If $f(x)=\left\{\begin{array}{ll}a x^2-b x+2, & x < 3 \\ b x^2-3, & x \geq 3\end{array}\right.$ is…
If $f(x)=\left\{\begin{array}{ll}a x^2-b x+2, & x < 3 \\ b x^2-3, & x \geq 3\end{array}\right.$ is differentiable at every $x \in \mathbb{R}$, then the area (in sq units) of the triangle formed by the line $\frac{x}{a}+\frac{y}{b}=1$ with the coordinate axes is
$\frac{175}{81}$
$\frac{175}{27}$
$\frac{35}{27}$
$\frac{125}{27}$
Solution
$\because \mathrm{f}(\mathrm{x})$ is differentiable at every $\mathrm{x} \in \mathrm{R}$.
$\therefore \mathrm{f}(\mathrm{x})$ will be differentiable at $\mathrm{x}=3$ also.
So, L.H.D = R.H.D.....(i)
$
\begin{aligned}
& \text { L.H.D }=\lim _{x \rightarrow 3}(2 a x-b)=6 a-b \\
& \text { R.H.D }=\lim _{x \rightarrow 3}(2 b x)=6 b \\
& \therefore 6 a-b=6 b \Rightarrow 6 a=7 b .....(ii)\\
& \because f(x) \text { is continuous on } x=3 \\
& \therefore \text { L.H.L }=\text { R.H.L } \\
& \Rightarrow 9 a-3 b+2=9 b-3 \\
& \Rightarrow 9 a-12 b=-5.....(iii)
\end{aligned}
$
Solving $\mathrm{eq}^{\mathrm{n}}$ (ii) \& (iii), we get :
$
\begin{aligned}
& a=\frac{35}{9}, b=\frac{10}{3} \\
& \therefore \text { Area }=\frac{1}{2} \times a b=\frac{1}{2} \times \frac{35}{9} \times \frac{10}{3}=\frac{175}{27}
\end{aligned}
$