Maths

$\sim \left(\sim q\wedge p\right)=\left(\sim \left(\sim q\right)\right)\vee \left(\sim p\right)$  

$=\sim p\vee q$            

$=p\to q$            

$\therefore contrapositiveis\sim q\to \sim p$            

 Hence, if $\sim$ q then $\sim$ p

A: runner succeeded exactly 3 times out of 5.
B: runner succeeds on the first trial.
P (B/A) = P (B ∩ A) / P (A) = (p (? C? p² (1−p)²) / (? C? p³ (1-p)²) = 3/5 = 0.6

Case-I : $a-2>0$

$F(-2)>0$

$4(a-2)+4a+a>0$

$9\mathrm{a}-8>0$

$a>\frac{8}{9}$

$F\left(1\right)>0$

$a-2-2a+a>0$

  $-2>0$  (not possible)

Case-II : $a-2<0;a<2$

$F(-2)<0\mathrm{}\Rightarrow \mathrm{}a<\frac{8}{9}$

$F\left(1\right)<0\mathrm{}\Rightarrow \mathrm{}a\in R$

$-2<\frac{-b}{2a}<1\mathrm{}a<\frac{4}{3}\mathrm{}a\in \left[0,\frac{8}{9}\right)\cup \left\{2\right\}$

$D\ge 0\mathrm{}a\ge 0$

|x|³/? = y
⇒ y² – 26y – 27 = 0 ⇒ y = −1 or 27
⇒ y = 27 ⇒ |x|³/? = 3³
|x| = 3? ⇒ x = ±3?
Product of roots = (3? ) (−3? ) = −3¹?

tan α = 1/x, tan β = 2/x, tan γ = 3/x
Now α + β + γ = 180°
⇒ tan α + tan β + tan γ = tan α tan β tan γ
1/x + 2/x + 3/x = (123)/ (xxx)
⇒ x² = 1

The equation of the line through P (x, y) making an angle with the x-axis which is supplementary to the angle made by the tangent at P (x, y) is

$Y-y=-\frac{dy}{dx}\left(X-x\right)$                    …. (1)

At the point where it meets the x-axis

Y = 0, X = x    $+\frac{y}{\frac{dy}{dx}}\Rightarrow OA=x+\frac{y}{\frac{dy}{dx}}$    …. (2)

The line through P (x, y) and perpendicular to (1) is

$Y-y=\frac{dx}{dy}\left(X-x\right)$           

At the point where it meets the y-axis

$X=0,Y=y=\frac{x}{\frac{dy}{dx}}\Rightarrow OB=y-\frac{x}{\frac{dy}{dx}}$           .

$\boxed{x^2+2xy-y^2=2}$            

RHL&LHL lim (x→0) (sin (2x²/a) + cos (3x/b)^ (ab/x²)
= e^ (lim (x→0) (sin (2x²/a) + cos (3x/b) - 1) (ab/x²) = e^ (4b²-9a)/2b)
f (0) = e³
For continuity at x = 0
Limit = f (0)
(4b² - 9a)/2b = 3 ⇒ 4b² – 6b – 9a = 0∀b ∈ R
⇒ D ≥ 0 ⇒ a ≥ -1/4
a? = -1/4
⇒ |1/a? | = 4.

x² + y² – 6x + 8y + 24 = 0 is circle having centre (3, −4) & r = √ (9+16-24) = 1
√x² + y² min. is min. distance from origin = 4
∴ minimum value of log? (x² + y²) = log? 16 = 4

A = {1,2,3,4,5} B = {0,1,2,3,4}
No. of elements common in A&B = 4.
∴ No. of elements common in A * B and B * A = 4 * 4 = 16

Centre of C₃ will lie on the radical axis of C₁ and C₂ which is 10x + 6y + 26 = 0.

Let center of C₃ is (h, k).

Equation of chord of contact through (h, k) to the C₁ may be given as hx + ky = 25 … (I)

Let the mid point of the chord is (x₁, y₁) the equation of the chord with the help of mid point may be given as  

$x{x}_1+y{y}_1=x_1^2+y_1^2$

Since (I) and (II) represents same straight line 10x + 6y + 26 = 0

$\Rightarrow \frac{h}{25}=\frac{x_1}{x_1^2+y_1^2},\frac{k}{25}=\frac{y_1}{x_1^2+y_1^2}$            

The locus of (x₁, y₁) is

$5x+3y+\frac{13}{25}\left(x^2+y^2\right)=0$