Q104 / 180 · KCET 2021 · Mathematics
If the parabola $y=\alpha x^{2}-6 x+\beta$ passes through the point $(0,2)$ and has its tangent at $x=\frac{3}{2}$ parallel to $X$-axis, then
- A$\alpha=2, \beta=-2$
- B$\alpha=-2, \beta=2$
- Correct answerC$\alpha=2, \beta=2$
- D$\alpha=-2, \beta=-2$
Solution
$y=\alpha x^{2}-6 x+\beta...(i)$
Equation of parabola passes through
$\begin{aligned}
&(0,2) \Rightarrow 2=0-0+\beta \\
&\Rightarrow \beta=2
\end{aligned}$
Differentiating Eq. (i) W.r.t. $x$,
$\frac{d y}{d x}=2 \alpha x-6$
$\begin{aligned}
\therefore \quad\left(\frac{d y}{d x}\right)_{x=\frac{3}{2}} &=2 \times \alpha \times \frac{3}{2}-6 \\
&=3 \alpha-6
\end{aligned}$
As, the tangent is parallel to $X$-axis,
$\begin{aligned}
& & \frac{d y}{d x} &=0 \\
\Rightarrow & & 3 \alpha-6 &=0 \\
\Rightarrow & \alpha &=2
\end{aligned}$
Hence, the correct answer is option: 'C'.