Class 12th

Photochemical smog occurs in warm, dry and sunny climate. The main components of photochemical smog result from the action of sunlight on unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories. Photochemical smog has high concentration of oxidising agents so called as oxidizing smog.

CaO – Basic Oxide

SO₃ & SiO₂ – Acidic oxides

Isotopes are the atom of same element with different atomic mass.

$N=M^2+M^4+.....+M^{98}$

$=\left(-{\alpha }^{2}I\right)+{\left(-{\alpha }^{2}I\right)}^2+....+{\left(-{\alpha }^{2}I\right)}^{49}$

$=I\left(-{\alpha }^2+{\alpha }^4-{\alpha }^6+....-{\alpha }^{98}\right)$

N = $-I\left({\alpha }^2-{\alpha }^4+{\alpha }^6.......+{\alpha }^{98}\right)$

$=-I\frac{{\alpha }^2\left(1-{\left(-{\alpha }^2\right)}^{49}\right)}{1-\left(-{\alpha }^2\right)}$

N = $\frac{-I{\alpha }^2\left(1+{\alpha }^{98}\right)}{1+{\alpha }^2}$

Now $\left(I-m^2\right)N=-2I$

$\left(I+{\alpha }^{2}I\right)\frac{\left(-I{\alpha }^2\cdot (1+{\alpha }^{98}\right)}{1+{\alpha }^2}=-2I$

? α¹⁰⁰ + α² = 2

? α = $\pm 1$

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$

So, g (x) has at least two roots in (2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$

It is same as number of roots of $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (2, 2)

Given a > b

Area common to x2 + y2 $\le a^{2}and\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1$

is $\pi a^2-\pi ab=30\pi ..............\left(i\right)$

Similarly $\pi ab-\pi b^2=18\pi .................\left(ii\right)$

Equation (i) and equation (ii) $\frac{a}{b}=\frac{5}{3}$

Equation (i) + equation (ii) $a^2-b^2=48$

a² = 75, b² = 27

 $\frac{dy}{dx}+\frac{2x}{x-1}\cdot y=\frac{1}{{\left(x-1\right)}^2}$

IF $=e^{{\displaystyle \int \frac{2x}{x-1}dx}}$

= $e^{2x}\cdot {\left(x-1\right)}^2$

$y\cdot e^{2x}{\left(x-1\right)}^2=\{e^{2x}\frac{{\left(x-1\right)}^2}{{\left(x-1\right)}^2}dx+C$

y = $\frac{e^{2x}}{2{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^2}$

y (2) = $\frac{1+e^4}{2e^4},\Rightarrow C=\frac{1}{2}$

y (3) = $\frac{e^{\alpha }+1}{\beta e^{\alpha }}=\frac{e^6+1}{8e^6}$

Data contradiction.

$\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=\left(\overrightarrow{a}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\overrightarrow{c}$