Let g(x) be a polynomial of degree one and f(x) be defined by `f(x)=[(g(x), xle0),([((1+x))/((2+x))]^(1//x), x gt0):}`. Find the continuous function `f(x)` satisfying `f'(1)=f(-1)`.

Options

A. `-(1)/(9)(1+6log_(e),3)x`
B. `(1)/(9)(1+6log_(e),3)`
C. `-(1)/(9)(1-6log_(e),3)x`
D. none of these

Answer

A

Explanation

Let g(x)=ax+b
Since f(x) is continuous at x=0
`therefore underset(x to 0^(+))lim f(x)=underset(x to 0^(-))lim f(x)=f(0)`
`Rightarrow underset(x to 0^(+))lim (1+x)/((2+x)^(1//x))=g(0) Rightarrow 0=b`
`"For "x gt 0,"we have "`
`f(x)=(1+x)/((2+x)^(1//x))`
`Rightarrow f(x)=(1+x)e^((-1//x)log(2+x))`
`Rightarrow f(x)=(2+x)e^(-1//x)+(1+x)(2+x)^(-1//x){(log(2+x))/(x^(2))-(1)/(x(2+x))}`
`Rightarrow f'(x)=(f(x))/(1+x)+f(x) {(log(2+x))/(x^(2))-(1)/(x(2+x))}`
`Rightarrow f'(1)=(f(1))/(2)+f(1){log3-(1)/(3)}`
`Rightarrow f'(1)=(1)/(3)+(2)/(3)log 3-(2)/(9)=(1)/(9)+(2)/(3)log3" "[therefore f(1)=2//3]`
Now,
`f(-1)=f'(1)`
`Rightarrow -a+b=(1)/(9)(1+6log3) Rightarrowa=-(1)/(9)(1+6log3)`
`"Hence", g(x)=-(1)/(9)(1+6log,3)x`