Step 1: Find the greatest value of the function f(x)
The given function is:
f(x)=x3+3x−9
First, find its derivative to check for critical points:
f′(x)=3x2+3
Since x2≥0 for all real numbers, 3x^2 + 3 > 0 for all x. This means f′(x) is strictly positive, implying that f(x) is a strictly increasing function on the entire interval [−2,3].
For a strictly increasing function, the greatest value occurs at the rightmost endpoint of the interval, which is x=3.
f(3)=(3)3+3(3)−9
f(3)=27+9−9=27
Thus, the sum of the infinite decreasing GP (S∞) is:
S∞=27
Step 2: Find the value of f′(0)
We have the derivative:
f′(x)=3x2+3
Substitute x=0:
f′(0)=3(0)2+3=3
According to the problem, the difference between the first two terms of the GP is equal to f′(0).
Let the first term be a and the common ratio be r. The first two terms are a and ar.
a−ar=3
a(1−r)=3
--- (Equation 1)
Step 3: Solve for the common ratio r
The formula for the sum of an infinite geometric progression is:
S∞=1−ra=27
From this, we can express a as:
a=27(1−r)
--- (Equation 2)
Substitute the value of a from Equation 2 into Equation 1:
[27(1−r)](1−r)=3
27(1−r)2=3
(1−r)2=273
(1−r)2=91
Taking the square root on both sides:
1−r=±31
This gives us two possible cases for r:
Case 1:
1−r=31⟹r=1−31=32
Case 2:
1−r=−31⟹r=1+31=34
Since the problem states that it is a decreasing infinite GP whose sum exists, the common ratio must satisfy |r| < 1.
Therefore, r=34 is rejected, and the valid value is r=32.
Correct Answer:
Correct Option: C
