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Correct answer is (b) INITIAL VELOCITY

Best explanation: We have, x = a + bt + ct^2……….(1)

Let, V and f be the velocity and acceleration of a PARTICLE at time t seconds.

Then, v = dx/dt = d(a + bt + ct^2)/dt= b + ct……….(2)

Now from (2) we get v = b, when t = 0.

Hence b represents the initial velocity of the particle.

Correct answer is (b) [v]max = 2a√a/3√b

The explanation: If v be the velocity of the moving PARTICLE at TIME t then its acceleration at time t will be dv/DT. By question,

dv/dt = a – bt^2

Integrating we GET, v = ∫ a – bt^2 dt = at – bt^3/3 + k……….(1)

where k is constant of integration.

Given, v = 0, when t = 0; hence from (1) we get,

0 = a(0) – b/3(0) + k

Or k = 0

Thus, v = at – bt^3/3……….(2)

Again, d^2v/dt^2 = d(a – bt^2)/dt = -2bt

Now, for minimum or maximum value of v we have,

dv/dt = 0

Or a – bt^2 = 0

Or t^2 = a/b

Or t = √a/√b[Since t > 0 and a, b are positive constants]

At t = √a/√b we have d^2v/dt^2 = -2b(√a/√b) < 0 [Since t < 0 and a, b are positive constants]

Putting t = √a/√b in (2),

We find, v is maximum at t = √a/√b and the minimum value of v is,

[v]max = 2a√a/3√b.

Correct option is (b) 120 cm

Easy explanation: LET, vcm/sec be the velocity and x cm be the distance of the PARTICLE from O and time t seconds.

Then the velocity of the particle and acceleration at time t seconds is, v = dx/dt and DV/dt respectively.

By the question, dv/dt = 5 + 6t

Or dv = (5 + 6t) dt

Or ∫dv = ∫(5 + 6t) dt

Or v = 5T + 6*(t^2)/2 + A……….(1)

By question v = 4, when t = 0;

Hence, from (1) we get, A = 4.

Thus, v = dx/dt = 5t + 3(t^2) + 4……….(2)

Or ∫dx = ∫(5t + 3(t^2) + 4) dt

Or x = 5t^2/2 + t^3 + 4t + B……….(3)

By question x = 0, when t = 0;

Hence, from (3) we get, B = 0

Thus, x = 5t^2/2 + t^3 + 4t

Thus, distance of the particle after 4 seconds,

= [x]t = 4 = (5/2*4^2 + 4^3 + 4*4)[putting t = 4 in (4)]

= 120 cm.

Correct answer is (c) 2

The explanation: LET, y = (x – 1)(3 – x) = 4X – x^2 – 3

Then, dy/dx = 0

Or 4 – 2x = 0

Or 2x = 4

Or x = 2

Now, [d^2y/dx^2] = -2 which is negative.

Therefore, (x – 1)(3 – x) will have its maximum at x = 2.

The CORRECT option is (c) 0.83 cm

Best explanation: Let, u cm/sec be the initial VELOCITY of the bullet.

By the question, the velocity of the bullet after pertaining 2.5 cm into the target will be u/2 cm/sec.

Now if the uniform RETARDATION to penetration be f cm/sec^2,

Then using the formula, v^2 = u^2 – 2fs, we get,

(u/2)^2 = u^2 – 2f(2.5)

Or f = 3u^2/20

Now let US assume that the bullet can penetrate x cm into the target.

Then the final velocity of the bullet will be zero after penetrating x cm into the target.

Hence, using formula v^2 = u^2 – 2fs we get,

0 = u^2 – 2(3u^2/20)(x)

Or 10 – 3X = 0

Or x = 10/3 = 3.33(approx).

Therefore, the required further penetration into the target will be

3.33 – 2.5 = 0.83 cm.

Right ANSWER is (a) 4(X – y) = 15

The best I can explain: The equation of the given parabola is, y^2 = 5x ……….(1)

Differentiating both SIDES of (1) with respect to y, we get,

2y = 5(dx/dy)

Or dx/dy = 2y/5

Take any point P((5/4)t^2, (5/2)t). Then, the normal to the curve (1) at P is,

-[dx/dy]P = -(2*5t/2)/5 = -t

By the question, SLOPE of the normal to the curve (1) at P is tan45°.

Thus, -t = 1

Or t = -1

So, the REQUIRED equation of normal is,

y – 5t/2 = -t(x – 5t^2/4)

Simplifying further we get,

4(x – y) = 15

The correct answer is (b) 176 cm

For EXPLANATION I would say: LET, the particle moving with a uniform acceleration of f cm/sec^2.

By question initial velocity of the particle is u = 10 cm/sec and its velocity after 8 SECONDS = v = 34 cm/sec.

Thus, using the formula v = u + ft we get

34 = 10 + f*8

Or 8f = 24

Or f = 3

Therefore, the required acceleration of the particle is 3 cm/sec^2.

Thus, the space described by the particle in 8 seconds,

= [10*8 + 1/2(3)(8*8)[using the formula s = ut +1/2(ft^2)]

= 80 + 96

= 176 cm.

The correct option is (B) x > 4 or x < -1

Explanation: Since F(x) = 2x^3 – 9x^2 – 24x + 5

Therefore, f’(x) = 6x^2 – 18x + 24

= 6(x – 4)(x + 1)

If x > 4, then, 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, then, 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

The correct answer is (c) At t = 4

The explanation: SUPPOSE, the two particle starts from rest at and move along the straight path OA.

Further assume that the distance between the particle is maximum after t minutes from start (before they MEET again).

If we be the position of the particle after 30 minutes from the start which moves with uniform ACCELERATION 5 m/min^2 and C that of the particle moving with uniform velocity 20 m/min, then we shall have,

OB = 1/2(5t^2) and OC = 20t

If the distance between the particle after t minutes from start be x m, then,

x = BC = OC – OB = 20t – (5/2)t^2

Now, dx/dt = 20 – 5t and d^2x/dt^2 = -5

For maximum or minimum values of x, we have,

dx/dt = 0

Or 20 – 5t = 0

Or t = 4

And [d^2y/dx^2] = -5 < 0

Thus, x is maximum at t = 4.

Correct answer is (b) 8 seconds

Explanation: Let the INITIAL velocity of the particle be u cm/second and its uniform retardation be f cm/sec^2.

Further ASSUME that the particle was in motion for t seconds.

By question, the particle comes to rest after t seconds.

Therefore, USING the formula, v = u – ft, we get,

0 = u – ft

Or u = ft

Again, the particle described 7cm in the 5th second. Therefore, using the formula

st = UT + 1/2(f)(2t – 1) we get,

7 = u – ½(f)(2.5 – 1)

Or u – 9f/2 = 7

Again, the distance described in the last second (i.e., in the t th second) of its motion

= 1/64 * (distance described by the particle in t seconds)

ut -1/2(f)(2t – 1) = 1/64(ut – ½(ft^2))

Putting u = ft, we get,

f/2 = 1/64((ft^2)/2)

Or t^2 = 64

Or t = 8 seconds.

The correct option is (a) 40 sec

For explanation: LET the particle projected from A with velocity u = 196 m/sec and P be its position at time t where, AP = X m, if v m/sec be its velocity at P,then the equation of motion of the particle is,

dv/dt = -g [where, v = dx/dt]……….(1)

Since v = u, when t = 0, hence from (1) we get,

u∫^vdv = -g 0∫^tdt

Or v – u = -GT

Or dx/dt = u – gt ……….(2)

Since x = 0, when t = 0, hence from (2) we get,

0∫^x dx = 0∫^x (u – gt)dt

Or x = UT – (1/2)gt^2 ……….(3)

If T(≠0) be the total time of flight, then x = 0, when t = T; hence, from (3) we get,

Or 0 = uT – (1/2)gT^2

Or gT = 2u

Or T = 2u/g = 2(196)/9.8[as, g = 9.8m/sec^2]

= 40 sec.

Correct choice is (b) 2

The explanation is: Let P and Q be the initial positions of the motor car and motorcycle B respectively, where PQ = 24m.

If possible, let us assume that B overtakes A after point R on the straight road after time t seconds from the start.

Then, considering the motion of motor car A, we get

PR = 1/2 (2) (t^2) = t^2

In this case when B runs at a uniform velocity 11 m/sec, we SHALL have, QR = 11t.

Therefore, in this case, QP + PR = QR gives,

Or 24 + t^2 = 11t

Or t^2 – 11t + 24 = 0

Or (t – 3)(t – 8) = 0

Or t = 3 or t = 8

Clearly, we are getting two real positive values of t.

Therefore, A and B will meet TWICE during the motion.

Correct ANSWER is (a) x = 0

Explanation: Equation of the given parabola is y^2 = 4ax ……….(1)

DIFFERENTIATING both SIDE of (1) with respect to x we get,

2y(dy/dx) = 4a

Or dy/dx = 2a/y

Clearly dy/dx does not EXIST at (0, 0). Hence, the tangent to the parabola (1) at (0, 0) is parallel to y axis.

Again, the tangent passes through (0, 0). Therefore, the REQUIRED tangent to the parabola (1) at (0, 0) is the y-axis and hence the required equation of the tangent is x = 0.

Right answer is (b) 31 cm

The BEST I can explain: 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 = ft^2 + 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 = t^2 + 5t + 7

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

= [x]t = 3 = (3^2 + 5*3 + 7)m = 31m

The correct answer is (c) (u1 – u2)^2 = 2x(f1 – f2)

For explanation I would say: Let A be the position of the express train and B, that of the goods train when each is seen from the other, where AB = x.

Evidently, to avoid a collision the express train will apply the greatest possible retardation f1 and the goods train will apply the greatest possible acceleration f2.

Now, it is just possible to avoid a collision,if the velocities of the two trains at a POINT C on the line are equal, when they just touch each other at C(because after the instant the velocity of the express train will decrease due to retardation f1 and that of the goods train will increase due to acceleration f2).

Let the two trains reach the point C, at TIME t after each is seen by the other and their equal velocities at C bev and BC = s.

Then the equation of motion of the express train is,

V = u1 – f1t……….(1)And x + s = u1t – (1/2)f1t^2 ……….(2)

And the equation of the motion of the goods train is,

v = u2 + f2t……….(3)And s = u2t – (1/2)f2t^2 ……….(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(t^2)(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)

Correct option is (a) 0

For explanation: Let, f be the UNIFORM ACCELERATION and u be the given initial VELOCITY of the moving particle.

From the conditions of problem we have the following equation of motion of the 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

Correct answer is (c) a – b = c – d

For explanation: We have, X^2/a + y^2/b = 1……….(1) and x^2/c + y^2/d = 1……….(2)

Let, US assume curves (1) and (2) intersect at (x1, y1). Then

x1^2/a + y1^2/b = 1……….(3) and x1^2/c + y1^2/d = 1……….(4)

Differentiating both side of (1) and (2) with respect to x we get,

2x/a + 2y/b(dy/dx) = 0

Or dy/dx = -xb/ya

Let, m1 and m2 be the slopes of the tangents to the curves (1) and (2) respectively at the point (x1, y1); then,

m1 = [dy/dx](x1, y1) = -(bx1/ay1) and m2 = [dy/dx](x1, y1) = -(dx1/cy1)

By QUESTION as the curves (1) and (2) intersects at right angle, so, m1m2 = -1

Or -(bx1/ay1)*-(dx1/cy1) = -1

Or bdx1^2 = -acy1^2 ……….(5)

Now, (3) – (4) gives,

bdx1^2(c – a) = acy1^2(d – b)……….(6)

Dividing (6) by (5) we get,

c – a = d – b

Or a – b = c – d.

The correct answer is (B) (5/4, 5/2)

Easiest explanation: The equation of the given PARABOLA is, y^2 = 5X ……….(1)

Differentiating both SIDES of (1) with respect to y, we get,

2y = 5(dx/dy)

Or dx/dy = 2y/5

Take any point P((5/4)t^2, (5/2)t). Then, the normal to the curve (1) at P is,

-[dx/dy]P = -(2*5t/2)/5 = -t

By the question, slope of the normal to the curve (1) at P is tan45°.

Thus, -t = 1

Or t = -1

So, the REQUIRED equation of normal is,

y – 5t/2 = -t(x – 5t^2/4)

Simplifying further we get,

4(x – y) = 15

The co-ordinates of the foot of the normal are, P((5/4)t^2, (5/2)t).

As t = 1, so putting the value of t = 1, we get,

P = (5/4, 5/2).

The CORRECT OPTION is (a) Decreases

For explanation: f(x) = 10 – 9x + 6x^2 – x^3

Thus, f’(x) =– 9 + 12X –3 x^2

= -3(x^2 – 4x + 3)

Or f’(x) = -3(x – 1)(x – 3) ……….(1)

If x > 3, then x – 1 > 0 and x – 3 > 0

Hence, (x – 1)(x – 3) > 0

Thus, from (1) it readily follows that, f’(x) < 0, when x > 3

So, f(x) decreases for VALUES of x > 3.

The correct answer is (a) -1 < X < 4

To explain: SINCE f(x) = 2x^3 – 9x^2 – 24x + 5

Therefore, f’(x) = 6x^2 – 18x + 24

= 6(x – 4)(x + 1)

If -1 < x < 4, then x – 4 < 0 and x + 1 > 0

Thus, (x – 4)(x + 1) < 0 i.e.,

f’(x) < 0, when -1 < x < 4

Therefore, f(x) decreases with x, when, -1 < x < 4.

Right choice is (a) x > log(x + 1)

The EXPLANATION is: ASSUME, f(x) = x – log(1 + x)

THEREFORE, f’(x) = 1 – (1/(1 + x))

= x/(x + 1) > 0, when x > 0

Hence, we must have, f(x) > f(0), when x > 0

But f(0) = 0 – log1 = 0

Thus, f(x) > 0, when x > 0 i.e.,

x – log(1 + x) > 0, when x > 0

Or x > (1 + x), when x > 0.

The CORRECT option is (a) 11m/sec

The explanation is: 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 = ft^2 + 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 = t^2 + 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.

Right choice is (d) 24 cm

The best explanation: Let, x be the DISTANCE travelled by the particle in time t seconds.

Then, v = dx/dt = 3t^2 – 4t + 5

Or ∫dx = ∫ (3t^2 – 4t + 5)dt

So, on integrating the above equation, we get,

x = t^3 – 2t^2 + 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

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

= 24 cm.

Correct answer is (a) 4/27 UNITS

Explanation: 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.

The correct option is (a) The retardation of the particle is the cube of its VELOCITY

The explanation is: We have, t = X^2/2 + 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 = -v^2*v[as, dx/dt = v]

= -v^3

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 = v^3 = cube of the particle.

Correct option is (c) 30 cm/sec

To ELABORATE: We have, s = 12T – 15t^2 + 4t^3 ……….(1)

Differentiating both SIDE of (1) with respect to t we get,

(ds/dt) = 12 – 30t + 12t^2

So, VELOCITY of the particle after 3 seconds is,

[ds/dt]t = 3 = 12 – 30(3) + 12(3)^2

= 30 cm/sec.

The CORRECT CHOICE is (c) 14cm/sec^2

Explanation: Let f be the acceleration of the particle in time t seconds. Then,

f = dv/DT = d(3t^2 – 4t + 5)/dt

= 6t – 4……….(1)

Therefore, the acceleration of the particle at the end of 3 seconds,

= [f]t = 3 = (6*3 – 4) cm/sec^2

= 14cm/sec^2

Correct answer is (B) 62

For EXPLANATION I would say: We have, f(X) = x^3 – 12x^2 + 45x + 8……….(1)

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

f’(x) = 3x^2 – 24x + 45

3x^2 – 24x + 45 = 0

Or x^2 – 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) = 3^3 – 12*3^2 + 45*3 + 8 = 62.

Correct answer is (c) 3 cm/sec^2

For EXPLANATION: LET, the particle moving with a UNIFORM acceleration of f cm/sec^2.

By question initial velocity of the particle is u = 10 cm/sec and its velocity after 8 seconds = v = 34 cm/sec.

Thus, using the formula v = u + ft we get

34 = 10 + f*8

Or 8f = 24

Or f = 3

Therefore, the required acceleration of the particle is 3 cm/sec^2.

Right option is (a) 4 SECONDS

To elaborate: Let us assume that the POINT takes t seconds to describe a distance 80 cm.

Then using the formula s = UT +1/2(ft^2) we get,

80 = 10*t + 1/2(5)(t^2)

Or 5t^2 + 20t – 160 = 0

Or t^2 + 4t – 32 = 0

Or (t – 4)(t + 8) = 0

Or t = 4 Or -8

Clearly, t = -8 is inadmissible.

Therefore, the required time = 4 seconds.

The CORRECT choice is (c) x – y + 2 = 0

The best explanation: Equation of the given parabola is, y^2 = 8x ……….(1)

Differentiating both sides with respect to x,

2y(dy/dx) = 8

Or dy/dx = 4/y

Thus, equation of the tangent to the parabola (1) at (x1, y1) = (2t^2, 4t) is,

y – y1 = [dy/dx](x1, y1) (x – 2t^2)

y – 4t = [dy/dx](2(t^2), 4t) (x – 2t^2)

Putting the value of y = 4t in the equation dy/dx = 4/y, we get,

y – 4t = 4/4t(x – 2t^2) ……….(2)

If the tangent to the parabola y^2 = 8x, which is INCLINED at an angle of 45° with the x axis,

Then, SLOPE of tangent (2) = tan 45° = 1

Thus, 4/4t = 1

Or t = 1

Thus, required equation of the tangent is,

y– 4 = 1(x – 2)

Putting, t = 1 in (2),

So, x – y + 2 = 0

Right choice is (c) 2

Easiest explanation: We have, x^4 –8x^3 + 22x^2 –24x + 8 ……….(1)

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

f’(x) = 4x^3 – 24x^2 + 44x – 24 and f”(x) = 12x^2 – 48x + 44……….(2)

At an extremum of f(x), we have f’(x) = 0

Or 4x^3 – 24x^2 + 44x – 24 = 0

Or x^2(x – 1) – 5X(x – 1) + 6(x – 1) = 0

Or (x – 1)(x^2 – 5x + 6) = 0

Or (x – 1)(x – 2)(x – 3) = 0

So, x = 1, 2, 3

Now, f”(x) = 12x^2 – 48x + 44

f”(1) = 8 > 0

f”(2) = -4 < 0

f”(3) = 8 < 0

So, f(x) has maximum at x = 2.

The correct choice is (d) After 25 sec

The best explanation: LET the particle projected from A with velocity U = 196 m/sec and P be its position at time t where, AP = x m, if v m/sec be its velocity at P,then the equation of motion of the particle is,

dv/dt = -g [where, v = dx/dt]……….(1)

SINCE v = u, when t = 0, hence from (1) we get,

u∫^vdv = -g 0∫^tdt

Or v – u = -gt

Or dx/dt = u – gt ……….(2)

Let the velocity of the particle be 49m/sec in the downward direction after T seconds from the instant of its PROJECTION. Then, v = -49 m/sec, when t = T; hence, from (2) we have,

-49 = 196 – (9.8)T[as, g = 9.8 m/sec^2]

Or 9.8(T) = 245

Or T = 25

Therefore, the particle will have a velocity of 49m/sec in the downward direction after 25 seconds from the instant of its projection.

Correct choice is (B) 1960 m

For explanation I would SAY: Let the PARTICLE projected from A with velocity u = 196 m/sec and P be its position at time t where, AP = x m, if v m/sec be its velocity at P,then the equation of motion of the particle is,

dv/dt = -g [where, v = dx/dt]……….(1)

Since v = u, when t = 0, hence from (1) we get,

u∫^vdv = -g 0∫^tdt

Or v – u = -gt

Or dx/dt = u – gt ……….(2)

Since x = 0, when t = 0, hence from (2) we get,

0∫^x dx = 0∫^x (u – gt)dt

Or x = ut – (1/2)gt^2 ……….(3)

Again, (1) can be written as,

dv/dx*dx/dt = -g

Or v(dv/dx) = -g……….(4)

Since v = u, when x = 0, hence, from (4) we get,

u∫^vvdv = -g 0∫^xdx

Or v^2 = u^2 – 2gx……….(5)

If H m be the greatest height of the particle, then v = 0, when x = H; hence, from (5) we get,

0 = u^2 – 2gH

Or H = u^2/2g

= (196*196)/(2*9.8)[as, g = 9.8 m/sec^2]

= 1960 m.

The CORRECT option is (a) A.P

The best I can EXPLAIN: When a = B, then the EQUATION of the motion of the particle from A to B is,

q^2 = p^2 + 2fa

Or q^2 – p^2 = 2fa……….(1)

Again, the equation of motion of the particle from B to C is,

r^2 = q^2 + 2fa

Or r^2 – q^2 = 2fa……….(2)

From (1) and (2) we get,

q^2 – p^2 = r^2 – q^2

Thus, p^2, q^2 and r^2 are in A.P.

The correct choice is (a) 2(bt1 – at2)/T1T2(t1 + t2)

Best explanation: Let the PARTICLE beam moving with UNIFORM acceleration f and its velocity at A be U.

Then, the equation of the motion of the particle from A to B is,

ut1 + ft1^2/2 = a[as, AB = a]……….(1)

Again, the equation of motion of the particle from A to C is,

u(t1 + t2) + f(t1 + t2)^2/2 = a + b[as, AC = AB + BC = a + b]……….(2)

Multiplying (1) by (t1 + t2) and (2) by t1 we get,

ut1 (t1 + t2) + ft1^2(t1 + t2)/2 = a(t1 + t2)……….(3)

And ut1(t1 + t2) + f(t1 + t2)^2/2 = (a + b)t1 ……….(4)

Subtracting (3) and (4) we get,

1/2(ft1)(t1 + t2)(t1 – t1 – t2) = at2 – bt1

Solving the above equation, we get,

f = 2(bt1 –at2)/t1t2(t1 + t2)

The correct choice is (c) X + 2y + 11 = 0

To explain I would SAY: The equation of any straight line PERPENDICULAR to the line 2x – y + 3 = 0 is,

x + 2y + k = 0……….(1)

Now, the co-ordinate of the center of the CIRCLE (3, -2) and its radius is,

√(9 + 4 – (-7) = 2√5

If straight line (1) be tangent to the given circle then, the perpendicular DISTANCE of the point (3, -2) from the line (1) = radius of the circle

Thus, ±(3 + 2(-2) + k)/√(1 + 4)

Or k – 1 = 2√5 * √5

So, k = 1 ± 10

= 11 or -9

Putting the value of k in (1) we get,

x + 2y + 11 = 0 and x + 2y – 9 = 0

Correct ANSWER is (a) (-2, -2)

Easy EXPLANATION: Equation of the given hyperbola is, xy = 4 ……….(1)

DIFFERENTIATING both side of (1) with respect to y, we get,

y*(dx/dy) + x(1) = 0

Or dx/dy = -(x/y)

Thus, the required equation of the normal to the hyperbola at (2, 2) is,

y – 2 = -[dx/dy](2, 2) (x – 2) = -(-2/2)(x – 2)

So, from here,

y – 2 = x – 2

Or x – y = 0 ……….(2)

Solving the equation (1) and (2) we get,

x = 2 and y = 2 or x = -2 and y = -2

Thus, the line (2) intersects the hyperbola (1) at (2, 2) and (-2, -2).

Hence, the EVIDENT is that the normal at (2, 2) to the hyperbola (1) again intersects it at (-2, -2).

Correct answer is (a) Decreases

For EXPLANATION: f(x) = 10 – 9x + 6x^2 – x^3

Thus, f’(x) =– 9 + 12x –3 x^2

= -3(x^2 – 4X + 3)

Or f’(x) = -3(x – 1)(x – 3)……….(1)

If x < 1, then x – 1 < 0 and x – 3 < 0

Hence, (x – 1)(x – 3) > 0

Thus, from (1) it readily follows that, f’(x) < 0, when x < 1

So, f(x) decreases for values of x < 1.

Correct choice is (c) -175/27

Explanation: We have, x = t^3 – t^2 – 5t ……….(1)

When x = 28, then from (1) we get,

t^3 – t^2 – 5t = 28

Or t^3 – t^2 – 5t – 28 = 0

Or (t – 4)(t^2 + 3T + 7) = 0

Thus, t = 4

Let v and f be the velocity and acceleration respectively of the particle at time t SECONDS. Then,

v = dx/DT = d(t^3 – t^2 – 5t)/dt

= 3t^2 – 2t – 5

And f = dv/dt = d(3t^2 – 2t – 5)/dt

= 6t – 2

Again the particle comes to rest when v = 0

Or 3t^2 – 2t – 5 = 0

Or (3t – 5)(t + 1) = 0

Or t = 5/3, -1

As, t > 0, so, t = 5/3

Therefore, the DISTANCE traversed by the particle before it comes to rest

= [(5/3)^3 – (5/3)^2 – 5(5/3)] m[putting t = 5/3 in (1)]

= -175/27

Correct answer is (a) 27/4 cm

The best I can EXPLAIN: We have, s = 12t – 15t^2 + 4t^3 ……….(1)

Differentiating both side of (1) with respect to t we get,

(ds/dt) = 12 – 30t + 12t^2

Clearly, the VELOCITY is instantaneously zero, when

(ds/dt) = 12 – 30t + 12t^2 = 0

Or 12 – 30t + 12t^2 = 0

Or (2t – 1)(t – 2) = 0

Thus, t = 2 or t = ½

Putting the value t = 2 and t = ½ in (1),

We get, when t = 2 then s = (s1) = 12(2) – 15(2)^2 + 4(2)^3 = -4.

When t = ½ we get, s = (s2) = 12(1/2) – 15(1/2)^2 + 4(1/2)^3 = 11/4.

Thus, the distance between the two positions of the particle at two times, when the velocity is instantaneously 0 = s2 – s1

= 11/4 – (-4)

= 27/4 cm.

The correct CHOICE is (b) 30 cm/sec^2

The explanation is: We have, x = 2t^3 – 12t + 11……….(1)

Let v and f be the velocity and acceleration respectively of the particle at time t seconds.

Then, v = dx/dt = d(2t^3 – 12t + 11)/dt

= 6t^2 – 12……….(2)

And f = dv/dt = d(6t^2 – 12)/dt

= 12t……….(3)

Putting the value of t = 2 in (3),

THEREFORE, the acceleration of the particle at the end of 2 seconds,

12t = 12(2)

= 24 cm/sec^2

Now putting the value of t = 2 in (2),

We get the displacement of the particle at the end of 2 seconds,

6t^2 – 12 = 6(2)^2 – 12

= 12 cm/sec……….(4)

And putting the value of t = 3 in (2),

We get the displacement of the particle at the end of 3 seconds,

6t^2 – 12 = 6(3)^2 – 12

= 42 cm/sec……….(5)

Thus, change in velocity is, (5) – (4),

=42 – 12

= 30cm/sec.

Thus, the average acceleration of the particle at the end of 3 seconds is,

= (change of velocity)/time

= (30 cm/sec)/1 sec

= 30 cm/sec^2

Right answer is (d) 30°

The best I can explain: Given, x^2 + 3y^2 = 12 Or x^2/12 + y^2/4 = 1

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

2x*(dx/dy) + 3*2y = 0

Or dx/dy = -3y/x

Suppose the normal to the ellipse (1) at the point P(√12cosθ, 2sinθ) makes an angle 60° with the major AXIS. Then, the SLOPE of the normal at P is tan60°

Or -[dx/dy]P = tan60°

Or -(-(3*2sinθ)/√12cosθ) = √3

Or √3tanθ = √3

Or tanθ = 1

Now the centre of the ellipse (1) is C(0, 0)

Therefore, the slope of the line CP is,

(2sinθ – 0)/(√12cosθ – 0) = (1/√3)tanθ = 1/√3 [as, tanθ = 1]

Therefore, the line CP is inclined at 30° to the major axis.