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 :
- -15
- -60
- 60
- 15
The Correct Option is A
Solution and Explanation
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
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Concepts Used:
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The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
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