Mathematics Test - 37

We have, t = x²/2 + x

Therefore, dt/dx = 2x/2 + 1 = x + 1

Thus, if v be the velocity of the particle at time t, then

v = dx/dt = 1/(dt/dx)

= 1/(x + 1) = (x + 1)^-1

Thus dv/dt = d((x + 1)^-1)/dt

= (-1)(x + 1)^-2 d(x + 1)/dt

= -1/(x + 1)² * dx/dt

As, 1/(x + 1) = dx/dt,

So, -(dx/dt)²(dx/dt)

Or dv/dt = -v²*v  [as, dx/dt = v]

= -v³

We know, dv/dt = acceleration of a particle.

As, dv/dt is negative, so there is a retardation of the particle.

Thus, the retardation of the particle = -dv/dt = v³ = cube of the particle.

We assume that the particle moves with uniform acceleration 2f m/sec.

Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.

Let, v be the velocity of the particle at time t seconds, then,

So, dv/dt = 2f

Or ∫dv = ∫2f dt

Or v = 2ft + b   ……….(1)

Or dx/dt = 2ft + b

Or ∫dx = 2f∫tdt + ∫b dt

Or x = ft² + bt + a   ……….(2)

Where, a and b are constants of integration.

Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.

Putting these values in (2) we get,

4f + 2b + a = 21   ……….(3)

16f + 4b + a = 43   ……….(4)

49f + 7b + a = 91  ……….(5)

Solving (3), (4) and (5) we get,

a = 7, b = 5 and f = 1

Therefore, from (2) we get,

x = t² + 5t + 7

Putting t = 3, f = 1 and b = 5 in (1),

We get, the velocity of the particle in 3 seconds,

= [v]_{t = 3} = (2*1*3 + 5)m/sec = 11m/sec.

We have, t = ax² + bx + c  ……….(1)

Differentiating both sides of (1) with respect to x we get,

dt/dx = d(ax² + bx + c)/dx = 2ax + b

Thus, v = velocity of the particle at time t

= dx/dt = 1/(dt/dx) = 1/(2ax + b) = (2ax + b)^-1  ……….(2)

Initially, when t = 0 and v = u, let x = x₀; hence, from (1) we get,

ax₀² + bx₀ + c = 0

Or ax₀² + bx₀ = -c   ……….(3)

And from (2) we get, u = 1/(2ax₀ + b)

Thus, 1/v² – 1/u² = (2ax + b)² – (2ax₀ + b)²

= 4a²x² + 4abx – 4a²x₀² – 4abx₀

= 4a²x² + 4abx – 4a(ax₀² – bx₀)

= 4a²x² + 4abx – 4a(-c)   [using (3)]

= 4a(ax² + bx + c)

Or 1/v² – 1/u² = 4at

Let, x be the distance travelled by the particle in time t seconds.

Then, v = dx/dt = 3t² – 4t + 5

Or ∫dx = ∫ (3t² – 4t + 5)dt

So, on integrating the above equation, we get,

x = t³ – 2t² + 5t + c where, c is a constant.  ……….(1)

Therefore, the distance travelled by the particle at the end of 3 seconds,

= [x]_{t = 3} – [x]_{t = 0}

= (3³ – 2*3² + 5*3 + c) – c [using (1)]

= 24 cm.

We assume that the particle moves with uniform acceleration 2f m/sec.

Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.

Let, v be the velocity of the particle at time t seconds, then,

So, dv/dt = 2f

Or ∫dv = ∫2f dt

Or v = 2ft + b  ……….(1)

Or dx/dt = 2ft + b

Or ∫dx = 2f∫tdt + ∫b dt

Or x = ft² + bt + a   ……….(2)

Where, a and b are constants of integration.

Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.

Putting these values in (2) we get,

4f + 2b + a = 21  ……….(3)

16f + 4b + a = 43  ……….(4)

49f + 7b + a = 91   ……….(5)

Solving (3), (4) and (5) we get,

a = 7, b = 5 and f = 1

Therefore, from (2) we get,

x = t² + 5t + 7

Therefore, the distance described by the particle in 3 seconds,

= [x]_{t = 3} = (3² + 5*3 + 7)m = 31m

Since the particle is moving in a straight line under a retardation kv³, hence, we have,

dv/dt = -kv³ ……….(1)

Or dv/v³ = -k dt

Or ∫v^-3 dv = -k∫dt

Or v^-3+1/(-3 + 1) = -kt – c [c = constant of integration]

Or 1/2v² = kt + c ……….(2)

Given, u = v when, t = 0; hence, from (2) we get,

1/2u² = c

Thus, putting c = 1/2u² in (2) we get,

1/2v² = kt + 1/2u²

Or 1/v² = 1/u² + 2kt

Let v be the velocity of the particle at time t seconds after start (that is at a distance s from O). Then,

v = ds/dt = d[(t – 1)²(t – 2)²]/dt

Or v = (t – 2)(3t – 4)

Clearly, v = 0, when (t – 2)(3t – 4) = 0

That is, when t = 2

Or 3t – 4 = 0 i.e., t = 4/3

Now, s = (t – 1)(t – 2)²

Therefore, when t = 4/3, then s = (4/3 – 1)(4/3 – 2)² = 4/27

And when t = 2, then s =(2 – 1)(2 – 2)² = 0

Therefore, the velocity of the particle is zero, when its distance from O is 4/27 units and when it is at O.

Let, y = sin(x + α)/sin(x + β)

Then,

dy/dx = [cos(x + α)sin(x + β) – sin(x + α)cos(x + β)]/sin²(x + β)

= sin(x+β – x-α)/sin²(x + β)

Or sin(β – α)/sin²(x + β)

So, for minimum or maximum value of x we have,

dy/dx = 0

Or sin(β – α)/sin²(x + β) = 0

Or sin(β – α) = 0 ……….(1)

Clearly, equation (1) is independent of x; hence, we cannot have a real value of x as root of equation (1).

Therefore, y has neither a maximum or minimum value.

We have, f(x) = x³ – 12x² + 45x + 8 ……….(1)

Differentiating both sides of (1) with respect to x we

f’(x) = 3x² – 24x + 45

3x² – 24x + 45 = 0

Or x² – 8x + 15 = 0

Or (x – 3)(x – 5) = 0

Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5

Therefore, f’(x) = 0 for x = 3 and x = 5.

If h be a positive quantity, however small, then,

f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.

f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.

Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.

So, f(x) has maximum at 3.

Putting, x = 3 in (1)

Thus, its maximum value is,

f(3) = 3³ – 12*3² + 45*3 + 8 = 62.