Class 12th

= IV cosf

$=\frac{20}{\sqrt[]{2}}*\frac{2}{\sqrt[]{2}}*$  cos 60°

= 10 W

Kindly go through the solution

p = i²R

P' = 0.64 i²R

By Brewsters law

m = tani

m =   $\sqrt[]{3}$

$1*\frac{\sqrt[]{3}}{2}=\sqrt[]{3}*sinr$

$sinr=\frac{1}{2}$    

r = 30°

|A| = 2

$\underset{2024times}{\underset{?}{adj\left(adj\left(adj....\left(a\right)\right)\right)}}={\left|A\right|}^{{\left(n-1\right)}^{2024}}$            

$={\left|A\right|}^{{\left(n-1\right)}^{2024}}$                                             

$=2^{2^{2024}}$                                          

$2^{2024}=\left(2^2\right)2^{2022}=4{\left(8\right)}^{674}=4{\left(9-1\right)}^{674}$             

-> $2^{2024}\equiv 4\left(mod9\right)$

->,  $2^{2024}\equiv 9m+4$  m ¬ even

$2^{9m+4}\equiv 16\cdot {\left(2^3\right)}^{3m}\equiv 16\left(mod9\right)$

7

  $?$  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)

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

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

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

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