Class 12th

f (x) = (c + 1)x² + (1 – c²)x + 2k

$f\text{'}\left(x\right)=\left(\left(c+1\right)\right)2x+\left(1-c^2\right).............\left(i\right)$

$f\text{'}\left(x\right)=\underset{y\to 0}{lim}I\frac{f\left(x+y\right)-f\left(x\right)}{x+y-x}$

$=\underset{y\to 0}{lim}\frac{f\left(y\right)-xy}{y}$

$f\text{'}\left(x\right)=\underset{y\to 0}{lim}\frac{f\left(y\right)}{y}-x$

$x+f\text{'}\left(x\right)=\underset{y\to 0}{lim}\frac{f\left(y\right)-f\left(0\right)}{y-0}asf\left(c\right)=0$

 $co{s}^{-1}\frac{1}{\sqrt[]{5}}=co{t}^{-1}\frac{1}{2}$

$tan\left(2\left(ta{n}^{-1}\frac{1}{2}+co{t}^{-1}\frac{1}{2}\right)\right)+ta{n}^{-1}\left(\frac{1}{2}\right)=tan\left(\pi +ta{n}^{-1}\frac{1}{2}\right)=\frac{1}{2}$

$tan2\alpha =2\sqrt[]{2}\Rightarrow \frac{2tan\alpha }{1-ta{n}^2\alpha }=2\sqrt[]{2}$

$\sqrt[]{2}ta{n}^2\alpha +tan\alpha -\sqrt[]{2}=0$

$tan\alpha =\frac{1}{\sqrt[]{2}},\left(-\sqrt[]{2}\right)\to rejected$

At maximum height velocity (v) = 0

So, momentum = mv = m * 0 = 0

$co{s}^{-1}\frac{1}{\sqrt[]{5}}=co{t}^{-1}\frac{1}{2}$

$tan\left(2\left(ta{n}^{-1}\frac{1}{2}+co{t}^{-1}\frac{1}{2}\right)\right)+ta{n}^{-1}\left(\frac{1}{2}\right)=tan\left(\pi +ta{n}^{-1}\frac{1}{2}\right)=\frac{1}{2}$

$tan2\alpha =2\sqrt[]{2}\Rightarrow \frac{2tan\alpha }{1-ta{n}^2\alpha }=2\sqrt[]{2}$

$\sqrt[]{2}ta{n}^2\alpha +tan\alpha -\sqrt[]{2}=0$

$tan\alpha =\frac{1}{\sqrt[]{2}},\left(-\sqrt[]{2}\right)\to rejected$

F.B.D of person w.r.t. elevator

mg – N = ma

N = mg – ma

N decreases so, person experiences weight loss.

$P=\frac{\alpha }{\beta }lo{g}_e\left(\frac{kt}{\beta x}\right)$

$\frac{kt}{\beta x}$ = Dimensionless

$\beta =\frac{kt}{x}=\frac{\left[M{L}^2{T}^{-2}{k}^{-1}\right]\left[k\right]}{\left[L\right]}$

$\frac{\alpha }{\beta }=$ dimensionless

= dimensionless of

= MLT²

(a) Sucrose  → α-D glucose and β-D fructose

(b) Lactose  → β-D- galactose and β-D glucose

(c) maltose  → α-D glucose and α-D-glucose

Consider the following image

It does not contains 2^nd most abundant element by weight in earth crust because that is Si Calgon

$\underset{}{\overset{Watersoluble}{\to }}2N{a}^{+}{\left[N{a}_4{\left(P{O}_3\right)}_6\right]}^{2-}$

$\underset{}{\overset{C{a}^{2+}}{\to }}2N{a}^{+}{\left[N{a}_{2}Ca{\left(P{O}_3\right)}_6\right]}^{2-}$

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