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1.The no of sol are in [0.2Pi] such that sin(2x) 4 =1/8. 2. What is the sum of greatest integer function of n 2 /2 from o to 40 i.e sigma o to 40 of [n 2 /2] 3.We want to find a polynomial p(x) of degree n such that f(1) = sq root of 2 and f(3) = pi. then prove that there are infinitely many such polynomia for each n or = 2 but not infinite for n=1.



Grade:11

1 Answers

Deepak Kumar
25 Points
8 years ago

1. For sin($) = (1/8) => $ = (1/8) (almost)
So, the general solution is (2x)^(4) = 2nπ+(1/8) or (2nπ+1)π-(1/8).
Now, for x in range [0,2π], let us check for the maximum limit, putting x=2π,
(2*2π)^4 = (2nπ+1)π-(1/8) => n = 3968 (the nearest small integer)
Therefore, the total number of solutions is 2*3968 = 7936 (considering the smaller solution as well that we have not checked for as it is small and hence comes under the range)

Tip:-Try seeing the problem (infact all such problems) on the curve of sine (corresponding) function.

2. For the variation of n from 0 to 40, whenever n is even the GIF is the same and whenever it is odd the GIF is (1/2) less.
So, the summation becomes the simple summation minus 20 times (1/2) as there are 20 odd terms that is (1/2){[0^2+1^2+2^2+.....40^2]-20} = (1/2)[{40*41*81}/6-20] = 11060.

3. If n = 1, p(x) = ax+b (in general)
P(1) = (2)^(1/2) and p(3) = π => a={(π-2^(1/2))/2} and b={(3*2^(1/2)-π)/2} which is unique.
And when n>or=2, then p(x) has more than 2 unknowns and we have only 2 equations. Hence, we will have infinite solutions. Let us see for n=2,
p(x)=ax^2+bx+c that is 3 unknowns as a,b and c while we have only 2 equations that is p(1)=2^(1/2) and p(3)=π. So, we have infinite solutions as we can not solve these set of equations for any particular solutions as we have an extra variable.

Thanks and regards,
Deepak Kumar,
AskIITians faculty,
B.Tech, IIT Delhi.

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