1cosine here as no signicance
x=1/t
lim 1 π t
____________
2t 5
t→0
= 1/10
1) if f(2x +y8,2x - y8)=xy
then f(x,y) +f(y,x)=???
2)the function f(x)=λmod(sinx) + λ2mod(cosx) +g(λ) has a period equalt to pie/2
then find λ
3)evaluate the limit \lim_{x\rightarrow \infty }(\frac{x}{2}cos(\frac{\pi ^{1000}}{9x})sin\left(\frac{\pi }{5x} \right) \right) )
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4 Answers
1f(2x +y8,2x - y8)=xy
= [(2x +y8) +(2x -y8) ] [(2x +y8) -(2x-y8)] / 64
so
f(x,y) = (x+y)(x-y) /64
f(y,x) = -(x+y)(x-y) /64
adding
f(x,y) + f(y,x) =0
1f(x)=λmod(sinx) + λ^2mod(cosx) +g(λ)
for f(x) to have a period pie/2
f(x+pie/2) = f(x)
i.e
λ|sin(pie/2 +x) | +λ2 |cos(pie/2 +x) +g(λ) = λ|sinx|+λ2 |cosx| +g(λ)
i.e
λ|cosx| +λ2 |sinx| = λ|sinx| +λ2 |cosx|
it implies
|cosx| +λ|sinx| = |sinx| +λ|cosx|
{ bcoz λ cant be zero as it wud yield a constant function }
so solving we get
λ =1
