Class 12th

The equation of a plane parallel to x - 2y + 2z - 3 = 0 is x - 2y + 2z + λ = 0.
The distance from the point (1, 2, 3) to this plane is 1.
|1 - 2 (2) + 2 (3) + λ| / √ (1² + (-2)² + 2²) = 1
|1 - 4 + 6 + λ| / √9 = 1
|3 + λ| / 3 = 1
|3 + λ| = 3
3 + λ = 3 or 3 + λ = -3
λ = 0 or λ = -6.

1 = (2-1)¹ (The n is likely 1).
3? = (7-4)³ (This seems to be a pattern matching (a-b)^c).
4²? = (12-8)? ! = 4²?
The blank space must be (5-3)² = 2² = 4.

Circle 1: x² + y² - 10x - 10y + 41 = 0
Center C? = (5,5).
Radius r? = √ (5² + 5² - 41) = √ (25+25-41) = √9 = 3.

Circle 2: x² + y² - 22x - 10y + 137 = 0
Center C? = (11,5).
Radius r? = √ (11² + 5² - 137) = √ (121+25-137) = √9 = 3.

Distance between centers d (C? , C? ) = √ (11-5)² + (5-5)²) = √ (6²) = 6.

Sum of radii r? + r? = 3 + 3 = 6.
Since the distance between the centers is equal to the sum of their radii, the circles touch externally at one point.

Linear: N? (azide ion)

Bent-shape: O? (ozone), NO? (nitrite ion), Cl? O (dichlorine monoxide)

1° amine: Forms a precipitate that is soluble in NaOH.

2° amine: Forms a precipitate that is insoluble in NaOH.

3° amine: No reaction.

A function f (x) is continuous at x=1, so lim (x→1? ) f (x) = lim (x→1? ) f (x) = f (1).
Assuming a piecewise function like f (x) = { -x, x<1; ax+b, x1 } (structure inferred from derivative).
Continuity at x=1: f (1) = 1. a (1)+b = 1 => a+b=1.

The function is differentiable at x=1. The derivative of f (x) at x=1 from the left is -1. The derivative from the right is a.
So, a = -1. (The image has 2a = -1, which would imply a function like -x and ax²+b). Let's assume f' (x) = 2a for x>1.
2a = -1 => a = -1/2.

From a+b=1, b = 1 - a = 1 - (-1/2) = 3/2.
So, a = -1/2 and b = 3/2.