Question

Let $P(x)$ be a polynomial of degree 4 having a local maximum at $x=2$ and $\lim_{x\to0} (3-\frac{P(x)}{x}) = 27.$ If $P(1)=-9$ and $P''(x)$ has a local maximum at $x=2$. If global maximum value of $y=P'(x)$ on the set $A=\{x:x^2+12\le7x\}$ is $k$, then the value of $k$ is

• A. 24
• B. 12
• C. 36
• D. 48

Answer 1

Answer: (A)

24

Answer 2

Let $P(x) = a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$. From $\lim_{x\to0} (3-\frac{P(x)}{x}) = 27$, we get $P(0)=0$ and $P'(0)=-24$. Thus, $a_0=0$ and $a_1=-24$. $P(1)=-9 \implies a_4+a_3+a_2+a_1+a_0 = -9 \implies a_4+a_3+a_2-24= -9 \implies a_4+a_3+a_2=15$. $P'(x) = 4a_4x^3 + 3a_3x^2 + 2a_2x + a_1$. $P'(2)=0 \implies 32a_4+12a_3+4a_2-24=0 \implies 8a_4+3a_3+a_2=6$. $P''(x) = 12a_4x^2 + 6a_3x + 2a_2$. For $P''(x)$ to have a local maximum at $x=2$, $P'''(2)=0$ and $P^{(4)}(2) \le 0$. $P'''(x) = 24a_4x + 6a_3$. $P'''(2)=0 \implies 48a_4+6a_3=0 \implies a_3=-8a_4$. $P^{(4)}(x) = 24a_4$. $P^{(4)}(2) \le 0 \implies 24a_4 \le 0 \implies a_4 \le 0$. Substituting $a_3=-8a_4$ into $a_4+a_3+a_2=15$ and $8a_4+3a_3+a_2=6$: $-7a_4+a_2=15$ and $-16a_4+a_2=6$. Subtracting the two equations: $9a_4=9 \implies a_4=1$. This contradicts $a_4 \le 0$. Assuming the question meant $P''(x)$ has a local minimum at $x=2$, then $a_4 \ge 0$, so $a_4=1$ is valid. If $a_4=1$, then $a_3=-8$, and $a_2=15+7a_4 = 15+7=22$. $P'(x) = 4x^3 - 24x^2 + 44x - 24$. The set $A=\{x:x^2+12\le7x\} \implies x^2-7x+12 \le 0 \implies (x-3)(x-4) \le 0 \implies x \in [3,4]$. We need to find the maximum of $P'(x)$ on $[3,4]$. $P''(x) = 12x^2 - 48x + 44$. The roots of $P''(x)=0$ are $x = 2 \pm \frac{\sqrt{3}}{3}$, which are not in $[3,4]$. Since $P''(3) = 12(9) - 48(3) + 44 = 108 - 144 + 44 = 8 > 0$, $P'(x)$ is increasing on $[3,4]$. The maximum value of $P'(x)$ occurs at $x=4$. $k = P'(4) = 4(4^3) - 24(4^2) + 44(4) - 24 = 256 - 384 + 176 - 24 = 24$.