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 $f(x)\cdot f^{\prime }(x)=x$ 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 $g^{\prime }(c)=0$
Now, $g^{\prime }(x)=2f(x)\cdot f^{\prime }(x)-2x=0$
$\therefore f(x)\cdot f^{\prime }(x)=x$
i.e., $f(c)\cdot f^{\prime }(c)-c=0$