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Correct answer is (c) Falls continuously

Explanation: We have y = 3x((X + a)/(x + B)) + 5

From the definition we GET,

MARGINAL cost = dy/dx = d/dx[3*((x^2 + xa)/(x+b))] + 0

Solving it further, we get,

= [((x^2 + xa)/(x+b))]d[3]/dx+ 3 d[((x^2 + xa)/(x+b))]/dx

= 0 + 3 [(x+b)d(x^2 + xa)/dx – (x^2 + xa)d(x+b)/dx]/(x+b)^2

= 3[1+(b(a – b)/(x+b)^2)]

By problem, a > 0, b > 0 and a > b; hence from the expression of dy/dx, it is evident that dy/dx decreases as x increases.

Hence, we conclude the marginal cost falls continuously as the output increases.

Right answer is (a) When they are equal distance from the junction

Explanation: Let OX and OY be two straight RAILWAY lines and they meet at O at right angles.

One train starts from the junction and moves with uniform velocity u km/hr along the line OY.

And at the same INSTANT, another train starts towards the junction O from station A on the line OX with same uniform velocity u km/hr.

Let C and B be the position of the two trains on lines OY and OX respectively after t hours from the start.

Then OC = AB = ut km. Join BC and let OA = a km and BC = x km.

Then OB = a – ut.

Now, from the right angled triangle BOC we get,

BC^2 = OB^2 + OC^2

Or x^2 = (a – ut)^2 + (ut)^2

Thus, d(x^2)/dt = 2(a – ut)(-u) + u^2(2t)

And d^2(x^2)/dt^2 = 2u^2 + 2u^2 = 4u^2

For MAXIMUM or minimum value of x^2(i.e., x) we must have,

d(x^2)/dt = 0

Or 2(a – ut)(-u) + u^2(2t) = 0

Or 2ut = a[Since u ≠ 0]

Or t = a/2u

Again at t = a/2u we have, d^2(x^2)/dt^2 = 4u^2 > 0

Therefore, x^2(i.e., x) is minimum at t = a/2u

Now when t= a/2u, then OC = ut = u(a/2u) = a/2 and OB =a – ut = a – u(a/2u) = a/2 that is at t = a/2u we have, OC = OB.

Therefore, the trains are nearest to each other when they are equally distant from the junction.

Correct answer is (a) 1/v = 1/u + kx

The best explanation: Since the PARTICLE is moving in a STRAIGHT line under a retardation kv^3, hence, we have,

dv/dt = -kv^3……….(1)

Or dv/dx*dx/dt = -kv^3

Or v(dv/dx) = -kv^3[as, dx/dt = v]

Or ∫v^-2 dv = -k∫dx

Or v^-2+1/(-2 + 1) = -kx – B, where B is a INTEGRATION constant

Or 1/v = kx + B……….(3)

Given, v = u, when x = 0; hence, from (3) we GET, B = 1/u

Thus, putting B = 1/u in (3) we get,

1/v = 1/u + kx.

The correct answer is (d) 5

Explanation: 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’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.

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

Clearly, f’(x) CHANGES sign from negative on the LEFT to positive on the right of the point x = 5.

So, f(x) has minimum at 5.

Correct answer is (c) 3

Best explanation: 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.

Right choice is (c) Does not possess a maximum or minimum value

Easiest EXPLANATION: We have, f(x) = 2/3(x^3) – 6x^2 + 20x – 5 ……….(1)

DIFFERENTIATING both side of (1) with RESPECT to x, we get,

f’(x) = 2X^2 – 12x + 20

Now, for a maximum and minimum value of f(x) we have,

f’(x) = 0

Or 2x^2 – 12x + 20 = 0

Or x^2 – 6x + 10= 0

So, x = [6 ± √(36 – 4*10)]/2

x = (6 ± √-4)/2, which is imaginary.

Hence, f’(x) does not vanishes at any point of x.

Thus, f(x) does not possess a maximum or minimum value.

Correct option is (B) -1

Easiest explanation: Let, f(X) = cosx

Then, f’(x) = -SINX and f”(x) = -cosx

At an extreme point of f(x) we must have,

f’(x) = 0

Or -sinx = 0

Or x = nπ where, n – any integer.

If n is an odd integer i.e., n = 2M + 1 where m is any integer, then at,

x = (2m + 1)π we have, f”(x) = [(2m + 1)π] = -cos(2mπ + π) = -cosπ = -1(-1) = 1

So, f”(x) is positive at x = (2m + 1)π

Hence, f(x) = cosx is MINIMUM at x = (2m + 1)π.

So, the minimum value of cosx is cos(2mπ + π) = cosπ = -1.

Correct answer is (b) 98 m/sec in the downward direction

The best I can explain: 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 v1 be the velocities of the particle after 10 SECONDS from the INSTANT of projection.

Then v = v1 when t = 10; hence from (2) we get,

v1 = 196 – (9.8)*30 = -98 m/sec[as, g = 9.8m/sec^2]

Since the upward direction is TAKEN as positive, hence after 30 seconds the velocity of the particle is -98 m/sec in the upward direction. So, the velocity will be 98 m/sec in the downward direction.

Correct option is (B) 22(2/9)m

The best I can EXPLAIN: Let f be the uniform retardation in m/sec^2 to the motion of the motor car due to application of brakes.

By question, the car is stopped by its brakes in 4 seconds, hence, the final velocity of the car after 4 seconds = 0.

THEREFORE, using the formula v = u – ft we get,

0 = ((40*1000)/(60*60) – f(4))[Since u = initial velocity of the motor car = 40 km/hr = (40*1000)/(60*60) m/sec]

Or f = 25/9

Let the car go through a distance s m from the point at which the brakes are first applied.

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

s = ((40*1000)/(60*60))*4 – 1/2(25/9)(4*4)

= 200/9

= 22(2/9)

Therefore, the required distance described by the car = 22(2/9)m.

The CORRECT choice is (c) Normal to hyperbola

The best explanation: Equation of the given hyperbola is, 3x^2 – 4y^2 = 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, the equation of the normal to the hyperbola (1) at the point (x1, y1) on it is,

y – y1 = -[dx/dy](x1, y1) (x – x1) = -4y1/3x1(x – x1)

Or 3x1y + 4y1x – 7x1y1 = 0

Now, if possible, LET US assume that the straight line

x + y + 7 = 0 ………..(2)

This line is normal to the hyperbola (1) at the point (x1, y1). Then, the equation (2) and (3) must be identical. Hence, we have,

3x1/1 = 4y1/1 = -7x1y1/7

So, x1 = -4 and y1 = -3

Now, 3x1^2 – 4y1^2 = 3(-4)^2 – 4(-3)^2 = 12

This shows the point (-4, -3) lies on the hyperbola (1).

Thus, it is evident that the straight line (3) is normal to the hyperbola (1).

The correct option is (d) (3/16, 3/4)

Easy explanation: GIVEN, y^2 = 3x ……….(1)and y = 2x + 4……….(2)

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

2y = 3(dx/dy)

Or dx/dy = 2y/3

Let P (X1, y1) be any point on the parabola (1). Then the slope of the NORMAL to the parabola (1) at point P is

-[dx/dy]P = -2y1/3

If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,

-2y1/3*2 = -1

Since the slope of the line (2) is 2

Or y1 = 3/4

Since the point P(x1, y1) LIES on (1) hence,

y1^2 = 3x1

As, y1 = 3/4, so, x1 = 3/16

Therefore, the required equation of the normal is

y – y1 = -(2y1)/3*(x – x1)

Putting the value of x1 and y1 in the above equation we get,

16x + 32y = 27

And the coordinates of the foot of the normal are (x1, y1) = (3/16, 3/4)

Correct answer is (c) 72 cm/sec

The BEST I can explain: 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 at time t seconds is, V = dx/dt

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)

Thus, velocity of the particle after 4 seconds,

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

= 72 cm/sec.

The correct choice is (c) -34

Explanation: Let y = 2x^3 + 3x^2 – 36x + 10……….(1)

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

dy/dx = 6x^2 + 6x – 36

And d^2y/dx^2 = 12x + 6

For MAXIMA or minima VALUE of y, we have,

dy/dx = 0

Or 6x^2 + 6x – 36 = 0

Or x^2 + x – 6 = 0

Or (x + 3)(x – 2) = 0

Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2

Now, d^2y/dx^2 = 12x + 6 = 12(2) + 6 = 30 > 0

Putting x = 2 in (1) we get its minimum value as,

2x^3 + 3x^2 – 36x + 10 = 2(2)^3 + 3(2)^2 – 36(2) + 10

= -34

Correct option is (c) 2

To explain: 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)

Suppose the particle ATTAINS the HEIGHT of 1254.4 m after t2 SECONDS from the instant of projection.

Then, x = 1254.4, where t = t2; hence, from (3) we get,

1254.4 = 196t2 – (1/2)(9.8)t2^2[as, g = 9.8 m/sec^2]

t2^2 – 40t2 + 256 = 0

Or (t2 – 8)(t2 – 32) = 0

Or t2 = 8, 32

Clearly, the particle attains the height of 1254.4m twice; once after 8 seconds from the instant of its projection during its upward motion and next, after 32 seconds from the same instant during its motion in the downward direction.

The correct answer is (a) No

To explain I WOULD say: 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

Again, considering the motion of motorcycle B we get QR = 9t.

Now,QP+PR = QR

Or 24 + t^2 = 9t

Or t^2 – 9t + 24 = 0

Or t = [9 ± √(81 – 4*24)]/2

Or t = (9 ± √-15)/2

Clearly, the values of t are imaginary.

Therefore, there is no real positive time when B overtake A.

Hence, the in this case B will never overtake A.

Correct option is (b) (u1^2f2 + u2^2f1) = 2f1f2(X)

Best explanation: By question the two motor cars approach each other along the same line.

Let P and Q be the positions of the cars on the line when each is seen from the other, where PQ = x.

It is evident that a COLLISION can be just avoided, if the two cars stop at the POINT somewhere between P and Q that is if velocities of both the motor cars at O are zero.

Since the initial velocity of the motor car is at P is u1 and it comes to rest at O, HENCE, its equation is

0 = u1^2 – 2f1(PO)

Or (PO) = u1^2/2f1

Again, the initial velocity of the motor car at Q is due to and it comes to rest at O; hence its equation of motion is,

0 = u2^2 – 2f2(QO)

Or (QO) = u2^2/2f2

Now, x = PQ = PO + OQ = u1^2/2f1 + u2^2/2f2

Or u1^2f2 + u2^2f1)/2f1f2 = x

Thus, (u1^2f2 + u2^2f1) = 2f1f2(x).

Correct choice is (B) Increases

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 1 < 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 1 < x < 3

So, f(x) increases for VALUES of 1 < x < 3.

The correct option is (a) The above statement is valid

For explanation: Let a particle be moving along the straight-line OX and P be its position at time t, when OP = x. CLEARLY, the velocity and ACCELERATION of the particle at P are dx/dt and d^2x/dt^2 respectively.

If possible, let US assume,

dx/dt α x

dx/dt = kx, where k is a constant variation.

Now, d^2x/dt^2 = d(kx)/dt = k(dx/dt)

= k^2x

At the starting POINT O, we have x = 0 and we see that dx/dt = 0 and d^2x/dt^2 = 0, when x = 0 that is, both velocity and acceleration of the particle are zero at x = 0 and the particle remains at rest at O.

Therefore, it is impossible for a particle to have any motion under the given CONDITION.

Right answer is (c) 1

The explanation: 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’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.

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

Clearly, f’(x) changes SIGN from negative on the left to positive on the right of the point x = 5.

So, f(x) has minimum at 5.

Putting, x = 5 in (1)

Thus, its maximum value is,

f(3) = 5^3 – 12*5^2 + 45*5 + 8 = 58.

The correct choice is (c) 3 CM

Easy EXPLANATION: We have, X = 2t^3 – 12t + 11 ……….(1)

Putting t = 2 in (1), we get,

Displacement = x = 2t^3 – 12t + 11

= 2(2)^3 – 12(2) + 11

= 3 cm.

Right answer is (a) (2m + 1)π

For explanation I WOULD say: LET, f(X) = cosx

Then, f’(x) = -sinx and f”(x) = -cosx

At an extreme point of f(x), we must have,

f’(x) = 0

Or -sinx = 0

Or x = nπ where, N is any integer.

If n is an odd integer i.e., n = 2m + 1 where m is any integer, then at,

x = (2m + 1)π we have, f”(x) = [(2m + 1)π] = -cos(2mπ + π) = -cosπ = -1(-1) = 1

So, f”(x) is positive at x = (2m + 1)π

Hence, f(x) = cosx is MINIMUM at x = (2m + 1)π.

Correct answer is (a) 98 m/sec in the upward direction

The 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 v1 be the velocities of the particle after 10 seconds from the instant of projection.

Then v = v1 when t = 10; hence from (2) we get,

v1 = 196 – (9.8)*10 = 98 m/sec[as, g = 9.8m/sec^2]

Since the upward direction is taken as positive, hence after 10 seconds the velocity of the particle is 98 m/sec in the upward direction.

The correct answer is (C) X^2 + y^2 + x + 2y = 0

Explanation: The equation of any circle through the points of INTERSECTION of the given circle is,

x^2 + y^2 + 2x + 4Y + 1 + K(x^2 + y^2 – 1) = 0

x^2 + y^2 + 2x(1/(k + 1)) + 2*2y/(k + 1) + (1 – k)/(1 + k) = 0

Clearly, the co-ordinates of the center of the circle (1) are, (-1/(1 + k), -2/(1 + k)) and its radius,

= √[(1/(1 + k))^2 + (2/(1 + k))^2 – ((1 – k)]/(1 + k))

= √(4 + k^2)/(1 + k)

Clearly, the line x + 2y + 5 = 0 is tangent to the circle (1), hence, the perpendicular distance of the line from the center of the circle = radius of the circle

± (-1/(1 + k))– 2(2/(1 + k)) + 5/ √(1^2 + 2^2) = √(4 + k^2)/(1 + k)

Or ±(5k/√5) = √(4 + k^2)

Or 5k^2 = 4 + k^2

Or 4k^2 = 4

Or k = 1 [as, k ≠ -1]

Putting k = 1 in (1), equation of the given circle is,

x^2 + y^2 + x + 2y = 0

The CORRECT ANSWER is (b) 20 sec

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)

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)

Let, t1 be the time of RISE of the particle; then v = 0, when t = t1.

Thus, from (2) we get,

0 = u – gt1

As, g = 9.8m/sec^2

Or t1 = u/g = 196/9.8 = 20 sec.

Right option is (a) 22 cm/sec^2

Explanation: 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 DISPLACEMENT of the particle at the end of 2 seconds,

12t = 12(2)

= 24 cm/sec^2

Right answer is (c) Uniform POSITIVE acceleration

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)

And f = dv/dt = d(b + ct)/dt = c……….(3)

CLEARLY, when c > 0,implies f > 0.

Hence in this case the particle moves withan uniform positive acceleration.

Right answer is (b) They are equal

To explain I would SAY: The given function is f(x) = ∣x − 1∣, x ∈ R.

It is known that a function f is differentiable at point x = c in its domain if both

\(\lim\limits_{h \rightarrow 0^-}\) hf(c + h) – f(c)

And

\(\lim\limits_{h \rightarrow 0^+}\) hf(c + h) – f(c) are FINITE and equal.

To check the differentiability of the function at x = 1,

LHS,

Consider the left hand limit of f at x=1

\(\lim\limits_{h \rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h}\)

= \(\lim\limits_{h \rightarrow 0^-}\frac{|h|}{h}\)

= \(\lim\limits_{h \rightarrow 0^-}\frac{-h}{h}\)

= −1

RHS,

Consider the right hand limit of f at x − 1

\(\lim\limits_{h \rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h}\)

= \(\lim\limits_{h \rightarrow 0^+}\frac{|h|}{h}\)

= 1

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1.

As, LHS = -1 and RHS = 1, it is clear that, f’(1) < 0 on the left of x = 1 and f’(x) > 0 on the right of the point x = 1.

Hence, f’(x) changes sign, from negative on the left to positive on the right of the point x = 1.

Therefore, f(x) has a local minima at x = 1.

The correct OPTION is (d) 3

For 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 minimum at x = 1 and 3.

Right option is (d) π/3

Easy explanation: X^2 – y^2 = 2a^2 ……….(1) and x^2 + y^2 = 4a^2 ……….(2)

Adding (1) and (2) we get, 2x^2 = 6a^2

Again, (2) – (1) gives,

2y^2 = 2a^2

Therefore, 2x^2 * 2y^2 = 6a^2 * 2a^2

4x^2y^2 = 12a^2

Or x^2y^2 = 3a^4

Or 2xy = ±2√3

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

2x – 2y(dy/dx) = 0

Or dy/dx = x/y

And 2x + 2y(dy/dx) = 0

Ordy/dx = -x/y

Let (x, y) be the point of intersection of the curves(1) and (2) and m1 and m2 be the slopes of the tangents to the curves (1) and (2) RESPECTIVELY at the point (x, y); then,

m1 = x/y and m2 = -x/y

Now the ANGLE between the curves (1) and (2) means the angle between the tangents to the CURVE at their point of intersection.

Therefore, if θ is the required angle between the curves (1) and (2), then

tanθ = |(m1 – m2)/(1 + m1m2)|

Putting the value of m1, m2 in the above equation we get,

tanθ = |2xy/(y^2 – x^2)|

As, 2xy = ±2√3a^2 and x^2 – y^2 = 2a^2

tanθ = |±2√3a^2/-2a^2|

Or tanθ = √3

Thus, θ = π/3.

Right answer is (C) 3

To explain: Assume that the velocity of the PARTICLE at time t SECOND is vcm/sec.

Then, v = dx/dt = 4t^3/12 – 6t^2/3 + 6t/2 + 1

So, v = dx/dt = t^3/3 – 2t^2/ + 3t + 1

Thus, dv/dt = t^2 – 4t + 3

And d^2v/dt^2 = 2t – 4

For maximum and minimum value of v we have,

dv/dt = 0

Or t^2 – 4t + 3 = 0

Or (t – 1)(t – 3) = 0

Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3

Now, [d^2v/dt^2]t = 3 = 2*3 – 4 = 2 > 0

Thus, v is minimum at t = 3.

The correct choice is (a) 1 cm/sec

The best explanation: ASSUME that the velocity of the particle at time t second is vcm/sec.

Then, V = dx/dt = 4T^3/12 – 6t^2/3 + 6t/2 + 1

So, v = dx/dt = t^3/3 – 2t^2/ + 3T + 1

Thus, dv/dt = t^2 – 4t + 3

And d^2v/dt^2 = 2t – 4

For maximum and minimum value of v we have,

dv/dt = 0

Or t^2 – 4t + 3 = 0

Or (t – 1)(t – 3) = 0

Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3

Now, [d^2v/dt^2]t = 3 = 2*3 – 4 = 2 > 0

Thus, v is minimum at t = 3.

Putting t = 3 in (1) we get,

3^3/3 – 2(3)^2/ + 3(3) + 1

= 1 cm/sec.

Right answer is (b) 2

For explanation I WOULD say: Let y = 2x^3 + 3x^2 – 36x + 10……….(1)

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

dy/dx = 6x^2 + 6x – 36

And d^2y/dx^2 = 12x + 6

For MAXIMA or minima value of y, we have,

dy/dx = 0

Or 6x^2 + 6x – 36 = 0

Or x^2 + x – 6 = 0

Or (x + 3)(x – 2) = 0

Therefore, EITHER x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2

Now, d^2y/dx^2 = 12x + 6 = 12(2) + 6 = 30, which is > 0.

Right OPTION is (b) 8 units

Explanation: The GIVEN equation of the circle is,

2x^2 + 2y^2 + 5 x + y – 15 = 0

Or x^2 + y^2 + 5/2 x + y/2 – 15/2 = 0 ………..(1)

The required length of the tangent drawn from the point (7, 2) to the circle (1) is,

√(7^2 + 2^2 + 5/2 (7) + 1/2 – 15/2)

= √(49 + 4 + 35/2 + 1 – 15/2)

= √64

= 8 units.

The CORRECT choice is (a) INITIAL position

The explanation is: We have, X = a + bt + ct^2……….(1)

Now from (1) we get x = a, when t = 0.

This means that the distance of the particle from O at time t = 0 is a ft.

Thus, a REPRESENTS the initial position of the particle.

The correct answer is (a) 16x + 32y = 27

Best EXPLANATION: GIVEN, y^2 = 3x ……….(1)and y = 2X + 4……….(2)

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

2y = 3(dx/dy)

Or dx/dy = 2y/3

Let P (x1, y1) be any point on the PARABOLA (1). Then the slope of the normal to the parabola (1) at point P is

-[dx/dy]P = -2y1/3

If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,

-2y1/3*2 = -1

Since the slope of the line (2) is 2

Or y1 = 3/4

Since the point P(x1, y1) lies on (1) hence,

y1^2 = 3x1

As, y1 = 3/4, so, x1 = 3/16

Therefore, the required equation of the normal is

y – y1 = -(2y1)/3*(x – x1)

Putting the value of x1 and y1 in the above equation we get,

16x + 32y = 27.

Right option is (c) -3

For explanation: Let y = 2x^3 + 3x^2 – 36x + 10……….(1)

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

dy/dx = 6x^2 + 6x – 36

And d^2y/dx^2 = 12x + 6

For maxima or minima value of y, we have,

dy/dx = 0

Or 6x^2 + 6x – 36 = 0

Or x^2 + x – 6 = 0

Or (x + 3)(x – 2) = 0

Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2

Now, d^2y/dx^2 = 12x + 6 = 12(-3) + 6 = -30, which is < 0.

The correct option is (a) UNIFORM retardation

Easy 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)

And f = dv/dt = d(b + ct)/dt = C……….(3)

Clearly, when c < 0,implies f < 0.

Hence in this case the particle moves withan uniform retardation.

Right option is (b) 22 m/sec^2

Best 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

Therefore, the acceleration of the particle at the end of 4 seconds i.e., when the particle is at a distance of 28 METRES from O,

[f]t = 4 = (6*4 – 2) m/sec^2

= 22 m/sec^2

Right choice is (d) A will again overtake B and they will never meet again

To explain: 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.

They will first meet after 3 seconds from start when the motorcycle B will overtake the motor car A.

But the velocity of A continuously increases, hence after a further period of (8 – 3) = 5 seconds A will again overtake B and they will never meet again.

The CORRECT OPTION is (B) 42 cm/sec^2

Easiest explanation: 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

And d^2s/dt^2 = -30 + 24t

So, acceleration of the particle after 3 SECONDS is,

[d^2s/dt^2]t = 3 = – 30 + 24(3)

= 42 cm/sec^2.

Right answer is (a) 1646.2 m

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)

Let, x = h when t = 12; then from (3) we get,

h = 196*12 – (1/2)*9.8*(12*12)[as, g = 9.8m/sec^2]

= 1646.4 m.

Right option is (a) After 3 seconds from start when the motorcycle B will overtake the motor car A

The EXPLANATION: 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 ata uniform velocity of 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.

They will first meet after 3 seconds from start when the motorcycle B will overtake the motor car A.

Correct answer is (B) x – y = 0

The best I can explain: 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

Right choice is (d) Uniform VELOCITY

For explanation I would say: 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)

And f = dv/dt = d(b + ct)/dt = C……….(3)

Clearly, when c = 0, then f = 0 that is, acceleration of the particle is zero.

Hence in this case the particle MOVES with an uniform velocity.

Right choice is (C) 91

The explanation is: Let y = 2x^3 + 3x^2 – 36x + 10……….(1)

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

dy/dx = 6X^2 + 6x – 36

And d^2y/dx^2 = 12x + 6

For maxima or minima value of y, we have,

dy/dx = 0

Or 6x^2 + 6x – 36 = 0

Or x^2 + x – 6 = 0

Or (x + 3)(x – 2) = 0

Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2

Now, d^2y/dx^2 = 12x + 6 = 12(-3) + 6 = -30 < 0

Putting x = -3 in (1) we get its maximum value as,

2x^3 + 3x^2 – 36x + 10 = 2(-3)^3 + 3(-3)^2 – 36(-3) + 10

= 91

Correct answer is (B) 3x1y + 4y1x – 7x1y1 = 0

The EXPLANATION: Equation of the given hyperbola is, 3x^2 – 4y^2 = 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, the equation of the normal to the hyperbola (1) at the point (x1, y1) on it is,

y – y1 = -[dx/dy](x1, y1) (x – x1) = -4y1/3x1(x – x1)

Or 3x1y + 4y1x – 7x1y1 = 0

Right option is (C) (x^2 – ay)X + (y^2 – ax)Y = axy

Explanation: Equation of the given curve is, x^3 – 3axy + y^3 = 0 ……….(1)

Differentiating both SIDES with respect to x we get,

3x^2 – 3A(x(dy/dx) + y) + 3y^2(dy/dx) = 0

Or dy/dx = (ay – x^2)/(y^2 – ax)

So, it is clear that this can be written as,

Y – y = (dy/dx)(X – x)

Or Y – y = [(ay – x^2)/(y^2 – ax)](X – x)

Simplifying the above equation by CROSS multiplication, we get,

(x^2 – ay)X + (y^2 – ax)Y = x^3 – 3axy + y^3 + axy

Using (1),

(x^2 – ay)X + (y^2 – ax)Y = axy