NIMCET 2019 Question Paper with solutions

Question : 41 of 120

Marks: +1, -0

The sum of infinite terms of decreasing GP is equal to the greatest values of the function f(x) =

\x^3

+ 3x - 9 in the interval [-2, 3] and the difference between the first two terms is f(0). Then the common ratio of the GP is

Solution:

We have,

f(x)=x^{3}+3x-9

f'(x)=3x^{2}+3>0∀x∈R

∴\;\;\;f(-2)=(-2)^{3}+3(-2)-9

=-8-6-9=-23

f(3)=(3)^3+3(3)-9

=27+9-9=27

∴\;\;\;\;\f_{\max}=f(3)=27

f'(0)=3

Let the GP is

a,ar,ar^2,..,|r| < 1

S_∞=a/{1-r}

27=a/{1-r}

[∵\;S_∞=f_{\max}=27]

and

a-ar=3

[∵\;f'(0)=3]

a(1-r)=3⇒a=3/{1-r}

∴\;27=3/(1-r)^2⇒(1-r)^2=1/9

⇒1-r=1/3,r=2/3

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