Class 11th

lim_ (x→0) (1/x? ) {1 - cos (x²/2) - cos (x²/4) + cos (x²/2)cos (x²/4)} = 2?
⇒ lim_ (x→0) ( (1 - cos (x²/2) (1 - cos (x²/4) / x? ) = 2?
⇒ 2? = 2? ⇒ k = 8

∴ center lies on x + y = 2 and in 1st quadrant center = (a, 2-a) where a > 0 and 2-a > 0 ⇒ 0 < a < 2
∴ circle touches x = 3 and y = 2
⇒ |3-a| = |2 - (2-a)| = radius
⇒ |3-a| = |a| ⇒ a = 3/2
∴ radius = a
⇒ Diameter = 2a = 3.

After burning, heat exchange occurs between helium and atmospheric. Hence, irreversible, isothermal process.

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!

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

The mass of the cylinder M = ρV = ρ (πR²L), where ρ is the density of the material.
We can express R² in terms of L: R² = M / (πρL).
The moment of inertia is I = M (R²/4 + L²/12).
Substitute R² into the equation for I:
I (L) = M * [ (M / (4πρL) + (L²/12) ]
To find the minimum possible I, we differentiate I with respect to L and set the derivative to zero.
dI/dL = M * [ -M/ (4πρL²) + 2L/12 ] = 0
M/ (4πρL²) = 2L/12 = L/6
M/ (πρ) = (4/6)L³ = (2/3)L³
Since R² = M/ (πρL), we have R²L = M/ (πρ).
Substitute M/ (πρ) into the differentiated equation:
R²L = (2/3)L³

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

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.