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If the tangent drawn at a point $P$ on the curve $y=3 x^2-5 x+7$ is parallel to its chord joining the points $\left(1, y_1\right)$ and $\left(2, y_2\right)$ on it, then the $x$-coordinates of the point $P$ is
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The correct answer is:
$\frac{3}{2}$
Let the point $P(h, k)$ on the given curve
$\begin{aligned} & y=3 x^2-5 x+7 \\ & \text { So, } \quad k=3 h^2-5 h+7 \\ & \text { and }\left.\quad \frac{d y}{d x}\right|_P=6 h-5=\frac{y_2-y_1}{2-1} \\ & \because \quad y_1=3-5+7=5 \text { and } y_2=12-10+7=9 \\ & \left.\therefore \quad \frac{d y}{d x}\right|_P=6 h-5=\frac{9-5}{1} \\ & \Rightarrow \quad 6 h-5=4 \\ & \Rightarrow \quad 6 h=9 \\ & \end{aligned}$
$\Rightarrow \quad h=\frac{3}{2}$
So, $x$-coordinate of point $P$ is $\frac{3}{2}$.
$\begin{aligned} & y=3 x^2-5 x+7 \\ & \text { So, } \quad k=3 h^2-5 h+7 \\ & \text { and }\left.\quad \frac{d y}{d x}\right|_P=6 h-5=\frac{y_2-y_1}{2-1} \\ & \because \quad y_1=3-5+7=5 \text { and } y_2=12-10+7=9 \\ & \left.\therefore \quad \frac{d y}{d x}\right|_P=6 h-5=\frac{9-5}{1} \\ & \Rightarrow \quad 6 h-5=4 \\ & \Rightarrow \quad 6 h=9 \\ & \end{aligned}$
$\Rightarrow \quad h=\frac{3}{2}$
So, $x$-coordinate of point $P$ is $\frac{3}{2}$.
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