Q110 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
$y=2 x^3-8 x^2+10 x-4$ is a function defined on $[1,2]$. If the tangent drawn at a point $(a, b)$ on the graph of this function is parallel to $X$-axis and $a \in(1,2)$, then $a=$
- A$0$
- B$5$
- C$1$
- Correct answerD$\frac{5}{3}$
Solution
$\frac{d y}{d x}=6 x^2-16 x+10$
Tangent is parallel to $X$ axis $\Rightarrow \frac{d y}{d x}=0$ $\begin{aligned} & 6 x^2+6 x+10=0 ; 3 x^2-8 x+5=0 \\ & (x-1)\left(x-\frac{5}{3}\right)=0 \quad \therefore a=\frac{5}{3} \in(1,2) . \end{aligned}$
Tangent is parallel to $X$ axis $\Rightarrow \frac{d y}{d x}=0$ $\begin{aligned} & 6 x^2+6 x+10=0 ; 3 x^2-8 x+5=0 \\ & (x-1)\left(x-\frac{5}{3}\right)=0 \quad \therefore a=\frac{5}{3} \in(1,2) . \end{aligned}$
Hence, the correct answer is option: 'D'.