosx}sinxdx}{\left(1+co{s}^{2}x\right)\left(e^{cosx}+e^{-cosx}\right)}}$

$2l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{sindx}{1+co{s}^{2}x}=2{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{sinxdx}{1+co{s}^{2}x}}}$

$l={\displaystyle \underset{1}{\overset{0}{\int }}\frac{-dt}{1+t^2}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+t^2}=\frac{\pi }{4}}}$

  ${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{8-sin2x}}=asi{n}^{-1}\left(\frac{sinx+cosx}{b}\right)+c}$

${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{9-{\left(sinx+cosx\right)}^2}}=}$           

Put (sin x + cos x) = t Þ (cos x – sin x) dx = dt

$={\displaystyle \int \frac{dt}{\sqrt[]{3^2-t^2}}=si{n}^{-1}\frac{t}{3}+c=si{n}^{-1}\left(\frac{sinx+cosx}{3}\right)+c}$        

= a = 1, b = 3

$\therefore \left(a,b\right)\equiv \left(1,3\right)$        

Temperature coefficient is 2 that means rate of reaction doubles at every 10°C rise in temperature.
Thus, k? °C = 2? = 64 k? °C

To decompose 1 mole of H? O, we need two moles of electrons i.e. 2 faraday
To decompose 4 moles of H? O, we need 8 faraday.
Now, Q = I * t? t = (8 * 96500) / 4 = 1.93*10? sec

  $\Delta -\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 2\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 0\hfill & \hfill -8\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|,\left[R_2-R_1-2R_3\right]$

= -8 (-3 + k)

For inconsistent $\Delta =0\Rightarrow k=3$  

${\Delta }_x=\left|\begin{array}{ccc}\hfill 10\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 6\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 5m\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=32-40m\ne 0\Rightarrow m\ne \frac{4}{5}$

y = x³

${\frac{dy}{dx}=3x^2\Rightarrow \frac{dy}{dx}|}_{\left(t,t^3\right)}=3t^2$           

Equation of tangent y – t³ = 3t² (x – t)  

Let again meet the curve at $Q\left(t_1,t_1^3\right)$  

$\Rightarrow t_1^3-t^3=3t^2\left(t_1-t\right)$           

$t_1^2+t{t}_1+t^2=3t^2\left[?t_1\ne t\right]$           

$t_1^2+t{t}_1-2t^2=0$            

->t₁ = -2t

Required ordinate = $\frac{2t^3+t_1^3}{3}=\frac{2t^3-8t^3}{3}=-2t^3$

The number of chiral carbons in open-chain aldohexose (such as glucose) is four, therefore, the number of stereoisomers = 2? = 16.