Question

Suppose $f$ is continuous on $[a, b]$, differentiable on $(a, b)$ and satisfies $f^{2}(a)-f^{2}(b)=a^{2}-b^{2} $ In $(a, b)$ the equation $f(x) \cdot f'(x)=x$ has

• A. no root
• B. at least one root
• C. exactly one root
• D. nothing definite can be said

Answer 1

Answer: (B)

Consider the function
$g(x)=f^{2}(x)-x^{2}, x \in[a, b] $
$g(a)=f^{2}(a)-a^{2} $
and $g(b)=f^{2}(b)-b^{2} $
Let $g(a)=g(b) $
$\therefore f^{2}(a)-a^{2}=f^{2}(b)-b^{2}$
$f^{2}(a)-f^{2}(b)=a^{2}-b^{2}$ which is true
Hence, Rolle’s theorem is applicable to $g(x)$
So, there exists at least one $c \in(a, b)$ where $g'(c)=0$
Now, $g'(x)=2 f(x) \cdot f'(x)-2 x=0$
$\therefore f(x) \cdot f'(x)=x$
i.e., $f(c) \cdot f'(c)-c=0$