The value of 'c' in Rolle's theorem for
in (a, b) where a>0 is
x²+ab
f(x) = log
x(a + b)
1) A.M. of a, b
2) G.M. of a, b
1 1
4)
a b
3) H.M. of a, b
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Step-by-step explanation:
We have,
The function f(x)=log
(a+b)x
x
2
+ab
in[a,b]
Now,
x∈(a,b)
∴
(a+b)x
x
2
+ab
∈R
+ve
We know that,
(1). log with the positive quantities as it domain is continuous.
∴f(x) is continuous on [a,b].
(2).f(x)=log(x
2
+ab)−log(a+b)x
=log(x
2
+ab)−log(a+b)x−logx
On differentiation and we get,
f
′
(x)=
x
2
+ab
2x
−0−
x
1
∵x∈(a,b)
So, f
′
(x) is exists on (a,b).
(3).f(a)=log(
(a+b)a
a
2
+ab
)=log(1)=0
f(b)=log(
(a+b)b
b
2
+ab
)=log(1)=0
∴f(a)=f(b)
Then,
There exists at least one real.
c∈(a,b) such that f
′
(c)=0.
Now,
f
′
(c)=0
⇒
c
2
+ab
2c
−
c
1
=0
⇒
c
2
+ab
2c
=
c
1
⇒2c
2
=c
2
+ab
⇒c
2
=ab
⇒c=
ab
∈R
+ve
Hence, this is the answer.
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