Let f be a differentiable function satisfying the functional rule f ( xy )= f ( x )+ f ( y )+ xy - x - y ∀ x , y 0 and f prime(1)=4 then ∫ (f(x)/x) d x is equal to [Note: ' C ' is constant of integration.]

Question

Let f be a differentiable function satisfying the functional rule f ( xy )= f ( x )+ f ( y )+ xy - x - y ∀ x , y >0 and f prime(1)=4 then ∫ (f(x)/x) d x is equal to [Note: ' C ' is constant of integration.] (A) 3 ln 2 x+x+C (B) 3 ln x + x + C (C) (3/2) ln x+x+C (D) (3/2) ln 2 x+x+C

Tardigrade Admin · Accepted Answer

Differentiable w.r.t. X text y f prime(x y)=f prime(x)+y-1 text Put x=1 y f prime(y)=f prime(1)+y-1 y f prime(y)=y+3 ∴ f prime(y)=1+(3/y) Integrating g ( y )= y +3 ln y + C text Also, f (1)=1 C =0 ∴ f ( x )=3 ln x + x ∫ ( f ( x )/ x ) dx =∫ (3 ln x / x ) dx +∫ dx =(3/2) ln 2 x + x + C

Q. Let $f$ be a differentiable function satisfying the functional rule $f (x y) = f (x) + f (y) + x y - x - y \forall x, y > 0$ and $f^{'} (1) = 4$ then $\int \frac{f ( x )}{x} d x$ is equal to
[Note: ' $C$ ' is constant of integration.]

$3 ln^{2} x + x + C$

$3 ln x + x + C$

$\frac{3}{2} ln x + x + C$

$\frac{3}{2} ln^{2} x + x + C$

Q. Let f be a differentiable function satisfying the functional rule f(xy)=f(x)+f(y)+xy−x−y∀x,y>0 and f′(1)=4 then ∫xf(x)​dx is equal to [Note: ' C ' is constant of integration.]

3ln2x+x+C

3lnx+x+C

23​lnx+x+C

23​ln2x+x+C

Differentiable w.r.t. X y f′(xy)=f′(x)+y−1 Put x=1 yf′(y)=f′(1)+y−1 yf′(y)=y+3 ∴f′(y)=1+y3​ Integrating g(y)=y+3lny+C Also, f(1)=1 C=0 ∴f(x)=3lnx+x ∫xf(x)​dx=∫x3lnx​dx+∫dx=23​ln2x+x+C

Q. Let $f$ be a differentiable function satisfying the functional rule $f (x y) = f (x) + f (y) + x y - x - y \forall x, y > 0$ and $f^{'} (1) = 4$ then $\int \frac{f ( x )}{x} d x$ is equal to
[Note: ' $C$ ' is constant of integration.]

$3 ln^{2} x + x + C$

$3 ln x + x + C$

$\frac{3}{2} ln x + x + C$

$\frac{3}{2} ln^{2} x + x + C$