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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 n2/2 from o to 40 i.e sigma o to 40 of [n2/2] 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.

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,