The sum of infinite terms of a decreasing G.P. is equal to the greatest value of the function f(x) = x3+3x–9 in the interval [–2, 3] and the difference between the first two terms is f1(0). The common ratio of the G.P. is

f(3) = 27 ⇒ S = 27 ⇒ f1(x) = 3x2 + 3⇒f1(0)=3⇒a−ar=3⇒a(1−r)=3a(1−r)×a1−r=81⇒a2=81,a=9⇒1−r=13⇒r=2/3S=91−r=27

The sum of infinite terms of a decreasing G.P. is equal to the greatest value of the function f(x) = x3+3x–9 in the interval [–2, 3] and the difference between the first two terms is f1(0). The common ratio of the G.P. is

f(3) = 27 ⇒ S = 27 ⇒ f1(x) = 3x2 + 3⇒f1(0)=3⇒a−ar=3⇒a(1−r)=3a(1−r)×a1−r=81⇒a2=81,a=9⇒1−r=13⇒r=2/3S=91−r=27

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The sum of infinite terms of a decreasing G.P. is equal to the greatest value of the function
f(x) = x³+3x–9 in the interval [–2, 3] and the difference between the first two terms is f¹(0). The common ratio of the G.P. is

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1/3

2/3

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4/3

answer is B.

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Detailed Solution

f(3) = 27 $\Rightarrow$ S = 27 $\Rightarrow$ f¹(x) = 3x² + 3
$\begin{array}{l} \Rightarrow f^{1} (0) = 3 \Rightarrow a - ar = 3 \Rightarrow a (1 - r) = 3 \\ a (1 - r) \times \frac{a}{1 - r} = 81 \Rightarrow a^{2} = 81, a = 9 \\ \Rightarrow 1 - r = \frac{1}{3} \Rightarrow r = 2 / 3 \\ S = \frac{9}{1 - r} = 27 \end{array}$