Maxima and Minima

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a month ago

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Vishal Baghel

Contributor-Level 10

Since p (x) has relative extreme at x=1 and 2
so p' (x)=0 at x=1 and 2
⇒ p' (x)=A (x-1) (x-2)
⇒ p (x)=∫A (x²-3x+2)dx
p (x)=A (x³/3 - 3x²/2 + 2x)+C
P (1)=8
From (1)
8=A (1/3-3/2+2)+C
⇒ 8=5A/6+C ⇒ 48=5A+6C
P (2)=4
⇒ 4=A (8/3-6+4)+C
⇒ 4=-2A/3+C ⇒ 12=-2A+3C
From 3 and 4, C=-12
So P (0)=C=-12

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

√3|a+b|+|a-b| ≤

2√ (3+1)=4.

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

f(x) = (3x²+ax-2-a)e?
f'(x) = (3x²+ax-2-a)e? + e?(6x+a)
= e?(3x²+(a+6)x-2)
∴ x=1 is a critical point
∴ f'(1)=0
∴ 3+a+6-2=0
a = -7
∴ f'(x) = e?(3x²-x-2)
= e?(3x+2)(x-1)


∴ maxima at x = -2/3
∴ minima at x = 1

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