Class 12th

Both (A) and (B) are correct but R is not correct explanation of A. In both oxidation state of metal is +1, also both have similar lattice structure.  

(a) Sucrose -> a-D glucose and b-D fructose

(b) Lactose -> b-D- galactose and b-D glucose

(c) maltose -> a-D glucose and a-D-glucose

$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2\left(1+4+2-16\right)}{1+2^2+2^2}$

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

$\left|\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z+1\hfill \\ \hfill 3\hfill & \hfill 6\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=0$ a

=> 2x – z = 1

option (B) satisfies.

Let AB   $\equiv x-2y+1=0$

AC $\equiv x-2y+1=0$  

So vertex A = (1, 1)

altitude from B is perpendicular to AC and passing through

orthocentre.

So, BH = x + 2y – 7 = 0

CH = 2x + y – 7 = 0

now solve AB & BH to get B (3, 2) similarly CH and AC to get C (2, 3) so centroid is at (2, 2)

Line $\perp$  to the normal

=>3p + 2q – 1 = 0

$\left(2,-1,-3\right)$ lies in the plane 2p + q = 8

From here p = 15, q = -22

Equation of plane 15x – 22y + z – 5 = 0

Distance from origin = $\left|\frac{5}{\sqrt[]{15^2+{\left(-22\right)}^2+1^2}}\right|=\sqrt[]{\frac{5}{142}}$  

Ceric ammonium nitrate is used to test alcohol while CHCl₃/alc. KOH is used to test 1° amine

$\Rightarrow {\displaystyle \underset{}{\overset{}{\int }}\frac{dy}{1+y^2}=-2{\displaystyle \underset{}{\overset{}{\int }}\frac{e^{x}dx}{1+{\left(e^x\right)}^2}+C}}$

$\Rightarrow ta{n}^{-1}y=-2\cdot ta{n}^{-1}{e}^x+C$

x = 0, y = 0

$\Rightarrow 0=2ta{n}^{-1}+C$

$C=+\frac{\pi }{2}$

now at x = $\mathcal{l}n\sqrt[]{3}$

$ta{n}^{-1}y=-2ta{n}^{-1}\left(e^{\mathcal{l}n\sqrt[]{3}}\right)+\frac{\pi }{2}$

$6\left(y\text{'}\left(0\right)+{\left(y\left(\mathcal{l}n\sqrt[]{3}\right)\right)}^2\right)=6\left(-1+\frac{1}{3}\right)=-4$

${\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2xdx}}-{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x-x^{2}dx}}={\displaystyle \underset{0}{\overset{1}{\int }}dy}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-y^2}dy-{\displaystyle \underset{0}{\overset{2}{\int }}\frac{y^2}{2}dy}+{\displaystyle \underset{1}{\overset{2}{\int }}2dy+I}}$

$\Rightarrow \frac{8}{3}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}=1-\frac{8}{6}+2+I$

$I=1-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}$

$FeC{l}_3\underset{}{\overset{hydrolysis}{\to }}Fe{\left(OH\right)}_3\downarrow \underset{Adsorption}{\overset{F{e}^{3+}}{\to }}Fe{\left(OH\right)}_3|F{e}^{3+}\left(colloidalparticle\right)$