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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!

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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

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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.

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

The internal energy of one mole of an ideal gas is given by U = (f/2)RT, where f is the number of degrees of freedom.
For a non-linear triatomic molecule (assumed to be a rigid rotator), there are:

3 translational degrees of freedom.

3 rotational degrees of freedom.
Vibrational modes are generally not considered at temperature T unless specified.
Total degrees of freedom f = 3 + 3 = 6.
U = (6/2)RT = 3RT.

The name of the element is derived from its atomic number using the IUPAC nomenclature for elements with Z > 100.
un = 1
nil = 0
enn = 9
So, unnilennium corresponds to the atomic number 109.

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∫ [-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