Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Continuity and Differentiability

Question:

If y = \( { x }^{ 4 } - k { x }^{ 3 } \) then for what value of x, we have \(\frac{d^3 y}{dx^3}\) = 0?

Options:

0

\[\frac{k}{4}\]

Both 0 and \(\frac{k}{2}\)

Neither 0 nor \(\frac{k}{2}\)

Correct Answer:

\[\frac{k}{4}\]

Explanation:

\(\frac{d^3 y}{dx^3}\) = 24x - 6kk