Maths

  $?$  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

After giving 2 apples to each child 15 apples left now 15 apples can be distributed in
^15+3–1C₂ = ¹⁷C₂ ways

  $=\frac{17*16}{2}=136$          

${\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{x^{2}sinx\cdot cosx}{si{n}^{4}x+co{s}^{4}x}dx}$

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}\left(x^2-{\left(\pi -x\right)}^2\right)dx}$

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinx\cdot cosx\left(2\pi x-{\pi }^2\right)}{si{n}^{4}x+co{s}^{4}x}x}$

$=2\pi \cdot \frac{\pi }{4}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx}-{\pi }^2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\frac{1}{2}sin2x}{1-\frac{1}{2}si{n}^{2}2x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sin2x}{2-si{n}^{2}2x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sin2x}{1+co{s}^{2}2x}dx}$

Let cot2x = t

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{1}{\overset{-1}{\int }}\frac{-\frac{1}{2}dt}{1+t^2}}$

$=-\frac{{\pi }^2}{4}{\displaystyle \underset{-1}{\overset{1}{\int }}\frac{dt}{1+t^2}}$

$=-\frac{{\pi }^2}{4}\cdot \frac{\pi }{2}=-\frac{{\pi }^2}{8}$

$\frac{120}{{\pi }^3}\left|-\frac{{\pi }^3}{8}\right|=15$

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)

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

$=8*7*6*5*4+\frac{9*8}{2}$

= 6756

(y – 2) = m (x – 8)

⇒   x-intercept

⇒     $\left(\frac{-2}{m}+8\right)$

⇒   y-intercept

⇒   (–8m + 2)

⇒   OA + OB = $\frac{-2}{m^2}$  + 8 – 8m + 2

$f\text{'}\left(m\right)=\frac{2}{m^2}-8=0$  

-> $m^2=\frac{1}{4}$

-> $m=\frac{-1}{2}$

-> $f\left(\frac{-1}{2}\right)=18$

->Minimum = 18

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.

z₁ + z₂ = 5

$z_1^3+z_2^3=20+15i$            

  $z_1^3+z_2^3={\left(z_1+z_2\right)}^3-3z_1{z}_2\left(z_1+z_2\right)$           

$z_1^3+z_2^3=125-3z_1\cdot z_2\left(5\right)$            

 ⇒ 20 + 15i = 125 – 15z₁z₂

⇒ 3z₁z₂ = 25 – 4 – 3i

3z₁z₂ = 21– 3i

z₁⋅z₂ = 7 – i

(z₁ + z₂)² = 25

$z_1^2+z_2^2=25-27\left(7-i\right)$     

= 11 + 2i

  ${\left(z_{\text{?}1}^2+z_2^2\right)}^2$         = 121 − 4 + 44i

⇒   $z_1^4+z_2^4+2{\left(7-i\right)}^2=117+44i$

⇒   $z_1^4+z_2^4$ = 117 + 44i − 2(49 −1−14i )

= 21 + 72i

⇒   $\left|Z_1^4+Z_2^4\right|=75$