Question

The normal to the curve y = f(x) at the points (3, 4) makes an angle \( \frac{3\pi}{4} \) with positive x-axis then \(f'(3) = \)

• A. 3
• B. 2
• C. 1
• D. 4

Hint:

The slope of a line making an angle \(\theta\) with the positive x-axis is \(\tan\theta\).
The derivative \(f'(x)\) at a point gives the slope of the tangent to the curve \(y=f(x)\) at that point.
If two lines are perpendicular, the product of their slopes is -1 (i.e., \(m_1 m_2 = -1\), or \(m_2 = -1/m_1\)), assuming neither slope is undefined.
\(\tan(3\pi/4) = \tan(135^\circ) = -1\).

Answer 1

The Correct Option is C

The curve is \(y = f(x)\). The point is \((3,4)\). The normal to the curve at \((3,4)\) makes an angle of \( \frac{3\pi}{4} \) with the positive x-axis. The slope of the normal (\(m_n\)) is the tangent of this angle: \( m_n = \tan\left(\frac{3\pi}{4}\right) \). \( \frac{3\pi}{4} \) is in the second quadrant (\(135^\circ\)). \( \tan\left(\frac{3\pi}{4}\right) = \tan(135^\circ) = \tan(180^\circ - 45^\circ) = -\tan(45^\circ) = -1 \). So, the slope of the normal is \(m_n = -1\). The slope of the tangent (\(m_t\)) at the point \((3,4)\) is \(f'(3)\). The tangent and the normal are perpendicular to each other. Therefore, the product of their slopes is -1 (if neither is vertical/horizontal). \( m_t \cdot m_n = -1 \) \( f'(3) \cdot (-1) = -1 \) \( -f'(3) = -1 \) \( f'(3) = 1 \). This matches option (c). \[ \boxed{1} \]