Maths

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

S = (2 . ¹P? - 3 . ²P? + 4 . ³P? upto 51 terms) + (1! - 2! + 3! - . upto 51 terms)
∴ [? ? P_ (n-1) = n!]
= (2! - 3! + 4! + 52!) + (1! - 2! + 3! - 4! + . . + (51)!)
= 1! + 52!

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

LHL : lim_ (x→0? ) |1-x-x|/|λ-x-1| = 1/|λ-1|
RHL: lim_ (x→0? ) |1-x+x|/|λ-x+0| = 1/|λ|
For existence of limit
LHL = RHL
⇒ 1/|λ-1| = 1/|λ| ⇒ λ = 1/2
∴ L = 1/|λ| = 2

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, ∞)

T_r+1 =? C_r (3)^ (n-r)/2) (5)^ (r/8) (n ≥ r)
Clearly r should be a multiple of 8.
∴ there are exactly 33 integral terms
Possible values of r can be
0,8,16, . . .,32 * 8
∴ least value of n = 256

Sum of 1st 25 terms = sum of its next 15 terms
? (T? + . + T? ) = (T? + . + T? )
? (T? + . + T? ) = 2 (T? + . + T? )
? 40/2 [2*3 + (39d)] = 2 * 25/2 [2*2 + 24 d]
? d = 1/6

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