Question

If $f\left(x\right)$ is twice differentiable function such that $f\left(a\right)=0,$ $f\left(b\right)=2,$ $f\left(c\right)=-1,$ $f\left(d\right)=2,$ $f\left(e\right)=0,$ where $a<b<c<d<e,$ then the minimum number of zeroes of $g\left(x\right)={\left(f^{\prime }\right(x\left)\right)}^2+f^{\prime \prime }\left(x\right)f\left(x\right)$ in the interval $[a,e]$ is

• A. $4$
• B. $2$
• C. $7$
• D. $6$

Answer 1

The correct option is D $6$

$g\left(x\right)={\left(f^{\prime }\right(x\left)\right)}^2+f^{\prime \prime }\left(x\right)f\left(x\right)g\left(x\right)=\frac{d}{dx}\left(f\right(x\left)f^{\prime }\right(x\left)\right)$
To get the zeroes of $g\left(x\right),$
let $h\left(x\right)=f\left(x\right)f^{\prime }\left(x\right)$
We know that between any two roots of $h\left(x\right)=0,$ there lies at least one root of $h^{\prime }\left(x\right)=0$
Here, $g\left(x\right)=h^{\prime }\left(x\right)$
$f\left(x\right)=0$ has minimum $4$ roots.
So, $f^{\prime }\left(x\right)$ has minimum $3$ roots.
Since $h\left(x\right)=f\left(x\right)f^{\prime }\left(x\right),$
$\therefore h\left(x\right)=0$ has minimum $7$ roots.
$\Rightarrow h^{\prime }\left(x\right)=g\left(x\right)$ has minimum $6$ roots.