## Thursday, February 26, 2009

If α denotes the magnitude of instantaneous acceleration of the particle, then which of the following is/are true?

(a) α cannot remain positive throughout

(b) α cannot exceed 2 at any point in its path

(c) α must be equal to 4 at some point or points in its path

(d) α must change sign during the motion, but no other assertion can be made with the information given

A)

(a) is obviously true, since for motion to initiate and end, acceleration and deceleration must be provided.

Now, for determining the least value of maximum acceleration in this motion, the particle can be considered to be moving in SHM (simple harmonic motion) with equilibrium position x = 1/2 and T (time period) = 2 seconds.

acceleration = -w^2 .x

Max acceleration in such a motion will be = w^2 . Amplitude = (2pi/2)^2 . (1/2) = pi^2 / 2 = 4.934 m/s^2

And max. acceleration would be higher if any other type of motion occurred (instead of SHM). So the least value of max. acceleration is 4.934 m/s^2 which means that 4m/s^2 acceleration HAS to occur at some point in the path of the particle's motion. (which implies that (c) is true too).

Hence, (a) and (c) are true.

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## Thursday, January 29, 2009

, if number of positive divisors of is odd

, if number of positive divisors of is even

(The positive divisors of include as well as .)Let be the real number whose decimal expansion contains in the -th place,.Determine,with proof,whether is rational or irrational.

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## Saturday, December 13, 2008

(From RMO - 2008)

A) Let x be the length of the isosceles side.

Perimeter = 2x + k = 2008

Using cosine rule:

k^2 = x^2 + x^2 - 2x^2. cos@ .......... where @ E (90, 180)

k^2 = 2x^2.[2.sin^2 (@/2)] ......... where (@/2) E (45, 90)

k E ( sqrt(2) x, 2x)

Also notice that 2x > k is satisfied automatically here.

Now 2x + k = 2008

Solving,

x E ( 2008/4, (2008)/(2+ sqrt2) )

x E ( 502, 588.something)

But only integral values are required, so x lies between 503 and 588, inclusive.

So, number of values

= 588 - 503 + 1

=

**86 solutions**

Thus, there are 86 such triangles.

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(From RMO - 2008)

A) The various possibilities of the digits chosen acc. to the conditions mentioned are:

1) Three 2s, One 3, One 1, One 0

2) One 2, Two 3s, Two 1s, One 0

3) One 4, One 2, One 3, One 1, Two 0

Number of permutations for the cases mentioned above:

1) Three 2, One 3, One 1, One 0

(6! / 3!) - (5! / 3!) -----> ie. net permutations - permutations having 0 as first digit since those won't be considered a 6 digit number. Similar for other cases too.

= 5.5! / 3!

= 100 ways

2) One 2, Two 3, Two 1, One 0

(6! / 2!.2!) - (5! / 2!.2!)

= 5.5! / 4

= 150 ways

3) One 4, One 2, One 3, One 1, Two 0

(6!/2!) - 5!

= 240 ways

Total = 100 + 150 + 240

= 490 such numbers

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A) Let the left pan have weight m1 and distance l1 from pivot.

Let the right pan have weight m2 and distance l2 from pivot.

Balancing the torques:

m1. l1 = m2.l2 (given that they are balanced when empty)

Also,

(X + m1).l1 = (6+m2). l2 -----------------------------(1)

(24 + m1).l1 = (X+m2).l2 ----------------------------(2)

where X is the unknown weight of the body.

Substituting m1.l1 = m2.l2 in both equations above, we get:

l1/l2 = X/24 = 6/X

so X^2 = 24*6

X = 12 gms

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( from RMO - 2008)

A) Let c = k.b ......... where k is a natural number

From the definition of a harmonic progression:

1/a + 1/c = 2/b

(1/20) + (1/bk) = 2/b

40 - (20/k) = b

Using values of k E [1, 20] we get b = 20, 30, 35, 36, 38, 39. But exclude 20 as we require strictly increasing values (since a = 20 already, b must not be 20).

[ Remember - use only those values of k which are factors of 20, else you won't get integral values... so k E {1, 2, 4, 5, 10, 20} ]

Then using :

c = 20 / [(40/b) - 1]

we get c = 60, 140, 180, 380, 780 respectively for the aforementioned values of b (excluding 20).

So there are 5 such Harmonic Progressions :

20, 30,60

20, 35, 140

20, 36, 180

20, 38, 380

20, 39, 780

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A) Consider any such arbitrary circle, with center (h,k).

It touches the x axis, so its radius is k units ( the ordinate).

Now, the distance of this circle's center from the given circle's center = 2 + k units

This distance is the same as the distance between center of the arbitrary circle and the line y = -2 .

Now, since the center of this arbitrary circle is equidistant from a fixed point (the center of given circle ie. (0,3) ) and the line y = -2, it follows the definition of the parabola and hence the locus of (h,k) is a parabola.

So the locus of the center of the circle is a parabola with focus (0,3) and directrix y = -2 .

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1)The triangles can not have more than one side common with the polygon;

2)The triangle can not have any side common with the polygon .

A) Total triangles possible in general = 10C3

Triangles having more than 1 side common with polygon = Triangles with 2 sides common with polygon = 10 (ten triangles with 2 sides as 2 consecutive sides of the polygon)

so

1) 10C3 - 10 triangles

2) Triangles with 1 side common with polygon = 10.(6) = 60

(Fix one side, then 8 vertexes are left, and also ignore 1 consecutive vertex on either side, so we are left with 6. And 10 such triangles are possible due to 10 vertexes)

So, triangles with no sides common with polygon = (10C3 - 10) - 60

= 10C3 - 70 triangles

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A)

Choose 1 green ball and 1 blue ball, + extra balls [ 0 to 10 in number]

5C1 . 4C1 . (10C0 + 10C1 + 10C2 ....10C10)

= 5*4* (1+1)^10

= (20). 2^10 groups

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A) Using apt equations of simple harmonic motion:

x = A.cos(wt)

(since the abcissa 1/2 is the midpoint of the given path, we measure amplitude and displacement by considering that as the central point)

At t=1 ---> 1/2 = 1/2* cos (w)

so, w = pi

And since acceleration (a) = -w^2. x

magnitude of a = (pi^2). (1/2) (a is max. at extremities)

= approx 5 units/s^2 --------> max. magnitude of acceleration

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A) Since he has already reported that it's a 6, either he is lying about it or it is actually a six.

That is, either it is truly a six & he's telling the truth, OR it's a non-six & he's lying by saying that it's a six.

Net Probability = (prob. of actual 6 & truth) / [prob. of actual 6 & truth + prob. of non 6 & lies ]

= (3/4)(1/6) / [3/4*1/6 + 1/4*5/6]

= 3/8

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A)

log_{7} 10 = 1 + log_{7} (10/7) = 1 + log_{7} 1.428

log_{11} 13 = 1 + log_{11} (13/11) = 1 + log_{11 } 1.18

Let A = log_{7} 1.428

and B = log_{11} 1.18

So, 7^{A} = 1.428

and 11^{B} = 1.18

Clearly A and B will both be fractional, and B will be smaller than A; (Since, in case of A, 7 is being reduced to 1.428 but with B, a larger number 11 is being reduced to a correspondingly less value than for A. )

Thus, 1 + log_{7} 1.428 > 1 + log_{11} 1.18

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## Sunday, November 2, 2008

A) [(1+2x)+3x^2)]^5

'r'th Term = 5Cr (1+2x)^r . (3x^2)^{5-r}

To form net effective 4th powers of x:

from the righthand term, we can only use x^0 x^2 and x^4

so from the lefthand term (1+2x)^r ... we use x^4, x^2 and x^0 respectively.

5C5.(3x^2)^0 . (1+2x)^5 -----------> 5C5. 5C4.(2x)^4

and

5C4.(3x^2).(1+2x)^4 --------> 5C4.(3x^2).[4C2.(2x)^2]

and

5C3.(3x^2)^2 .(1+2x)^3 ---------> 5C3.(3x^2)^2 [3C0. (2x)^0]

So, coefficient of x^4 :

5C5.5C4.2^4 + 5C4.3.4C2. 2^2 + 5C3.3^2. 1

= 5. 16 + 5.3.6.4 + 10.9

= 80+ 360 + 90

= 530

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1 + 1/2 + 1/3 + 1/4 .... up to infinite terms.

A)

1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...

= ( 1/1 ) + ( 1/2 ) + ( 1/3 + 1/4 ) + ( 1/5 + 1/6 + 1/7 + 1/8 ) + ...

(where each set of parenthesis stops at a reciprocal of a power of two.) Replacing the terms in each set of parenthesis by the smallest term (in those parenthesis) we get a smaller sum:

1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...

> ( 1/1 ) + ( 1/2 ) + ( 1/4 + 1/4 ) + ( 1/8 + 1/8 + 1/8 + 1/8 ) + ...

= 1/1 + 1/2 + 1/2 + 1/2 + 1/2 + ...

which clearly gets arbitrarily large! So it is a diverging series and will not have a definite value since it is not converging. The sum will go on approaching infinity.

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**and**

**.**

A) y = 2xy

y (1-2x) = 0..... So, either y=0 or x = 1/2

For y = 0,

x = x^3

x^3 - x = 0 = (x - x^1/3) (x^2 + x^{2/3} + x^{4/3} )

x = -1,1, 0 for y = 0 ............ 3 real roots.

AND for x = 1/2

1/2 - 1/8 = y^4

0.375 = y^4

y^4 - 0.375 = 0 = (y^2 + rt(0.375) ) (y^2 - rt(0.375) )

which has 2 real and 2 imaginary roots.

So, total 3 + 2 = 5 real solutions.

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Q) Find the integral of :

A)

=

=

Now take out the 1 and you get:

e^x + e^x [ f(x) + f'(x) ] ..... where f(x) = -4/(x+4)

And we know that integral of e^x [ f(x) + f'(x) ] = e^x. f(x) + C

So the integral is :

e^x + e^x.[-4 / (x+4) ] + C

OR

e^x . [x/ (x+4)] + C

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A) Number of ways of selecting x and y: 9C2 . 2!

Out of these, 1/2 will be such that x than y, and 1/2 such that x greater than y

So for x less than y: 9C2 selections

And z can be any number from 0 to 9, so 10 ways.

So total number of such numbers is 9C2 . 10

= 360 numbers

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## Saturday, November 1, 2008

A) AM = n. HM

(a+b) / 2 = n. [2ab/(a+b)]

a^2 + b^2 + 2ab = 4*n*ab

(a+b)^2 = 4*n*ab

(a/b + 1)^2 = 4*n*a/b

Let a/b = x

(x + 1)^2 = 4nx

x^2 + 1 + 2x - 4nx = 0

x^2 + x(2-4n) + 1 = 0

x = 4n-2 +- sqrt(16n^2 + 4 - 16n -4) / 2

x = 4n-2 +- 4.sqrt(n^2 - n) / 2

x = 2n - 1 +- 2.sqrt (n^2 - n)

OR

a/b = 2n - 1 +- 2.sqrt (n^2 - n)

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A)

v = u + at

0 = u + -(g + 2)t

t = u / (g + 2)

0 - u^2 = 2as

- u^2 = 2 *-(g + 2)s

s = u^2 / 2(g + 2)

s = 1/2 aT^2

u^2 / 2(g+2) = (1/2) * (g-2)T^2

u / sqrt(g^2 - 4) = T

So t/T = sqrt (g-2) / sqrt (g+2) = sqrt (8/12) = sqrt (2/3)

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## Tuesday, September 16, 2008

S1 + S2 = 200

[v^2 - u^2 / 2a] + [ V^2 - U^2 / 2A ] = 200

v^2 / 2 + (-v^2 / 8) = 200

v^2 = 320

v = sqrt(320) m/s

Also,

t = sqrt(320)

And,

V - U = AT

T = sqrt (320) / 4

Total time = t + T = sqrt (320) (5/4)

Total time = 10 sqrt(5) seconds

= 22.36 seconds

This is of the form : | 2ab / (a^2 + b^2) | where a is 2x and b is 3. And we know: (a - b)^2 > = 0a^2 + b^2 > = 2ab so | 2ab / (a^2 + b^2) | < = 1 so y < = 1 and y > = 0 ....(since it is absolute value of something) So the two boundary lines are y = 1 and y = 0 . |

A) 7^23 = (7^22) . 7

= [ (7^2)^11] . 7

= [(50 - 1)^11] .7

Now In expansion of (50 - 1)^11 , all terms except the last one will be divisible by 25 since they contain an exponent of 50 in them.

Last term = 11C11 * (-1)

So we are left with

(-1)(7)

= -7

so 25 - 7 = 18 which is the remainder

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A) .7 + .77 + .777 ... n terms

= 7 (0.1 + 0.11 + 0.111....)

= (7/9)*(0.9 + 0.99 + 0.999 ...)

= (7/9)*(1 - 0.1 + 1 - 0.01 + 1 - 0.001 ....)

= (7/9)* [n - (0.1*(1 - (0.1)^n) / (1 - 0.1) ) ]

= (7/9)* [n - (0.1)(1 - (0.1)^n) / 0.9]

= (7/9)*[ n - (1 - 0.1^n)/9]

= 7n/9 - 7(1 - (0.1)^n) / 81

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Q) A fly wheel rotates about its central axis . Due to friction at the axis, it experiences angular retardation proportional to its angular velocity. If its angular velocity falls to half its initial value while its makes n revolutions , how many more revolutions will it make before coming to rest?

A) - a = k.w .........(k is a constant of proportionality)

(dw/dQ)(dQ/dt) = k.(dw/dt)

dw = - k.dQ

w = -kQ + c

Initially Q = 0, and w = W0

W0 = C

w = -kQ + W0

After n revolutions,

W0 / 2 = -k(2.n.pi) + W0

W0 / 2 = 2.k.n.pi

k = W0 / 4.n.pi

For final stop, w = 0

w = -kQ + W0

0 = -(W0 / 4.n.pi). Q + W0

Q = 4.n.pi

means 2n revolutions.

And given that n revolutions have already occurred. So number of revolutions that will further occur before the flywheel stops = 2n - n = n revolutions.

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Q) The figure below shows the acceleration-time graph of a particle moving along a straight line. After what time does the particle acquire its initial velocity?

A) Let initial velocity be u

At t =1, velocity = u + at = u + 2

For t >= 1, da/dt = -2 ....(slope)

Using point-slope form of equation of straight line:

(a-2) = (-2)(t-1)

a = 4 - 2t

dv/dt = 4 -2t

V = 4t -t^2 + C

at t = 1, we have V = u+2, so C= u-1

so V = 4t - t^2 + U -1

So for V to equal U:

t^2 - 4t + 1 = 0

t = 2 + root(3) seconds.

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A)

(A phone number doesn't begin with 0.)

Number of phone numbers with at least one repeated digit

= (Total ways in which a phone number can be made) - (number of phone numbers with no repeated digits)

= 9*10*10*10*10 - 9*9*8*7*6

= 9.(10^4) - 9.(9P4)

= 62784

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^{-1}in the direction A to B notices that a bus goes past him every 18 minutes in the direction of his motion,and every 6 minutes in the opposite direction. Find the period T of the bus service and the speed with which the buses ply between A and B. Assume the speed of the bus to be constant.

A)

Let D = distance between A and B Let velocity of the buses = v Relative velocity of buses with respect to the cyclist (going from A to B) v-20 km/hr for buses going from A to B v+20 km/hr for Buses going from B to A Now use Time = distance/velocity Use the 2 equations to get d and v D/(v-20) = 18/60 ---------------------1 D/(v+20) = 6/60 -----------------------2 =>v=40 km/hr and D=6 km T = 6/40 hr = 9 mins Hence the speed of the buses is 40km/hr & the period of bus service is 9 mins . (A new bus is sent after every 9 mins) |

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A) Rewrite the expression as

(a-1 + 1/b + b - b) (b - 1 + 1/c + c - c) (c - 1 + 1/a + a - a)

Max value of the terms will be obtained when 1/b + b, 1/c+ c and 1/a +a are max, since a, b , c are all positive.

And we know that 1/n + n >= 2

(and for max value, n = 1)

So a=b=c=1 for the max value :

(a -b +1)(b- c+1)(c - a+1)

= 1*1*1

= max value --> 1

so the expression is less than or equal to 1.

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A) Use the fact that center of mass remains stationary in absence of any external forces.

So the length of the rope can be divided in inverse ratio of the masses.

M1 / M2 = L2 / L1

L2 / L1 = 2/3

And the sum of lengths is 5.

Therefore, they meet at 3m from initial position of A.

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A) New momentum = m (1.0001 v)

So new velocity = 1.0001 v

So new kinetic energy = 1/2 * m * V^2

= 1/2 * m * (1.0001 v)^2

= 1/2 * m * 1.00020001 v^2

= 1.0002 x original momentum

So the increase is 0.02 %.

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A) When it collides with the inclined plane at height h, the body's velocity is sqrt(2g(H-h) ).

And time taken to cover this distance is sqrt(2(H-h) / g).

(By using ---> s = (1/2) g.t^2 )

Now this becomes the horizontal velocity. So now time taken to reach the ground will be :

t = sqrt (2h/g) since downward velocity initially now is 0.

So total time taken = sqrt(2(H-h) / g) + sqrt (2h/g)

Time taken will be max when the two terms are equal (symmetry), so 2(H-h) = 2h

2H = 4h

h/H = 1/2

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A)

The 3 groups are A, B and C. The prob. we have to find is of getting BGG where B is boy and G is girl. Case 1 - boy is selected from A, and the girls from B and C. prob. of BGG = 1/4*1/2*1/4 = 1/32 Case 2 - boy is selected from B, and girls from A and C. prob. of BGG = 1/2*3/4*1/4 = 3/32 Case 3 - boy is selected from C, and girls from A and B. prob. of BGG = 3/4*3/4*1/2 = 9/32 Total prob. = Sum of prob.= 1/32 + 3/32 + 9/32 = 13/32 |

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A) Let the GP be : a, ar, ar^2 ....a.r^n

a + a.r^n = 66 = a(1 + r^n)

ar.a.r^(n-1) = 128 = a^2.r^n

So 66 = a (1 + 128/a^2)

66a = a^2 + 128

a^2 - 66a + 128 = 0

a = 66 +- rt(66^2 - 4(128)) / 2

a = 66 +- 62 / 2

a = 2, 64

Since, 66 = a(1 + r^n)

66 = 2(1 + r^n) OR 66 = 64 (1 + r^n)

Since r>1 (increasing GP) so we take

66 = 2(1 + r^n)

r^n = 32

And,

a(r^(n+1) - 1) / (r - 1) = 126

2(32r - 1) / (r-1) = 126

63r - 63 = 32r - 1

31r = 62

r = 2

and r^n = 32

2^n = 32

so

n = 5 .

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^{4 _ }y

^{4}= 3789108 .

A) Let A = x^2 and B = y^2

A^2 - B^2 = 3789108

(A + B)(A - B) = 3789108

Let A + B = T

(T)(T-2B) = 3789108

T^2 - 2BT - 3789108 = 0

T = 2B +- root (4B^2 + 4*3789108) / 2

T = B +- root (B^2 + 3789108)

A + B = B +- root (B^2 + 3789108)

A = +- root (B^2 + 3789108)

Now since A = x^2 so A cannot be negative, thus

A = root (B^2 + 3789108)

Now A is x^2, and B is y^2, where x and y are integers.

Squares of integers (ie. x^2 and y^2) end in 0, 1, 4, 5, 9, 6 and squares of these perfect squares (ie. A^2 and B^2) end in 0, 1, 5,6.

Now,

last digit of B^2 + last digit of 3789108 should end in 0,1,5,6

the various sum of last digits possible are 8, 9, 3, 4 none of which is 0, 1, 5 or 6...

So no integral solution exists.

The equation has 0 integral solutions.

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^{2}+ 2

^{2}+ 3

^{2}+ ..... + 2003

^{2}= (2003)(4007)(334) and (1)(2003)+(2)(2002)+(3)(2001)+....(2003)(1) = (2003)(334)(x), then find x .

A)

So for **2003 terms** the summation is = (2003)(2004)(4007) / 6 = **(2003)(4007)(334)**

Now,

(1)(2003)+(2)(2002)+(3)(2001)+....(2003)(1) =

= [ 2004(2003)(2003 + 1) / 2 ] - [ (2003)(4007)(334) ]

= [ (1002)(2003)(2004) ] - [ (2003)(4007)(334) ]

= 3 [(334)(2003)(2004) ] - [ (2003)(4007)(334)]

= (2003)(334)( 3*2004 - 4007)

= (2003)(334)(2005)

So the answer is 2005.

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A)

1 + (1+2) + (1+2+4) + ...... till n terms = 1 + [1*(2^2 -1)/(2-1)] + [1*(2^3 - 1)/(2 - 1)] + ....(1 + 2 + 4 + ...2^(n-1) ) ---->( Concept of sum of infinite Geometric Progression ) = 1 + [1*(2^2 -1)/(2-1)] + [1*(2^3 - 1)/(2 - 1)] + ...[1*(2^n - 1) / (2 -1) ] = 1 + [2^2 - 1 + 2^3 - 1 + 2^4 - 1 .... + 2^n - 1] = 1 + [ 2^2.(2^(n-1) - 1) / (2-1) - (n-1) ] = 1 + [ 4. 2^(n-1) - 4 - n + 1] = 2^(n+1) - 2 - n |

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A) Let Vf = velocity of flow & V = velocity of boat

Distance covered by boat in 60 mins = (V + Vf) x 1 hour = V + Vf

Also

6 km = Vf * (60 min + T) where T is time taken for the boat to return to the bottle again.

6 = Vf [1 hour + (V + Vf - 6) / (V - Vf) ] ........... V - Vf is net speed for upstream motion

6 = Vf (2V - 6) / V - Vf

6V - 6Vf = 2V.Vf - 6Vf

6V = 2V.Vf

Vf = 3

So the flow speed = 3 km/hr

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#### Q) If satisfies the equation x^{2} - 28x + 27 = 0, find the value of . Given that x lies between 0 and pi/2 .

#### A) The given expression can be simply rewritten as:

= 3^ (sin^2 x + sin^4 x + sin^6 x ....infinity)

= 3^[ (sin^2 x) / (1 - sin^2x)] (Using concept of sum of infinite GP)

= 3^(tan^2 x)

Now,

x^2 - 28x + 27 = 0

x^2 - 27x - x + 27 = 0

(x-1)(x-27) = 0

roots are 1, 27

Given that 3^(tan^2 x) satisfies the equation so it = 1, 27

3 ^(tan^2 x) = 1, 27

tan^2 x = 0, 3

tan x = 0, +- rt(3)

x = 0, +- pi/3

Since for second part of the question x lies between 0 and pi/2, so we take x = pi/3

So

1/ (cosx + sinx) = 1 / (cos60 + sin60)

= 1/ (1/2 + sqrt(3)/2)

= 2 / ( 1 + sqrt(3) )

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So

1/ (cosx + sinx) = 1 / (cos60 + sin60)

= 1/ (1/2 + sqrt(3)/2)

= 2 / ( 1 + sqrt(3) )

A) v = | t-2 | m/s

The critical point is t = 2 (At which the modulus becomes 0)

so from t=0 to t=2 , velocity is (t-2) m/s

and from t=2 to t= 4, velocity is (2 - t) m/s

Total distance covered = mod of ( integral (0 to 2) [ t-2] + integral (2 to 4) [ 2 - t] )

= [t^2/2 - 2t] 0 to 2 + [ 2t - t^2/2] 2 to 4

= | -2 - 2 |

= |-4 |

= 4 m

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A)

f(101) + f(102) + f(103) .... f(200)

= 99(1) for hundreds places + 10(1+2+3+4+5+6+7+8+9) for tens places + 10(1+2+3+4+5+6+7+8+9) for units places + 2 (for 200)

= 99 + 20(45) + 2

= 101 + 900

= 1001

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A)

x^2 - 6x + y^2 + 12y = 0

Form perfect squares:

x^2 - 6x + 9 - 9 + y^2 + 12y + 36 - 36 = 0

(x + 3)^2 + (y + 6)^2 = 45

(x + 3)^2 + (y + 6)^2 = [sqrt(45)]^2

This is the equation of a circle, with radius sqrt(45)

so area = pi.r^2

= 45.pi

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**Consider a sequence { } such that**.

**How many distinct pairs chosen from this sequence have g.c.d.= 6 ?**

A) a1 = 2

a2 = 3

a3 = 7

a4 = 43

Now since

a(n+1) = a.n^2 - an + 1

and after a1, the next terms are odd...

so terms will be of the type

odd^2 - odd + 1

odd^2 is always odd... and odd^2 + 1 will become even.

then even - odd = odd always....

So there won't be any distinct pair with GCD = 6... 2 will be the only even number in the sequence.

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