Class 12th

$\underset{x\to 0}{lim}\frac{{\displaystyle \underset{0}{\overset{x^2}{\int }}\left(sin\sqrt[]{t}\right)dt}}{x^3}$ $\left(\frac{0}{0}\right)$   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.

All monosaccharides which differ in configuration at C? and C? gives the same osazone. Since, glucose and fructose differ from each other only in configuration at C? and C? therefore, they give the same osazone. All other options given in the questions do not satisfy this condition and hence, do not from the same osazone.