Let f(x)=x amd g(x)=|x| for all `x in R`. Then the function `phi(x)"satisfying"{phi(x)-f(x)}^(2)+{phi(x)-g(x)}^(2)` =0 is

Let phi(a)=f(x)+f(2ax) and f'(x)>0 for all x in[0,a]. Then ,phi(x)

Let f(x)=x+1 and phi(x)=x-2. Then the value of x satisfying |f(x)+phi(x)|=|f(x)|+|phi(x)| are :

Let f(x)=x+1 and phi(x)=x-2. Then the value of x satisfying |f(x)+phi(x)|=|f(x)|+|phi(x)| are :

int (f(x)phi'(x)-phi(x)f'(x))/(f(x)phi(x)) {logphi(x) - logf(x)} dx is

Let f(x+y)=f(x)f(y) for all x,y in R and f(0)!=0 . Let phi(x)=f(x)/(1+f(x)^2) . Then prove that phi(x)-phi(-x)=0

Let f(x+y)=f(x)f(y) for all x,y in R and f(0)!=0. Let phi(x)=(f(x))/(1+f(x)^(2)). Then prove that phi(x)-phi(-x)=0

f'(x)=phi(x) and phi'(x)=f(x) for all x.Also,f(3)=5 and f'(3)=4. Then the value of [f(10)]^(2)-[phi(10)]^(2) is

Let f(x+y)=f(x)*f(y)AA x,y in R and f(0)!=0* Then the function phi defined by .Then the function phi(x)=(f(x))/(1+(f(x))^(2)) is

Let f(x)=|x-a| phi(x), where is phi continuous function and phi(a)!=0. Then