Let g(x) be a polynomial of degree one and f(x) be defined by `f(x)=-{g(x), x<=0 and |x|^sinx, x>0` If f(x) is continuous satisfying `f'(1)=f(-1)`, then g(x) is

Answer

`f(x) = {{:(2/3(log(2/3) - 1/6)x", " x le 0),(((1+x)/(2+x))^(1//x)", "xgt0):}`

Explanation

Let g(x) = ax + b be a polynomial of degree one.
`rArr" " f(x) = {{:(ax+ b", " x le 0),(((1+x)/(2+x))^(1//x)" , "xgt0):}`
Since, f(x) is continuous and f'(1) = f(-1)
`:." "("LHL at "x = 0 ) = ("RHL at " x = 0) `
` rArr" " underset( x to 0) lim (ax + b) = underset( x to 0) lim ((x+1)/(x + 2))^(1//x) `
` rArr" " b = 0` ...(i)
Also, `f'(1) = f(-1) `
` rArr" " f(x) = ((1+x)/(2+x))^(1//x) , x gt 0 `
` rArr" " logf(x) = 1/x [log (1+x) - log(2+x)]` On differentiating both sides, we get
`(f'(x))/(f(x)) = (x[1/(1+x) -1/(2+x)]-1[log((1+x)/(2+x))])/x^(2) `
`:." " f'(x) = ((1+x)/(2+x)) ^(1//x)[(x/((1+x)(2+x))-log((1+x)/(2+x)))/x^(2)]`
` rArr" " f'(1) = 2/3 {1/6-log(2/3)}`
and ` f(-1) =- a+b=-a` [from Eq. (i)]
`:." "-a= 2/3 (1/6 - log (2/3) )`
Thus, `f(x) = {{:(2/3(log(2/3)-1/6)x", " x le 0),(((1+x)/(2+x))^(1//x)" , " x gt0):}`
Now, to check continuity of f(x) (at x = 0) .
RHL = ` underset(x to 0) lim ((1+x)/(2+x))^(1//x) = 0`
` :." "LHL = underset( x to 0) lim 2/3 [log(2/3) - 1/6] x = 0`
Hence, f(x) is continuous for all x.