The slope of normal to the curve y = kx2 − 3x + 2 at \(x = \displaystyle\frac{1}{2}\) is 5. The value of ‘k’ is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{14}{5}\)
The slope of the tangent is
\( \displaystyle \frac{dy}{dx} = 2kx - 3 \)
At \(x = \frac{1}{2}\):
\( \displaystyle \frac{dy}{dx} = 2k \times \frac{1}{2} - 3 = k - 3 \)
The slope of the normal is the negative reciprocal of the slope of the tangent:
\( \displaystyle -\frac{1}{k-3} = 5 \)
\( \displaystyle k - 3 = -\frac{1}{5} \)
\( \displaystyle k = 3 - \frac{1}{5} = \frac{14}{5} \)