We have, x = t³ ? t² ? 5t  ???.(1)
When x = 28, n from (1) we get,
t³ ? t² ? 5t = 28
Or t³ ? t² ? 5t ? 28 = 0
Or (t ? 4)(t² + 3t +7) = 0
Thus, t = 4
Let v and f be velocity and acceleration respectively of particle at time t seconds. Then,
v = dx/dt = d(t³ ? t² ? 5t)/dt
= 3t² ? 2t ? 5
And f = dv/dt = d(3t² ? 2t ? 5)/dt
= 6t ? 2
Therefore, acceleration of particle at end of 4 seconds i.e., when particle is at a distance of 28 metres from O,
[f]_{t = 4} = (6*4 ? 2) m/sec²
= 22 m/sec²

We have, x = t³ ? t² ? 5t ???.(1)
When x = 28, n from (1) we get,
t³ ? t² ? 5t = 28
Or t³ ? t² ? 5t ? 28 = 0
Or (t ? 4)(t² + 3t + 7) = 0
Thus, t = 4
Let v and f be velocity and acceleration respectively of particle at time t seconds. Then,
v = dx/dt = d(t³ ? t² ? 5t)/dt
= 3t² ? 2t ? 5
And f = dv/dt = d(3t² ? 2t ? 5)/dt
= 6t ? 2
Again particle comes to rest when v = 0
Or 3t² ? 2t ? 5 = 0
Or (3t ? 5)(t + 1) = 0
Or t = 5/3, -1
As, t > 0, so, t = 5/3
Therefore, distance traversed by particle before it comes to rest
= [(5/3)³ ? (5/3)² ? 5(5/3)] m   [putting t = 5/3 in (1)]
= -175/27

Equation of given hyperbola is, 3x² ? 4y² = 12 ???.(1)
Differentiating both sides of (1) with respect to y we get,
3*2x(dy/dx) ? 4*(2y) = 0
Or dx/dy = 4y/3x
Therefore, equation of normal to hyperbola (1) at point (x₁, y₁) on it is,
y ? y₁ = -[dx/dy]_{(x1, y1)} (x ? x₁) = -4y₁/3x₁(x ? x₁)
Or 3x₁y + 4y₁x – 7x₁y₁ = 0

We have, x⁴ ?8x³ + 22x² ?24x + 8 ???.(1)
Differentiating both sides of (1) with respect to x, we get,
f?(x) = 4x³ ? 24x² + 44x ? 24 and f?(x) = 12x² ? 48x + 44 ???.(2)
At an extremum of f(x), we have f?(x) = 0
Or 4x³ ? 24x² + 44x ? 24 = 0
Or x²(x ? 1) ? 5x(x ? 1) + 6(x ? 1) = 0
Or (x ? 1)(x² ? 5x + 6) = 0
Or (x ? 1)(x ? 2)(x ? 3) = 0
So, x = 1, 2, 3
Now, f?(x) = 12x² ? 48x + 44
f?(1) = 8 > 0
f?(2) = -4 < 0
f?(3) = 8 < 0
So, f(x) has maximum at x = 2.

Let, vcm/sec be velocity and x cm be distance of particle from O and time t seconds.
Then velocity of particle at time t seconds is, v = dx/dt
By question, dv/dt = 5 + 6t
Or dv = (5 + 6t) dt
Or ∫dv = ?(5 + 6t) dt
Or v = 5t + 6*(t²)/2 + A   ???.(1)
By question v = 4, when t = 0;
Hence, from (1) we get, A = 4.
Thus, v = dx/dt = 5t + 3(t²) + 4  ???.(2)
Thus, velocity of particle after 4 seconds,
= [v]_{t = 4} = (5*4 + 3*4² + 4) [putting t = 4 in (2)]
= 72 cm/sec.

Let us assume that point takes t seconds to describe a distance 80 cm.
Then using formula s = ut +1/2(ft²) we get,
80 = 10*t + 1/2(5)(t²)
Or 5t² + 20t ? 160 = 0
Or t² + 4t ? 32 = 0
Or (t ? 4)(t + 8) = 0
Or t = 4 Or -8
Clearly, t = -8 is inadmissible.
Therefore, required time = 4 seconds.

Let A be position of express train and B, that of goods train when each is seen from or, where AB = x.
Evidently, to avoid a collision express train will apply greatest possible retardation f₁ and goods train will apply greatest possible acceleration f₂.
Now, it is just possible to avoid a collision,if velocities of two trains at a point C on line are equal, when y just touch each or at C(because after instant velocity of express train will decrease due to retardation f₁ and that of goods train will increase due to acceleration f₂).
Let two trains reach point C, at time t after each is seen by or and ir equal velocities at C bev and BC = s.
Then equation of motion of express train is,
v = u₁ ? f₁t ???.(1)   And x + s = u₁t ? (1/2)f₁t² ???.(2)
And equation of motion of goods train is,
v = u₂ + f₂t ???.(3)   And s = u₂t ? (1/2)f₂t² ???.(4)
From (1) and (3) we get,
u₁ ? f₁t = u₂ + f₂t
Or (f₁+ f₂)t = u₁ ? u₂
Or t = (u₁ ? u₂)/(f₁ + f₂)
Again, (2)?(4) gives,
x = (u₁ ? u₂)t ? ½(t²)(f₁ + f₂)
= t[(u₁ ? u₂) ? (1/2)(t)(f₁ + f₂)]
As, (f₁ + f₂)t = (u₁ ? u₂)
So, x = (u₁ ? u₂)/(f₁ + f₂)[(u₁ ? u₂) ? (1/2)(u₁ ? u₂)]
So, (u₁ ? u₂)² = 2x(f₁ ? f₂)

Let, f be uniform acceleration and u be given initial velocity of moving particle.
From conditions of problem we have following equation of motion of particle:
u + ½(f)(2p ? 1) = a ???.(1)
u + ½(f)(2q ? 1) = b ???.(2)
u + ½(f)(2r ? 1) = c ???.(1)
Thus, a(q ? r) + b(r ? p) + c(p ? q) = [u + ½(f)(2p ? 1)](q ? r) + [u + ½(f)(2q ? 1)](r ? p) + [u + ½(f)(2r ? 1)](p ? q)
= u (q ? r + r ? p + p ? q) + f/2[(2p ? 1)(q ? r) + (2q ? 1)(r ? p) + (2r ? 1)(p ? q)] = u*0 + f/2[2(pq ? rp + qr ? pq + rp ? qr) ? q + r ? r + p ? p + q]
= f/2*0
= 0

Since f(x) = 2x³ ? 9x² ? 24x + 5
Therefore, f?(x) = 6x² ? 18x + 24
= 6(x ? 4)(x + 1)
If x > 4, n, x ? 4 > 0 and x + 1 > 0
Thus, (x – 4)(x + 1) > 0 i.e., f?(x) > 0, when x > 4
Again, if x < -1, n, x ? 4 < 0 and x + 1 < 0
So, from here,
(x ? 4)(x + 1) > 0 i.e., f?(x) > 0, when x < -1
Hence, f?(x) > 0, when x > 4 Or x < -1
Therefore, f(x) increases with x when, x > 4 or x < -1

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, eir 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, n,
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 left to negative on right of point x = 3.
So, f(x) has maximum at 3.