and f(0) = 0 . Then f(1) = ?A)
=
=
By Applying partial fractions -
=
=
=
= 
BY using the statement that f(0) = 0 ........We get C = 0
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Saturday, December 13, 2008
Q) Find the number of all integer-sided isosceles obtuse-angled triangles with perimeter 2008 units.
(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) 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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Q) Find the number of all 6-digit natural numbers such that the sum of their digits is 10 and each of the digits 0, 1, 2, 3 occurs at least once in them.
(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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(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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Q) A body weighs 6 grams when placed on one pan and 24 grams when placed on the other pan of a defective beam balance according to the reading shown on the instrument. If the beam is horizontal when both pans are empty, what is the true weight of the body?
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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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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Q) Find all three-term Harmonic Progressions a, b, c of strictly increasing positive integers in which a = 20, and b divides c .
( 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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( 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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Q) A circle touches the x axis and also touches another fixed circle with center (0,3) and radius = 2 units. What is the locus of the center of the (first) circle?
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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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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Q) Find the number of triangles that can be formed with the vertexes of a 10-sided-polygon (decagon) as its vertexes ,if
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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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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Q) How many groups can be made from 5 different green balls, 4 different blue balls and 3 different red balls, if at least 1 green ball and 1 blue ball are to be included?
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)
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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Q) A particle starts from rest x=0 at t=0, moves on a straight-line path and again comes to rest at x=1 at t=1. If the motion is simple harmonic, find the approx. maximum acceleration.
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) 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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Q) A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually six.
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) 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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Q) Two equal resistances are connected in the gaps of a meter bridge. If the resistance in the left gap is increased by 10%, by what percent does the balance-point shift (and in which direction) ?
A) Since a Metre-Bridge has wire of 100 cm, so if the 2 resistances are equal, the balance point is at 50 cm.
Now if the left resistance is increased by 10%, it becomes 1.1 times its original value.
So balance point L :
1.1 R/ R = L / (100 - L)
110 - 1.1 L = L
110 = 2.1 L
L = 52.4 cm approx.
So the balance point L has shifted right by 2.4 cm
So in terms of %, it has shifted right by (2.4 / 50 x 100) %
= 4.8 % to the right
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Q) Prove that
.
A)
.A)
log7 10 = 1 + log7 (10/7) = 1 + log7 1.428
log11 13 = 1 + log11 (13/11) = 1 + log11 1.18
Let A = log7 1.428
and B = log11 1.18
So, 7A = 1.428
and 11B = 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 + log7 1.428 > 1 + log11 1.18
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