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The sum of infinite terms of a decreasing GP 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 f'(0). Then the common ratio of GP is
1. \(-\frac 2 3\)
2. \(\frac 4 3\)
3. \(\frac 2 3\)
4. \(-\frac 4 3\)

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Correct Answer - Option 3 : \(\frac 2 3\)

Concept:         

GP is represented as \(a,ar,a{r^2},a{r^3},............,a{r^n}\),  where a is the first term, r is the common ratio.

The sum of infinite terms of a decreasing GP\( = \frac{a}{{1 - r}}\)

Calculation:   

  f(x) = x3 + 3x - 9

\(f'(x) = 3{x^2} + 3 = 3({x^2} + 1) \ge 3\) for all x ∈ R

f(x) is strictly increasing function for all x ∈ R

so the greatest value of f(x) in the interval is [-2, 3] is at x = 3 So, let's find f(3)

⇒ f(3) = 27

According to question we know that the sum of infinite terms of a decreasing GP is equal to the greatest value of the function f(x)

\(⇒ \frac{a}{{1 - r}} = 27\)   

⇒ a = 27 × (1 - r)        -----------(1)

According to question, the difference between the first two terms is f'(0)

⇒ a - ar = f'(0) = 3

⇒ a(1 - r) = 3 ------------(2)

So, using equation (1) and (2) we get,

⇒ 27 × (1 - r) × (1 - r) = 3

⇒ (1 - r)2 = 1/9

⇒ (1 - r) = ± 1/3

\(\Rightarrow r = \frac{2}{3}\ or\ \frac{4}{3}\)

GP is decreasing series so common ratio \(r = \frac{2}{3}\)

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