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

The Correct Option is A

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