1.

Let f(x) be a non negative continuous and bounded function for all xge0 .If (cos x)f(x) lt (sin x- cosx)f(x) forall x ge 0, then which of the following is/are correct?

Answer»

`f(6)+f(5)gt0`
`x^(2)-3x+2+f(7)=0` has 2 distinct solution
f(5)f(7)-f(5)=0
`underset(xrarr6)lim (f(x)-sin (PIX))/(x-6)=1`

Solution :Let `g(x)=e^(x)cosxf(x)`
`therefore g(x)=e^(x)cosxf(x)+e^(x)(cosx-sinx)f(x)`
`=e^(x)(cosxf(x)+(cosx-sinx)f(x)`
`le0`
`therefore g(X)` is a NON increasing function
`therefore f(6)=e^(6)cos 6 f(6)le0`
`therefore f(6)le0`
But given that f(x) is non negative ltbegt `therefore f(6)=0`
with similar reasons f(5),f(7)=0
Thus `x^(2)-3x+2+f(7)=0`
or `x^(2)-3x+2=0` l has 2 distinct solution
`underset(xrarr6)lim(f(x)-sin(pix))/(x-6)`
=`underset(xrarr6)limf(x)-picos(pix)/(1)` (applying L hospital rule)
`=f(6)-pilt0`


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