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
\[ \left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \cdots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 \, {}^{14}C_1 \cdots {}^{14}C_{12}} \]
Then \[ 30\alpha = \underline{\hspace{1cm}} \]
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