Class 12th

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

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.

$|\mathrm{A}\mathrm{d}\mathrm{j}?A|=64=|A{|}^{n-1}\Rightarrow |A|=\pm 8$ $\alpha =\pm 8$

$\beta =|\mathrm{a}\mathrm{d}\mathrm{j}?(\mathrm{a}\mathrm{d}\mathrm{j}?A\left)\right|=|A{|}^{(n-1{)}^2}=8^4\mathrm{}\text{Also}\left|\frac{\sqrt[]{\beta }}{\alpha }\right|=\frac{8^2}{8}=8$

Volume = |a b c; b c a; c a b| = |3abc – a³ – b³ – c³|
= | (a + b + c) (a² + b² + c² – ab – bc – ca)| = | (a + b + c) (a + b + c)² – 3 (ab + bc + ca)|
= |9 (81-3 * 15)| = 324

lim (x→∞) (∫? ^ (√x²+1) tan? ¹t dt) / x = lim (x→∞) (tan? ¹ (√x²+1) * (x/√ (x²+1) = lim (x→∞) (tan? ¹ x) * (x/√ (x²+1) = π/2

I? = ∫? ¹ (1 − x? )? · 1 dx
= (1 - x? )? x|? ¹ - ∫? ¹ 7 (1 - x? )? (-4x³)xdx = -28∫? ¹ (1 − x? )? (1 − x? − 1)dx
I? = −28∫? ¹ (1 − x? )? dx + 28∫? ¹ (1 − x? )? dx
I? = -28I? + 28I?
29I? = 28I?
I? /I? = 28/29 ⇒ (29/4) * (I? /I? ) = (29/4) * (28/29) = 7

A = ∫? ² lnx dx = 2ln2 – 1
A' = 4 - 2 (2ln2 – 1) = 6 – 4ln2

  $\left(\lambda ,2\lambda ,3\lambda \right)$

$cos\theta =\frac{6}{\sqrt[]{42}}\Rightarrow sin\theta =\frac{\sqrt[]{6}}{\sqrt[]{42}}$ (where q is the angle between L₁ & L₂)

$Area\Delta OAB=\frac{1}{2}\left(OA\right)\left(OB\right)sin\theta$

$=\frac{1}{2}\left(\sqrt[]{3}\right)\left|\lambda \right|\left(\sqrt[]{14}\right)\frac{\sqrt[]{6}}{\sqrt[]{42}}=\sqrt[]{6}\Rightarrow \lambda =\pm 2$

${\mathrm{M}\mathrm{a}}_2{\mathrm{b}}_2{\mathrm{c}}_2=5$ Geometrical Isomers

M (AA)b2c2:3  $\mathrm{M}\left(\mathrm{A}\mathrm{A}\right){\mathrm{b}}_2{\mathrm{c}}_2:3$  Geometrical Isomers