Question

Let $f\left(x\right)$ be a function such that $f\left(a_1\right)=0,f\left(a_2\right)=1,$ $f\left(a_3\right)=-2,f\left(a_4\right)=3$ and $f\left(a_5\right)=0;$ where $a_i\in R$ and $a_i<a_j\forall i<j.$ Let $g\left(x\right)$ be a function defined as $g\left(x\right)=f^{\prime }(x{)}^2+f(x\left)f^{\prime \prime }\right(x)$ on $[a_1,a_5].$ If $f\left(x\right)$ be thrice differentiable, then $g\left(x\right)=0$ has at least

• A. $4$ real roots
• B. $5$ real roots
• C. $6$ real roots
• D. $7$ real roots

Answer 1

The correct option is C $6$ real roots

$g\left(x\right)=f^{\prime }(x{)}^2+f(x\left)f^{\prime \prime }\right(x)=\frac{d}{dx}(f(x\left)f^{\prime }\right(x))$

Since, $f\left(a_1\right)=0,f\left(a_2\right)=1,f\left(a_3\right)=-2,f\left(a_4\right)=3,f\left(a_5\right)=0$
$\Rightarrow f\left(x\right)=0$ has at least $4$ roots in $[a_1,a_5].$


Let roots are $a_1,m,n,a_5;$ where $a_2<m<a_3$ and $a_3<n<a_4.$

And $f^{\prime }\left(x\right)=0$ has at least three roots, say $b_1,b_2,b_3;$ where $a_1<b_1<m,$ $m<b_2<n$ and $n<b_3<a_5.$

$\therefore$ There are at least $7$ roots of $f\left(x\right)f^{\prime }\left(x\right)=0$
$\Rightarrow$ There are at least 6 roots of $\frac{d}{dx}(f\left(x\right)f^{\prime }\left(x\right))=0$
$\Rightarrow$ There are at least $6$ roots of $g\left(x\right)=0$