1.the range of f9x)=cos(sin(log(x2+e2/x2+1))) + sin(cos(log(x2+e2/x2+1))) is (a)[-2,2] (b)[c0s(sin10+sin(cos1),1+sin1] 2.the period of |sinx|+|cosx 3.let f(x)=x-[x]/1+x-[x], x belongs to R, then the range of f is (a)[0,1/2] (b0[0,1]

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yudhisthir narvaria · Answer

the period of |sin x|+ |cos x| is &prod;

Aniket Patra · Answer

1.The range is most probably the option (b)..Here f(x)=cos(sin p) + sin(sin(pi/2- p)..where p=log(x^2+e^2/x^2+1)..And so the value of cos(sin p) and sin(cos p) cant be 1 for the same value of p..becoz they are complementary functions..

2.The period of mod (sinx) + mod (cosx) is definitely pi/2...Both of them are even functions..And their period is 1/2 of the LCM of pi and pi...becoz period of mod (sinx) and mod (cosx) is pi..

3.The range is option (a)..Becoz the range of x-[x] i.e.{x} is 0 < x < 1.

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1.the range of f9x)=cos(sin(log(x2+e2/x2+1))) + sin(cos(log(x2+e2/x2+1))) is (a)[-2,2] (b)[c0s(sin10+sin(cos1),1+sin1] 2.the period of |sinx|+|cosx 3.let f(x)=x-[x]/1+x-[x], x belongs to R, then the range of f is (a)[0,1/2] (b0[0,1]

1.the range of f9x)=cos(sin(log(x2+e2/x2+1))) + sin(cos(log(x2+e2/x2+1))) is (a)[-2,2] (b)[c0s(sin10+sin(cos1),1+sin1] 2.the period of |sinx|+|cosx 3.let f(x)=x-[x]/1+x-[x], x belongs to R, then the range of f is (a)[0,1/2] (b0[0,1]

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