Relations and Functions

number of elements in $A\cap B=5$ which is

(0, 0) (1, 0) (1, 1) (1, -1) (2, 0)

Similarly number of elements in $A\cap C=5$ which is

(2, 0) (2, 2) (1, 1) (2, 1) (3, 1)

Hence number of relation from

$\left(A\cap B\right)to\left(A\cap C\right)=2^{5*5}=2^{25}$            

->P = 25

 $-1\le \frac{x^2-3x+2}{x^2+2x+7}\le 1$

$\Rightarrow 0\le \frac{2x^2-x+9}{x^2+2x+7}\&\frac{-5x-5}{x^2+2x+7}\le 0$

$x\in R\&-1\le x<\infty$

a₀ = 0, a₁ = 0

a_n+2 = 3a_n+1 – 2a_n + 1

a₂₅ a₂₃ – 2a₂₅a₂₂ – a₂₃a₂₄ + 4a₂₂a₂₄ =?

a₂ = 3a₁ – 2a₀ + 1

a₃ = 3a₂ – 2a₁ + 1

a₄ = 3a₃ – 2a₂ + 1

a₅ = 3a₄ – 2a₃ + 1

a_n+2  = 3a_n+1 – 2a_n + 1

$\left(+\right)\Rightarrow \left(a_2+a_3+a_4+....+a_{n+1}+a_{n+2}\right)$                

⇒ a_n+2 = 2 (a₂ + a₃ + …. + a_n + a_n+1) –2 (a₁ + a₂ + ….+ a_n) + n + 1

a_n+2 = 2a_n+1 + n + 1

a₂₅ a₂₃ -2a₂₅ a₂₂ -a₂₃ a₂₄ + 4a₂₂ a₂₄

= a₂₅ (a₂₃ – 2a₂₂) -2a₂₄ (a₂₃ – 2a₂₂)

As a_n+2 = 2a_n+1 + n + 1

⇒ a_n+2 – 2a_n+1 = n + 1

⇒ a_n+1 -2a_n = n

⇒ 24 * 22 = 528

y = 2x $\left|3x^2-5x+2\right|,0\le x\le 1$

$=\{\begin{array}{cc}\hfill -3x^2+7x-2,\hfill & \hfill 3x^2-3x+2,\hfill \end{array}\begin{array}{cc}\hfill 0\le x\le \frac{2}{3}\hfill & \hfill \frac{2}{3}<x\le 1\hfill \end{array}$

3x² – 5x + 2 = 0

$x=\frac{+5\pm \sqrt[]{25-24}}{6}$

= $\frac{5+1}{6}=1,\frac{2}{3}$

3x² – 7x + 3 = 0

x = $\frac{7\pm \sqrt[]{49-39}}{6}=\frac{7\pm \sqrt[]{13}}{6}$

$-3x^2+7x-2$

$\frac{-7\pm \sqrt[]{49-24}}{-6}=\frac{7+5}{6}=2,\frac{1}{3}$

3x² + 7x – 2 = 1

3x² – 7x + 1 = 0

x = $\frac{7\pm \sqrt[]{49-12}}{6}=\frac{7\pm \sqrt[]{37}}{6}$

I = ${\displaystyle \underset{0}{\overset{6}{\int }}\left(\begin{array}{cc}\hfill \left(-2\right)\hfill & \hfill +1\hfill \end{array}\right)dx+{\displaystyle \underset{\frac{7-\sqrt[]{37}}{6}}{\overset{\frac{1}{3}}{\int }}\left(\begin{array}{cc}\hfill \left(-1\right)\hfill & \hfill +1\hfill \end{array}\right)dx}+{\displaystyle \underset{-\frac{1}{3}}{\overset{\frac{7-\sqrt[]{13}}{6}}{\int }}\left(\begin{array}{cc}\hfill 0\hfill & \hfill +1\hfill \end{array}\right)dx+{\displaystyle \underset{\frac{7-\sqrt[]{13}}{6}}{\overset{\frac{2}{3}}{\int }}\left(1+1\right)dx}}}+{\displaystyle \underset{\frac{2}{3}}{\overset{1}{\int }}\left(1+1\right)dx}$

$=\frac{\sqrt[]{37}+\sqrt[]{13}-4}{6}$

For R₁, take a = 1, b = 0, c = -1

$ab\ge 0,bc\ge 0$ but AC < 0

So R₁ is not an equivalence relation.

For R₂, it will not be symmetric.

So R₂ is also not an equivalence relation.

Kindly consider the following figure

  $x^2-10x+9\le 0$

  $\left(x-1\right)\left(x-9\right)\le 0$              

A = {1, 2, 3, ……, 9}

for set B,   $f\left(x\right)\le {\left(x-3\right)}^2+1$

total number of such function = 2 * 1 * 1 * 1 * 2 * 3 * 4 * 5 * 6 = 2 * 6! = 1440

$\left(x^2+1\right)*1=x*2$

$\Rightarrow {\left(x^2+1\right)}^2+1=x^2+8\Rightarrow x^2=2$

$2si{n}^{-1}\left(\frac{x^4+x^2-2}{x^4+x^2+2}\right)=2si{n}^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$