Maths

Find RHL, x = -1 + h

$\underset{h\to 0}{lim}a.sin\left(\frac{\pi \left[-1+h\right]}{2}\right)+\left[2+1-h\right]=-a+2$               

 Find LHL, x = -1 – h

$\underset{h\to 0}{lim}asin\frac{\left(\pi \left[-1-h\right]\right)}{2}=\left[2+1+h\right]=3$               

${\displaystyle \underset{0}{\overset{4}{\int }}f\left(x\right)dx={\displaystyle \underset{0}{\overset{1}{\int }}1dx+{\displaystyle \underset{1}{\overset{2}{\int }}\left(-1\right)dx+{\displaystyle \underset{2}{\overset{3}{\int }}\left(-1\right)dx+{\displaystyle \underset{3}{\overset{4}{\int }}\left(-1\right)dx=-2}}}}}$               

$\frac{S_5}{S_9}=\frac{5}{17}\Rightarrow d=4a$  

110 < a₁₅ < 120

110 < a + 14d < 120

110 < 57a < 120

->a = 2, d = 8

Use binomial theorem

2023 = 7 * 289

$2021^{2022}+2022^{2021}={\left(2022-2\right)}^{2022}+{\left(2023-1\right)}^{2021}$

$=7P_1+2^{2022}+7P_2-1$

$=7\left(P_1+P_2\right)+1+7P_3-1$                          

Kindly go through the solution

Use characteristic equation = 0

$V={\left|z\right|}^2+{\left|z-3\right|}^2+{\left|z-6i\right|}^2$  

Put z = x + iy

$V=3\left[{\left(x-1\right)}^2+{\left(y-2\right)}^2+10\right]$                

For minimum x = 1, y = 2, V₀ = 30

So, z₀ = 1 + 2i

f (a) = a, a is max of powers of prime P such that P^a divides a.

f (2) = 1; g (2) = 3

f (3) = 1; g (3) = 4

f (4) = 2; g (4) = 5

f (5) = 1; g (5) = 6

f (2) + g (2) = 4

f (3) + g (3) = 5

f (4) + g (4) = 7

f (5) + g (5) = 7

So, f (x) + g (x) is many-one-and an into function. We won't get 'I' in the range.

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.

  $\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-2\left(3a-4b+12c+19\right)}{3^2+{\left(-4\right)}^2+12^2}$

$\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-6a+8b-24c-38}{169}$               

$\left(x,y,z\right)\equiv \left(a-6,\beta ,\gamma \right)$               

  $\frac{\left(a-b\right)-a}{3}=\frac{\beta -b}{-4}=\frac{\gamma -c}{12}=\frac{-6a+8b-24c-38}{169}$              

  $\frac{\beta -b}{-4}=-2$              

$\Rightarrow \beta =8+b$               

$\Rightarrow 3a-4b+12c=150$                   ….(i)

a + b + c = 5

$\Rightarrow 3a+3b+3c=15$ ….(ii)

Applying (i) – (ii), we get :

= 56 + 216 + 7b – 9c = 56 + 216 – 135 = 137

Let the equation of circle be

$x\left(x-\frac{1}{2}\right)+y^2+\lambda y=0$               

$\Rightarrow x^2+y^2-\frac{1}{2}x+\lambda y=0$

Radius = $\sqrt[]{\frac{1}{16}+\frac{{\lambda }^2}{4}}=2$  

$\Rightarrow {\lambda }^2=\frac{63}{4}\Rightarrow {\left(x-\frac{1}{4}\right)}^2+{\left(y+\frac{\lambda }{2}\right)}^2=4$   

$?$ This circle and parabola

$y-\alpha ={\left(x-\frac{1}{4}\right)}^2$ touch each other, so

$\alpha =-\frac{\lambda }{2}+2\Rightarrow \alpha -2=-\frac{\lambda }{2}\Rightarrow {\left(\alpha -2\right)}^2=\frac{{\lambda }^2}{4}=\frac{63}{16}$  

${\left(4\alpha -8\right)}^2=63$  

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side

  $?a=\sqrt[]{{\left(3?1\right)}^2+{\left(6?2\right)}^2}=\sqrt[]{20}\text{?}\text{?}and$

$\frac{b}{2}=\frac{4}{\sqrt[]{5}}?b=\frac{8}{\sqrt[]{5}}$

Area

$ab=2\sqrt[]{5}.\frac{8}{\sqrt[]{5}}=16$