Mathematics Asked on March 8, 2021

The sum of all the integral value(s) of ‘k’ so that the curve $x^4+3kx^3+6x^2+5$ is not situated below any of its tangent line is

(A) 5

(B) 2

(C) 0

(D) -2

I have no idea of how to proceed this question

The curve $y=x^4 + 3 k x^3 + 6 x^2 + 5$ is concave up if $y''>0$ that is

$$6 left(2x^2+3 k x+2right)>0 ;;forall xinmathbb{R}$$ this happens when the discriminant $(9 k^2-16)$ is negative, that is when $-frac{4}{3}<k<frac{4}{3}$ and the sum of the integer values in this interval is $0$.

Correct answer by Raffaele on March 8, 2021

This property is called concave up, so we just solve for $f''(x)>0$ for all $x$ where $f(x)$ is the given curve.

Answered by user600016 on March 8, 2021

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