Class 12th

Kindly consider the following figure

$l={\displaystyle {\int }_{-3}^{103}\left(\left[sin\left(\pi x\right)\right]+e^{\left[cos\left(2\pi x\right)\right]}\right)}dx$

$\left[sin\pi x\right]$ is periodic with period 2 and

$e^{\left[cos2\pi x\right]}$ is periodic with period 1.

So,

$I=52{\displaystyle {\int }_0^2\left(\left[sin\left(\pi x\right)\right]+e^{\left[cos2\pi x\right]}\right)}dx$

= $\frac{52}{e}$

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

$\Rightarrow \frac{1}{2}=\frac{1}{3}+\frac{1}{5}-P\left(A\cap B\right)$

$P\left(\frac{A}{B\text{'}}\right)+P\left(\frac{B}{A\text{'}}\right)$

$=\frac{P\left(A\right)-P\left(A\cap B\right)}{1-P\left(B\right)}+\frac{P\left(B\right)-P\left(A\cap B\right)}{1-P\left(A\right)}=\frac{5}{8}$

Following are the $\Delta H°$  hydration of given ions

Cr²⁺ - 1926 KJ/mole

Mn²⁺ - 1860 KJ/mole

Fe²⁺ - 2000 KJ/mole

Co²⁺ - 2078 KJ/mole

Cobalt (II) meta borate is a blue colour of Bead.

Here 'X' is the Gypsum (CaSO₄.2H₂O) which is used to enhance the setting of time

In leaching suitable compound is used as solvent.

$A{\mathcal{l}}_2{O}_3+2NaOH+3H_{2}O\to 2Na\left[A\mathcal{l}{\left(OH\right)}_4\right]$

Stannane is the tetrahedral shape of molecule.

Hybridisation is- sp³, shape and structure are tetrahedral

$11\equiv 2mod\left(9\right)$

${\left(11\right)}^{1011}\equiv 2^{1011}mod\left(9\right)$

Again 23 $\equiv$ 1 mod (9)

$\Rightarrow {\left(2^3\right)}^{337}\equiv {\left(-1\right)}^{337}mod\left(9\right)$

$\therefore {\left(11\right)}^{1011}+{\left(1011\right)}^{11}\equiv 8mod\left(9\right)$

$\therefore$ Remainder = 8

f : {1, 3, 5, 7, ….,99} {2, 4, 6, 8, …….100}

$f\left(3\right)\ge f\left(9\right)\ge f\left(15\right)......\ge f\left(99\right)$

$3+\left(n-1\right)6=99\Rightarrow n=17$

cases f (3) > f (9) > f (15) ……. > f (99)

$\therefore$ from the set {2, 4, 6, …., 100}

17 distinct numbers can be selected in 50C17 ways again remaining {1, 5, 7, 11, ….} can map in 33! ways

$\therefore$ total number of such required functions

${=}^{50}{C}_{17}*33!$

$=\frac{50!}{33!17!}*33!{=}^{50}{P}_{33}$