Question

Suppose that the equation $f\left(x\right)=x^2+bx+c=0$ has two distinct real roots $\alpha$ and $\beta$. The angle between the tangent to the curve $y=f\left(x\right)$ at the point $(\frac{\alpha +\beta }{2},f\left(\frac{\alpha +\beta }{2}\right))$ and the positive direction of the $x$-axis is

• A. $0^{}$
• B. ${30}^{}$
• C. ${60}^{}$
• D. ${90}^{}$

Answer 1

Answer: (A)

$0^{}$

Since, $\alpha$ and $\beta$ are the roots of
$f\left(x\right)=x^2+bx+c$
$\therefore \alpha +\beta =-b$ and $\alpha \beta =c$
$\therefore f\left(\frac{\alpha +\beta }{2}\right)={\left(\frac{\alpha +\beta }{2}\right)}^2+b\left(\frac{\alpha +\beta }{2}\right)+c$
$={(-\frac{b}{2})}^2+b(-\frac{b}{2})+c$
$=\frac{b^2}{4}-\frac{b^2}{2}+c=-\frac{b^2}{4}+c$

Now, $\frac{dy}{dx}=f^{\prime }\left(x\right)=2x+b$

At point, $(\frac{\alpha +\beta }{2},f\left(\frac{\alpha +\beta }{2}\right))$, i.e., $(-\frac{b}{2},-\frac{b^2}{4}+c)$,

$\frac{dy}{dx}=2(-\frac{b}{2})+b=0$

Hence, the slope of the tangent to the curve and the positive direction of $x$-axis is $0^{}$.