[" (2) "f(x)={[(x^(2)-2x-3)/(x+1)," (riti "f(x!=-1],[ln1," wit "5x=-1]]

f(x)=sqrt(log((3x-x^(2))/(x-1)))

f(x)=sqrt(log((3x-x^(2))/(x-1)))

f(x)=sqrt(log((3x-x^(2))/(x-1)))

f(x) = log(x^(2)+2)-log3 in [-1,1]

f(x) = log(x^(2)+2)-log3 in [-1,1]

If f(x+1/x)=x^2+1/x^2+8 then f(3), f(5), f(-1)

If f(x) is continuous at x=1 , where f(x)=((x+3x^(2)+5x^(3)+...+(2n-1)x^(n)-n^(2))/(x-1)) , for x !=1 , then f(1)=

If f(x) = {:{((x -1)/(2x^(2) - 7 x +5) , " for " x ne 1) , (-1/3 , " for " x = 1):} then f' (1) is equal to

If f(x) = {:{((x -1)/(2x^(2) - 7 x +5) , " for " x ne 1) , (-1/3 , " for " x = 1):} then f (1) is equal to

Let f(x) = ln(x-1)(x-3) and g(x) = ln(x-1) + ln(x-3) then,