Q107 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics
Let $f(x)=x^3+2 x^2-x$ be a real valued function. Then, the value of Lagrange's constant $C$ in $(-1,2)$ is
- A$\frac{-4+\sqrt{76}}{3}$
- Correct answerB$\frac{-2+\sqrt{19}}{3}$
- C$\frac{-4+\sqrt{19}}{6}$
- D$\frac{-2+\sqrt{19}}{6}$
Solution
We have, $f(x)=x^3+2 x^2-x$
Clearly, $f(x)$ is continuous in $[-1,2]$ and differentiable in $(-1,4)$
$[\because$ polynomial functions are everywhere continuous and differentiable.]
$\therefore$ By Lagrange's mean value theorem, there exist $C \in(-1,4)$ such that $f^{\prime}(c)=\frac{f(2)-f(-1)}{2-(-1)}$
$$
\begin{aligned}
& \Rightarrow 3 c^2+4 c-1=\frac{(8+8-2)-(-1+2+1)}{3} \\
& \Rightarrow 3 c^2+4 c-1=\frac{14-2}{3}=4 \\
& \Rightarrow \quad 3 c^2+4 c-5=0 \\
& \Rightarrow \quad c=\frac{-4 \pm \sqrt{16+60}}{6} \\
& \Rightarrow \quad c=\frac{-4 \pm \sqrt{76}}{6}
\end{aligned}
$$
Now, as $8 < \sqrt{76} < 9$
$$
\begin{aligned}
& \therefore 4 < -4+\sqrt{76} < 5 \Rightarrow \frac{4}{6} < -\frac{4+\sqrt{76}}{6} < \frac{5}{6} \\
& \therefore \quad c=\frac{-4+\sqrt{76}}{6}=\frac{-2+\sqrt{19}}{3} \in(-1,2)
\end{aligned}
$$
Hence, the correct answer is option: 'B'.