Question

Let $f$ and $g$ be differentiable on the interval $I$ and let $a,b\in I,a<b$. Then

• A. If $f\left(a\right)=0=f\left(b\right)$, the equation $f^{\prime }\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)=0$ is solvable in $(a,b)$
• B. If $f\left(a\right)=0=f\left(b\right)$, the equation $f^{\prime }\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)=0$ may not be solvable in $(a,b)$
• C. If $g\left(a\right)=0=g\left(b\right)$, the equation $g^{\prime }\left(x\right)+kg\left(x\right)=0$ is solvable in $(a,b),k\in R$
• D. If $g\left(a\right)=0=g\left(b\right)$, the equation $g^{\prime }\left(x\right)+kg\left(x\right)=0$ may not be solvable in $(a,b),k\in R$

Answer 1

Answer: (A), (C)

The correct options are
A If $f\left(a\right)=0=f\left(b\right)$, the equation $f^{\prime }\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)=0$ is solvable in $(a,b)$
C If $g\left(a\right)=0=g\left(b\right)$, the equation $g^{\prime }\left(x\right)+kg\left(x\right)=0$ is solvable in $(a,b),k\in R$

$f^{\prime }\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)=0$
Considering this as a Differential Equation in terms of $f\left(x\right).$
Integrating factor $e^{\int g^{\prime }\left(x\right)dx}=e^{g\left(x\right)}$

Let $h\left(x\right)=f\left(x\right)e^{g\left(x\right)}$
$h\left(x\right)$ is a composition of $2$ continuous function and differentiable on $I$. So $h\left(x\right)$ is continuous and differentiable in $(a,b).$
Also, $h\left(a\right)=0=h\left(b\right)$
By Rolle's theorem, we can conclude that $h^{\prime }\left(x\right)=0$ has atleast one root in $(a,b).$
$\Rightarrow e^{g\left(x\right)}\left(f^{\prime }\right(x)+f(x\left)g^{\prime }\right(x\left)\right)=0$ has atleast one root in $(a,b).$
$\Rightarrow f^{\prime }\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)=0$ has atleast one root in $(a,b).$
So, it is solvable in $(a,b).$

Similarly, for $g^{\prime }\left(x\right)+kg\left(x\right)$
Let $j\left(x\right)=e^{kx}g\left(x\right),k\in R$
$j\left(a\right)=0=j\left(b\right)$
By Rolle's theorem, we can conclude that $j^{\prime }\left(x\right)=0$ has atleast one root in $(a,b).$
$\Rightarrow g^{\prime }\left(x\right)+kg\left(x\right)=0$ has atleast one root in $(a,b).$
Hence, $g^{\prime }\left(x\right)+kg\left(x\right)=0$ is solvable in $(a,b),k\in R$