Maths

f (x) = λ (x-2)²
⇒ 12 = λ (2)² ⇒ λ = 3
f (x) = 3 (x-2)² f (6) = 3 * 4² = 48

z - 2² = z + 2²
⇒ x=0
Hence minimum '0'

∫ (from 0 to 3) (x-1) (x-2) (x-3) dx = ∫ (from 0 to 3) (x³ - 6x² + 11x - 6) dx

A = ∫ (from 0 to 1) (x³ - 6x² + 11x - 6) dx - ∫ (from 1 to 2) (x³ - 6x² + 11x - 6) dx + ∫ (from 2 to 3) (x³ - 6x² + 11x - 6) dx
A = 11/4

Let P (B₁) = a      P (B₂) = b            P (B₃) = c

Given a (1 – b) (1 – c) = a . (i)

b (1 – a) (1 – c) = b              . (ii)

c (1 – b) (1 – a) =  $\gamma$             . (iii)

(1 – a) (1 – b) (1 – c) = p     . (iv)

$\left(\alpha -2\beta \right)p=\alpha \beta$  

->a – ab – 2b + 2ab = ab Þ a = 2b . (v)

Again $\left(\beta -3\gamma \right)p=2\beta \gamma$  

-> b – bc – 3c + 3bc = 2bc Þ b = 3c   . (vi)

$\Rightarrow \frac{P\left(B_1\right)}{P\left(B_3\right)}=\frac{a}{c}=\frac{2b}{b/3}=6$        

(Var)new = k² (Var)old
= 4 * 5 = 20

  $x^2+y^2-2x-6y+6=0$ centre (1, 3)

$\begin{array}{l}r=\sqrt[]{1+9-6}=2\\ CM=\sqrt[]{1+4}=\sqrt[]{5}\end{array}$           

$r=\sqrt[]{5+4}=3$

y = (3/2)sin (2θ)
x = e^θ sinθ
dy/dθ = 3cos (2θ)
dx/dθ = e^θ (cosθ + sinθ)
dy/dx = (3cos (2θ) / (e^θ (cosθ + sinθ) = (3 (cosθ - sinθ) / e^θ

~ (p v (~ p v q)
= p ^ ( (~p v q) = ~p ^ (p ^ ~q)

Sum of elements $A\cap \left(B\cup C\right)=274*400$  

In set B numbers of the form 9k + 2 are {101, 109, .992}

  $\therefore sum=\frac{100}{2}\left(101+992\right)=\frac{100*1093}{2}......\left(i\right)$         

Another possible number is 9k + 5 forms are {104, .995}

  $\therefore sum=\frac{100}{2}\left(104+995\right)=\frac{100}{2}*1099.........\left(ii\right)$

$\therefore Total=\frac{100}{2}*\left[1093+1099\right]=100*1096=274*4*100=274*400$           

$\therefore$ possible value of $\mathcal{l}$ = 5

  $\frac{4}{sinx}+\frac{1}{1-sinx}=a\forall x\in \left(0,\frac{\pi }{2}\right)$

Let sin x = t, t  $\in$ (0, 1)

$g\left(t\right)=\frac{4}{t}+\frac{1}{1-t}$         

$g\text{'}\left(t\right)=0\Rightarrow t=\frac{2}{3}$           

$g\text{'}\text{'}\left(\frac{2}{3}\right)>0$           

$\therefore g{\left(t\right)}_{minimum}=\frac{4}{2/3}+\frac{1}{1-2/3}=9$          

Minimum value of a for which solution exist = 9