Most Important Selected Qs for JEE AdvancedMathematicsDifferential Equations
Let f:(0, ) R be a differentiable function satisfying the equation f(x y)=e^ x y-x-y (e^y f(x)+e^x f(y) ) x, y>0 . If f^ (1)=e , then:
Options
- A_ x e [ f(x)-e^x x-e ]=e^ (e-1)
- Bnumber of roots of the equation f(x)=x e^x in (0, ) is 2 .
- C₁^e f(x) d x < e^e(e-1)
- Df(x) is a strictly increasing function in (0, ) .
Correct answer
D. f(x) is a strictly increasing function in (0, ) .
Step-by-step solution
f(x y)=e^ x y-x f(x)+e^ x y-y f(y) Differentiating w.r.t. xy f^ (x y)=e^ x y-x f^ (x)+f(x) e^ x y-x (y-1)+f(y) e^ x y-y y Put x=1y f^ (y)=e^ y-1 f^ (1)+f(1) e^ y-1 (y-1)+y f(y) Now put x=y=1 ; f(1)=0 y f^ (y)=e^y+y f(y) x (f^ (x)-f(x) )=e^xe^ -x (f^ (x)-f(x) )= 1 x Integrate both sides e^ -x f(x)= x+Cf(1)=0 C=0 f(x)=e^x x . Now verify.