$\mathrm{f}(\mathrm{x})$ is a continuous function on $\mathbb{R}$ and $\mathrm{y}=\mathrm{f}(\mathrm{x})$ is…
$\mathrm{f}(\mathrm{x})$ is a continuous function on $\mathbb{R}$ and $\mathrm{y}=\mathrm{f}(\mathrm{x})$ is a curve. If $(\alpha, \beta)$ is a point such that $\beta=f(\alpha)$ and $p \alpha+m \beta+n=0$ $(\mathrm{p} \neq 0, \mathrm{~m} \neq 0)$, then which one of the following is True?
When $\mathrm{p}+\mathrm{mf}^{\prime}(\alpha)=0, \mathrm{px}+\mathrm{my}+\mathrm{n}=0$ intersects the curve $y=f(x)$
$\mathrm{px}+\mathrm{my}+\mathrm{n}=0$ is always a tangent to the curve $y=f(x)$
When $\mathrm{p}+\mathrm{mf}^{\prime}(\alpha) \neq 0, \mathrm{px}+\mathrm{my}+\mathrm{n}=0$ intersects the curve $y=f(x)$
$\mathrm{px}+\mathrm{my}+\mathrm{n}=0$ is never a tangent to the curve $y=f(x)$
Solution
Since $f(x)$ is a curve contains $(\alpha, \beta)$ and
$\mathrm{p} \alpha+\mathrm{m} \beta+\mathrm{n}=0$ ... (i)
So, curve intersect $\mathrm{px}+\mathrm{my}+\mathrm{n}=0$
Let equation of curve $y=f(x)=a x^2+b x+c$
$\begin{aligned}
& \therefore \beta=a \alpha^2+b \alpha+c \text { putting in (i), we get } \\
& P \alpha+m a \alpha^2+m b \alpha+c+n=0 ... (ii)\\
& \therefore f(x)=a x^2+b x+c \Rightarrow f^{\prime}(x)=2 a x+b
\end{aligned}$
Now, $p+m f^{\prime}(\alpha)=p+m(2 a \alpha+b)$ from (i)
$=\frac{\mathrm{c}}{\alpha}+\frac{\mathrm{n}}{\alpha}-\mathrm{ma} \alpha \neq 0$