Question

Let $f$ be a differentiable function on $R$ defined by $f\left(x\right)=5-(x+1{)}^2$. Let $A$ be the point of intersection where the tangent line drawn to the graph of $y=f\left(x\right)$ at the point $P(x,f(x\left)\right)$ intersects with $x$-axis and $B$ be the intersection point where the tangent line at $P(x,f(x\left)\right)$intersects with $y$-axis. If $S\left(x\right)$ denotes the area of $\Delta AOB$, where $O$ is the origin, then the minimum value of $S\left(x\right)$ in the interval $[0,1]$ is equal to $\frac{p}{q}$, $p$ and $q$ being relatively prime. The value of $p+q$ is

Answer 1

Equation of tangent at $P(x,f(x\left)\right)$ is
$Y-f\left(x\right)=f^{\prime }\left(x\right)(X-x)$
So, $A\equiv (x-\frac{f\left(x\right)}{f^{\prime }\left(x\right)},0)$
and $B\equiv (0,f(x)-x{f}^{\prime }(x\left)\right)$

Let $S\left(x\right)$ be the area of $\Delta AOB$.
$S\left(x\right)=\frac{1}{2}\times OA\times OB$
$\Rightarrow S\left(x\right)=\frac{1}{2}[x-\frac{f\left(x\right)}{f^{\prime }\left(x\right)}]\left[f\right(x)-x{f}^{\prime }(x\left)\right]$
$=-\frac{1}{2}\frac{{\left[f\right(x)-x{f}^{\prime }(x\left)\right]}^2}{f^{\prime }\left(x\right)}$
$=\frac{{[5-x^2-2x-1+2x^2+2x]}^2}{4(x+1)}$
$=\frac{(x^2+4{)}^2}{4(x+1)}$

$S^{\prime }\left(x\right)=\frac{2\times 4(x+1)(x^2+4)\cdot 2x-4(x^2+4{)}^2}{16(x+1{)}^2}$
Putting $S^{\prime }\left(x\right)=0$, we get
$3x^2+4x-4=0$
$\Rightarrow (x+2)(3x-2)=0$
$\Rightarrow x=\frac{2}{3}(\because 0\le x\le 1)$
$S\left(\frac{2}{3}\right)=\frac{80}{27},S\left(0\right)=4,S\left(1\right)=\frac{25}{8}$
$\therefore$ Minimum value of $S\left(x\right)$ in $[0,1]$ is $\frac{80}{27}$
Hence, $p+q=80+27=107$