Class 12th

? 2 adj (3 A adj (2A)|
= 23.? 3 A adj (2A)|
|2
= 23 ⋅ (33)2 ⋅ | A|2 ⋅ |adj (2 A)|2
= 23 ⋅ 36 ⋅ | A|2 ⋅ (|2 A|2)2
= 23 ⋅ 36 ⋅ | A|2 [ (23)2 ⋅ | A|2]2
= 23. 36. |A|2. 212. |A|4
= 215. 36. |A|6
= 215 ⋅ 36 ⋅ 56 = 2? ⋅ 3? ⋅ 5?
? ? = 15? = 6? = 6
? +? +? = 27

Kindly go through the solution

A = [ x 1 ]
[ 1 0 ]
A² = [ x 1 ] [ x 1 ] = [ x²+1 x ]
[ 1 0 ] [ 1 0 ] [ x 1 ]
A? = [ x²+1 x ] [ x²+1 x ]
[ x 1 ] [ x 1 ]
= [ (x²+1)²+x² x (x²+1)+x ]
[ x (x²+1)+x²+1 ]
a? = (x² + 1)² + x² = 109
⇒ x = ±3
a? = x² + 1 = 10

0 ≤ y ≤ x² + 1, 0 ≤ y ≤ x + 1, 1/2 ≤ x ≤ 2
Required area
= 19/24 + 5/2 = 79/24

2π - (sin? ¹ (4/5) + sin? ¹ (5/13) + sin? ¹ (16/65)
= 2π - (tan? ¹ (4/3) + tan? ¹ (5/12) + tan? ¹ (16/63)
= 2π - (tan? ¹ (63/16) + tan? ¹ (16/63)
= 2π - π/2 = 3π/2

f (x) = (3x - 7)x²/³
⇒ f (x) = 3x? /³ - 7x²/³
⇒ f' (x) = 5x²/³ - 14/ (3x¹/³)
= (15x - 14) / (3x¹/³) > 0

∴ f' (x) > 0 ∀x ∈ (-∞, 0) U (14/15, ∞)

When the two conducting spheres are connected by a wire, they will reach the same electric potential, V.
The total charge Q_total = 12µC + (-3µC) = 9µC. This total charge will redistribute.
Let the final charges be q? and q? + q? = 9µC.
The potential of a sphere is V = kq/r.
V? = V?
k q? /R? = k q? /R? ⇒ q? /R? = q? /R?
q? / (2R/3) = q? / (R/3) ⇒ q? /2 = q? ⇒ q? = 2q?
Substitute this into the charge conservation equation:
2q? + q? = 9µC ⇒ 3q? = 9µC ⇒ q? = 3µC.
Then, q? = 2 * 3µC = 6µC.
The final charges are 6µC and 3µC.

Sum obtained is a multiple of 4.
A = { (1,3), (2,2), (3,1), (2,6), (3,5), (4,4), (5,3) (6,2), (6,6)}
B: Score of 4 has appeared at least once.
B = { (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (4,1), (4,2), (4,3), (4,5), (4,6)}
Required probability = P (B/A) = P (B? A)/P (A)
= (1/36) / (9/36) = 1/9

∫ [-π to π] |π - |x|dx = 2∫ [0 to π] |π - x|dx
= 2∫ [0 to π] (π - x)dx
= 2 [πx - x²/2] (from 0 to π) = π²