Class 12th

  $?$  R is symmetric relation

⇒   (y, x) ∈ R V (x, y) ∈ R

(x, y) ∈ R ⇒ 2x = 3y and (y, x) ∈ R ⇒ 3x = 2y

Which holds only for (0, 0)

Which does not belongs to R.

∴ Value of n = 0

Take $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$  

x = 2λ + 1, y = 3λ + 2, z = 4λ + 3

  $\overrightarrow{AB}$  = (α − 2)  $\widehat{i}$ + (β − 3) $\widehat{j}$ + (γ − 4) $\widehat{k}$  

Now,

(α − 2) ⋅ 2 + (β − 3) ⋅3 + (γ − 4) ⋅ 4 = 0

2α − 4 + 3β − 9 + 4γ −16 = 0

⇒ 2α + 3β + 4γ = 29

  $f\left(x\right)=e^{x^3-3x+1}$

$f\text{'}\left(x\right)=e^{x^3-3x+1}$  (3x² − 3)

$e^{x^3-3x+1}$ = ⋅ 3(x −1)(x +1)

For x ∈ (−∞, −1], f '(x) ≥ 0

∴     f(x) is increasing function

∴     a = e^–∞ = 0 = f (−∞)

b = e⁻¹⁺³⁺¹ = e³ = f (−1)

∴     P(4, e³ + 2)

$d=\frac{\left(e^3+2\right)\left(e^{-3}\right)}{\sqrt[]{1+e^{-6}}}=\frac{1+2e^{-3}}{\sqrt[]{1+e^{-6}}}=\frac{1+2e^{-3}}{\sqrt[]{1+e^{-6}}}=\frac{e^3+2}{\sqrt[]{e^6+1}}$

Take e^sinx = t (t > 0)

⇒ $t-\frac{2}{t}=2$

⇒   $\frac{t^2-2}{t}=2$

->t² – 2t – 2 = 0

->t² – 2t + 1 = 3

⇒   (t −1)² = 3

⇒   t = 1 ± $\sqrt[]{3}$  

⇒   t = 1 ± 1.73

⇒   t = 2.73 or –0.73 (rejected as t > 0)

⇒   e^{sin x} = 2.73

->log_e e^{sin x} = log_e 2.73

⇒ sin x = log_e 2.73 > 1

So no solution.

f is increasing function

x < 5x < 7x

f (x) < f (5x) < f (7x)

  $\frac{f\left(x\right)}{f\left(x\right)}<\frac{f\left(5x\right)}{f\left(x\right)}<\frac{f\left(7x\right)}{f\left(x\right)}$           

$\underset{x\to \infty }{lim}\frac{f\left(x\right)}{f\left(x\right)}<\underset{x\to \infty }{lim}\frac{f\left(5x\right)}{f\left(x\right)}<\underset{x\to \infty }{lim}\frac{f\left(7x\right)}{f\left(x\right)}$             

-> $1<\underset{x\to \infty }{lim}\frac{f\left(5x\right)}{f\left(x\right)}<1$ $\underset{x\to \infty }{lim}\frac{f\left(5x\right)}{f\left(x\right)}=1$

$\underset{x\to \infty }{lim}\left(\frac{f\left(5x\right)}{f\left(x\right)}-1\right)=0$            

Area =   $\left|{\displaystyle \underset{1}{\overset{4}{\int }}\left[{\left(4-x\right)}^2-\frac{{\left(x-4\right)}^2}{3}\right]}\right|dx$

Area =   $=\left|\left(16-\frac{64}{3}\right)-\left(4-\frac{1}{3}+\frac{27}{9}\right)\right|$

$=\left|\left(16-\frac{64}{3}\right)-\left(4-\frac{1}{3}+\frac{27}{9}\right)\right|$

$=\left|16-\frac{64}{3}-4+\frac{1}{3}+3\right|$

$=\left|15-2\right|=6$

Let probability of tail is   $\frac{1}{3}$

⇒ Probability of getting head = $\frac{2}{3}$  

∴ Probability of getting 2 heads and 1 tail

$=\left(\frac{2}{3}*\frac{2}{3}*\frac{1}{3}\right)*3$

$=\frac{4}{27}*3$

$=\frac{4}{9}$                  

This is an example of chromyl chloride test

K₂Cr₂O₇ + 4KCl + 6H₂SO₄? 6KHSO₄ + 2CrO₂Cl₂ + 3H₂O

Oxidation state of Cr is +6.

The balanced reaction is as follows

$2Mn{O}_4^{-}+5C_2{O}_4^{2+}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$            

2 mole  $Mn{O}_4^{-}$ react with 16 mole H⁺

1 mole  $Mn{O}_4^{-}$ will react with 8 mole H⁺