Question

If the curves \( y = x^3 - 3x^2 - 8x - 4 \) and \( y = 3x^2 + 7x + 4 \) touch each other at a point \( P \), then the equation of the common tangent at \( P \) is

• A. \( x - y + 1 = 0 \)
• B. \( 2x - y + 1 = 0 \)
• C. \( x + y + 1 = 0 \)
• D. \( 2x + y + 1 = 0 \)

Hint:

Equal values and equal slopes at the same point means curves are tangent. Use this to find common tangents.

Answer 1

The Correct Option is A

If two curves touch at a point \( P \), they must have the same coordinates and equal derivatives at \( P \).
Set \( x^3 - 3x^2 - 8x - 4 = 3x^2 + 7x + 4 \).
Simplify to find the x-coordinate of point \( P \), then compute the corresponding y.
Differentiate both functions:
\( y_1' = 3x^2 - 6x - 8 \), \quad \( y_2' = 6x + 7 \).
Set slopes equal to get common tangent slope, and use point-slope form to derive the tangent line.
Final tangent: \( x - y + 1 = 0 \).