Let f : [0, 1] → R be a continuous function such that f(x) = f( (3) √3 x), ∀x ∈ [0, 1]. Show that f is a constant function.
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Answer:
Correct options are B) and C)
f(x)=1−2x+∫
o
x
e
x−t
f(t)dt
⇒e
−x
f(x)=e
−x
(1−2x)+∫
o
x
e
−t
f(t)dt
Differentiate w.r.t.x.
−e
−x
f(x)+e
−x
f
′
(x)=−e
−x
(1−2x)+e
−x
(−2)+e
−x
f(x)
⇒−f(x)+f
′
(x)=−(1−2x)−2+f(x)
⇒f
′
(x)−2f(x)=2x−3
Integrating factor =e
−2x
f(x)⋅e
−2x
=∫e
−2x
(2x−3)dx
=(2x−3)∫e
−2x
dx−∫((2)∫e
−2x
dx)dx
=
−2
(2x−3)e
−2x
−
2
e
−2x
+c
f(x)=
−2
2x−3
−
2
1
+ce
2x
f(0)=
2
3
−
2
1
+c=1⇒c=0
∴f(x)=1−x
Area =
4
π
−
2
1
=
4
π−2
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