The function 
is such that 
. Consider two statements.
(S1) there exists 
such that 
and 
(S2) there exists 
, such that f is decreasing in 
, increasing in 
and 
Then
Text Solution
Verified by ExpertsA
f(x) = x 3 − 6x 2 + ax + b
f(b) = 8 − 24 + 2a + b = 0
2a + b = 16…
f(d) = 64 − 96 + 4a + b = 0
4a + b = 32…
Solving and
a = 8, b = 0
f(x) = x 3 − 6x 2 + 8x
f(x) = x 3 − 6x 2 + 8x
f’ (x) = 3x 2 − 12x + 8
f” (x) = 6x − 12
⇒ f” (x) ↑ x > 2 f’ (x) ↓ x < 2
f’ = 12 − 24 + 8 = −4
f’ = 48 − 48 + 8 = 8
f’ (x) = 3x2 − 12x + 8
vertex (2,−4)
f’ = −4, f’ = 8, f’ = 27 − 36 + 8
f ′ (x1) = −1 then x 1 = 3
f ′ (x 2 ) = 0
Again
f ′(x) < 0 for x ∈ (2, x 4 )
f ′(x) > 0 for x ∈ (x 4 , 4)
x 4 ∈ (3, 4)
f(x) = x 3 − 6x 2 + 8x
f(c) = 27 − 54 + 24 = −3
f(d) = 64 − 96 + 32 = 0
For x 4 (3, 4)
f (x 4 ) < −3√3
and f ′ (x 3 ) > −4
2f ′ (x 3 ) > −8
So, 2f ′ (x 3 ) = √3f (x 4 )
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