Question

Let $ƒ(x)$ be a polynomial function such that $ƒ(x) + ƒ^′(x) + ƒ^{′′}(x) = x^5 + 64$. Then, the value of $\lim\limits_{x \to 1}\frac {f(x)}{x−1}$ is equal to :

• A. -15
• B. -60
• C. 60
• D. 15

Answer 1

Answer: (A)

-15

Answer 2

ƒ(x) + ƒ′(x) + ƒ′′(x) = x5 + 64
Let ƒ(x) = x5 + ax4 + bx3 + cx2 + dx + e
ƒ′(x) = 5x4 + 4ax3 + 3bx2 + 2cx + d
ƒ′′(x) = 20x3 + 12ax2 + 6bx + 2c
x5 + (a + 5)x4 + (b + 4a + 20) x3 + (c + 3b + 12a) x2 + (d + 2c + 6b) x + e + d + 2c = x5 + 64
a + 5 = 0
b + 4a + 20 = 0
c + 3b + 12a = 0
d + 2c + 6b = 0
e + d + 2c = 64

∴ a = – 5, b = 0, c = 60, d = –120,e = 64
∴ ƒ(x) = x5 – 5x4 + 60x2 – 120x + 64

Now,
$\lim\limits_{x \to 1}$ $\frac {x^5−5x^4+60x^2−120x+64}{x−1}$ is ($\frac 00$ from)
According to the L′ Hopital rule:
$\lim\limits_{x \to 1}$ $\frac {5x^4−20x^3+120x−120}{1}$ = –15

So, the correct option is (A): -15