Q92 / 200 · WBJEE 2014 · Mathematics
Suppose that the equation $f(x)=x^{2}+b x+c=0$ has two distinct real roots $\alpha$ and $\beta$. The angle between the tangent to the curve $y=f(x)$ at the point $\left(\frac{\alpha+\beta}{2}, f\left(\frac{\alpha+\beta}{2}\right)\right)$ and the positive direction of the $x$ -axis is
- Correct answerA$0^{\circ}$
- B$30^{\circ}$
- C$60^{\circ}$
- D$90^{\circ}$
Solution
Since, $\alpha$ and $\beta$ are the roots of
$$
\begin{aligned}
f(x) &=x^{2}+b x+c \\
\therefore \quad \alpha+\beta &=-b \text { and } \alpha \beta=c \\
\therefore \quad f\left(\frac{\alpha+\beta}{2}\right) &=\left(\frac{\alpha+\beta}{2}\right)^{2}+b\left(\frac{\alpha+\beta}{2}\right)+c \\
&=\left(-\frac{b}{2}\right)^{2}+b\left(-\frac{b}{2}\right)+c \\
&=\frac{b^{2}}{4}-\frac{b^{2}}{2}+c=-\frac{b^{2}}{4}+c
\end{aligned}
$$
Now, $\quad \frac{d y}{d x}=f^{\prime}(x)=2 x+b$
At point $\left(\frac{\alpha+\beta}{2}, f\left(\frac{\alpha+\beta}{2}\right)\right)$
i.e., $\begin{aligned}\left(-\frac{b}{2},-\frac{b^{2}}{4}+c\right) \\ \frac{d y}{d x} &=2\left(-\frac{b}{2}\right)+b=0 \end{aligned}$
Hence, the slope of the tangent to the curve and the positive direction of $x$ -axis is $0^{\circ}$.
Hence, the correct answer is option: 'A'.