What is the condition that f(x) = x³ + x² + kx has no local extremum?

Correct Answer is : 3k>13k > 13k>1

Correct Answer is : 3k>13k > 13k>1

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UPSC NDA 2 2021 Mathematics Solved Paper

Question : 95 of 120

Marks: +1, -0

f(x) = x^{3} + x^{2} + kx

has no local extremum?

Solution:

Concept:

No local extremum means the derivative has no real roots, so the quadratic equation from

f'(x)

has a negative discriminant.

Explanation:

Differentiate

f(x) = x^3 + x^2 + kx

f'(x) = 3x^2 + 2x + k

For no local extremum,

f'(x) \neq 0

for any real

x

This requires the quadratic

3x^2 + 2x + k

to have no real roots, so its discriminant must be negative.

Compute discriminant

D = b^2 - 4ac = 2^2 - 4(3)(k) = 4 - 12k

Set

D < 0

4 - 12k < 0

\Rightarrow

4 < 12k

\Rightarrow

1 < 3k

\Rightarrow

3k > 1

Thus, the condition is

3k > 1

Answer:

3k > 1

(Option B)

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