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(5) Evaluate

Submission: Closed

Cracked by:

1. **Tarun Arora** (The first to crack), NIT Surathkal. Tarun has cleared mains 2003, congrats!

2.Ravi Subramanian, IIT Bombay, B.Tech. 1'st Year.

3.Tushar Jain, Meerut Public School, Meerut U.P.

4.Sankalp Guha, Vivekananda Kendra Vidyalaya, Port Blair.

5.M. Chetan Reddy, New Generation Junior College, Class XII, Hyderabad.

(4)

A particle of mass m is placed on a frictionless horizontal surface. It is attached to the ends of three horizontal springs S1, S2 and S3 of equal stiffness k.

The particle is in equilibrium with S2 and S3 at right angles and S1 making equal angles with S2 and S3. If the particle is slightly pushed against S1 and released find its period of oscillation.

Submission: Closed

Cracked by:

1. Mayank Gaur.

2.Tushar Jain, Meerut Public School, Meerut U.P.

3.Nishant Kumar, Class XII, Modern School, Nagpur.

4.Anupam Narayan, Class XII, Swami Vivekanand Junior College, Mumbai.

5.Umang Merwana, Satna (M.P.).

(3) No three diagonals of a convex decagon intersect at one point. The vertices of the decagon constitute the set A and the intersection points of the diagonals constitute the set B. 3 elements are picked at random from the set A U B to form a set C. What is the chance that C is a subset of B?

Submission: Closed

Cracked by:

Nachiket Vasant Vaidya, 12'th, Sathaye College, Vile Parle, Mumbai.

Solution Provided by Nachiket

As no three diagonals intersect each other, If we consider any for vertices of the polygon there will be only one intersection of the vertices the other drawn diagonals will not correspond to intersections.

Hence the number of intersection points are nC4 i.e. 10C4 = 210.
Hence, n(A)=10 and n(B)=210 and n(A U B)=220
Hence the required Probability is
P= 210C3/220C3= 7*19*52/73*109 =6916/7957=0.869

(2) If a + b + g =90°, prove that sin2a + sin2b + sin2g + 2 sina sinb sing =1.

Submission: Closed

Cracked by:

1. **Dhanush Kandathil** (first to crack), Trivandrum.

2. Ramesh Nidadavolu, Hyderabad.

3. Sachin Aggarwal, Delhi.

Here is the solution provided by Dhanush Kandathil:

a + b + g= 90.
This means sin
g is the same as cos(a + b).

Substituting in Sigma(sin2
a) + 2PI(sin a) = sin2a + sin2b + cos2(a + b) + 2sinasinbcos(a+b)
but cos(
a+b)=cosacosb -sinasinb
cos2(
a+b)= cos2acos2b +sin2asin2b - 2sinasinbcosacosb

Also We have
2sin
asinbcos(a+b)=2sinasinbcosacosb-2sin2asin2b

On adding the 2sin
asinbcosacosb terms cancel out.
and expression reduces to sin2
a+sin2b+ (cosacosb)2-(sinasinb)2

But(cos
acosb)2-(sinasinb)2 = cos(a+b)cos(a-b)=cos2b-sin2a +cos2b = 1, which is the required answer.

(1) The first person to crack the first problem was Rahul Deshpande from Bhopal. The problem was:

f(x) = cos8(x), integrate f(x) w.r.t. x.

Submission: Closed

Here is the solution provided by him:

f(x) = cos8(x) [the question was to integrate f(x)]

Using the method of integration by DeMoivre's theorem.

from DeMoivre's theorem,
z+{1\over z}=2\cos x
also, z^n +{1\over {z^n }}=2\cos nx --> (1), where z is a complex number

therefore,
(2\cos x)^8 =\left({z+{1\over z}}\right)^8
256\cos ^8 x =z^8 +{1\over {z^8 }}+8\left({z^6 +{1\over {z^6 }}}\right)+28\left({z^4 + {1\over {z^4 }}}\right)+56\left({z^2 +{1\over {z^2 }}}\right)+70

therefore, final answer after using (1)\int {\cos ^8 xdx={1\over {128}}\left[{{1\over 8}\sin 8x+{4\over 3}\sin 6x+7\sin 4x+28\sin 2x+35x}\right]}+C
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