Clearly specify the limit , is it x tends to 'a' or x tends to infinity??

 

If it is x tends to infinity.

L = _{[x][infinity]} a^x sin(b/a^x)

 

If |a| < 1 , a^x tends to 0 as x tends to infinity.

In that case , L = (something tending to 0)x(a value between 1 and -1) = 0

 

If |a| = 1 , L = sinb

 

If |a| > 1 , a^x tends to infinity as x tends to infinity.

Hence b/a^x tends to 0 as x tends to infinity.

So sin(b/a^x) tends to b/a^x.

Hence L = _{[x][infinity]} a^x sin(b/a^x) = _{[x][infinity]} a^x (b/a^x) = b

 

Hence , limit = 0       if  |a| < 1

 

                   = sinb   if  |a| = 1 

 

                   = b       if  |a| > 1 

 

Let M be the mid-point of AB.

Then M is {(3+5)/2,(4-2)/2} = (4,1).

Since PA = PB , triangle is isoceles whose base is AB and altitude is PM.

AB = {(3-5)² + (4+2)²}^1/2 = 40^1/2

 

Now let P be (h,k)

PA = PB

{(h-3)² + (k-4)²}^1/2 = {(h-5)² + (k+2)²}^1/2

h - 3k = 1.............(1)Area = (1/2)(PM)(AB) = 10

 

(1/2)(40^1/2)(PM) = 10

PM = 10^1/2

PM² = 10

(h-4)² + (k-1)² = 10

(1+3k-4)² + (k-1)² = 10  { from(1) : h = 1+3k}

(3k-3)² + (k-1)² = 10

9(k-1)² + (k-1)² = 10

10(k-1)² = 10

(k-1)² = 1

k-1 = 1

k = 0 , 2

 

Hence h = 1+3k = 1 , 7

 

Hence possible P are (1,0) and (7,2).

Let in equilibrium position the string make an angle of x with the vetical.

Then sinx = 1/20

Clearly draw free body diagram and everything would be clear.

 

Balance vertical forces :

Tcosx = mg

T = mgsecx.............(1)

 

Balance horizontal forces :

Tsinx = qE

(mgsecx)sinx = qE

mgtanx = (1/4₀)q²/3²

 

This equation contains only m as unknown and hence m can be found out.

After m is known , put m in (1) to get the value of T.

Divide nemerator and denominator by x.

 

L = _{[x][infinity]} {6a(1+ a/x + a²/x²) - (1+a²/x²)} / {a(6 + 2a/x - 1/ax²) - e^1/x}

L = (6a-1)/(6a-1)

 

L = 1

Let the velocity at the top of the window be u.

Journey through the window :

1 = u(0.1) + (1/2)g(0.1)²

u = 9.5 m/s

 

Hence distance travelled upto the top of the window, x₁ = u²/2g = 4.5125 m

Distance travelled through the window, x₂ = 1 m

 

Velocity at the bottom of the window, v = u + g(0.1) = 10.5 m/s

Time taken by the ball to return to the window from the ground = 2 sec

Hence time taken to reach the ground from the bottom of window = 1 sec

Distance travelled from bottom of window to the ground, x₃ =(10.5)1+(1/2)(10)1²

x₃ = 15.5

 

Hence height = x₁ + x₂ + x₃ = 4.5125 + 1 + 15.5 = 21.0125 m

 

x = vt/2

 

We know : v = dx/dt

 

x = (dx/dt).t/2

2dt/t = dx/x

 

Integrating both the sides :

2lnt + c = lnx (where c is the constant of integration)

ln(e^c.t²) = ln(x)

 

x = e^c.t²

dx/dt = e^c.(2t)

acceleration = d²x/dt² = d(dx/dt)/dt = d{e^c.(2t)}/dt = 2e^c (a constant)

 

Hence acceleration is constant.

L = _[x][0] (36^x - 9^x - 4^x + 1)/{2 - (1+cosx)} which is in the form of 0/0 .

 

Numerator , N = (36^x - 9^x - 4^x + 1) = (9^x.4^x - 9^x - 4^x + 1) = (9^x - 1)(4^x - 1)

 

Denominator, D = {2 - (1+cosx)}.{2 + (1+cosx)}/{2 + (1+cosx)} (rationalize)

                       = (1 - cosx)/{2 + (1+cosx)}

                       = {2sin²(x/2)}/{2 + (1+cosx)}

As x tends to zero sin²(x/2) tends to (x/2)²

Hence _[x][0] D = 2(x/2)²}/{2 + (1+1)} = x²/42

 

Hence L = _[x][0] N/D

 

L = _[x][0] (9^x - 1)(4^x - 1)/{x²/42}

 

L = _[x][0] {(9^x - 1)/x}.{(4^x - 1)/x}.(42)

 

Now use _[x][0] (a^x - 1)/x = loga

 

L = (42).(log4).(log9)

L = (162).(log2).(log3)

  

 

 

 

When the two spherical shells are combined let the mass be M and radius be R of new spherical shell.

Let the surface density be d.

Then d = M₁/(4R₁²) = M₂/(4R₂²) = M/(4R²).....................(1)

 

Potentials at the centre :

V₁ = -GM₁/R₁

V₂ = -GM₂/R₂

V₁ : V₂ = (M₁/R₁)/(M₂/R₂) = 2/3..................................(2)

 

M = M₁ + M₂ ............................................(3)

(4R²)d = (4R₁²)d + (4R₂²)d

R² = R₁² + R₂² ............................................(4)

 

From (1) and (2) :

(R₁/R₂) = 2/3.....................................(5)

(M₁/M₂) = (R₁/R₂)² = 4/9.....................(6)

 

Potential : V = -GM/R   = -G(M₁ + M₂)/(R₁² + R₂²)^1/2

V = -G((4/9)M₂ + M₂)/((4/9)R₂² + R₂²)^1/2 = (13^1/2/3)M₂/R₂

 

Hence V₁ : V₂ : V = (2/3) : 1 : (13^1/2/3) = 2 : 3 : 13^1/2