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Question

If the value of a+baΨ(x)dx does not depend upon a, then b can be

A
1
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B
1
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C
2
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D
None of these
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Solution

The correct option is C 2
Given Ψ(x)=f(g(x)) ...(1)
f(mx+ny)=(f(x))m(f(x))nx,y,m,nR ...(2)
f(0)=ln2f(0) ...(3)
g(x) is periodic function given by g(x)=|x1|12;0x2 ...(4)
h(x)=g(x)+sinπx is periodic with period 2
h(x+2)=h(x)g(x+2)+sinπ(x+2)=g(x)+sinπx
g(x+2)+sinπx=g(x)+sinπxg(x+2)=g(x)
g(x) must be a periodic function with period 2
g(x)=⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪|x1|12;0x2|x3|12;2x4|x5|12;4x6andsoon...
Now, f(mx+hy)=(f(x))m(f(y))n x,y,m,nR
for m=n=0
f(0)=1; for m=n=1;f(x+y)=f(x).f(y)
f(a)=limh0f(x+h)f(x)h=limh0f(x).f(h)f(x)h
=limh0f(x)[f(h)1h]=limh0f(x)[f(h)f(0)h]
(f(0)=1)
=f(x)f(0)=f(x)(ln2.f(0))=(ln2)(f(x))
f(x)f(x)=ln2lnf(x)+C=(ln2)xlnf(x)+xln2=Cln2x.f(x)=C
for x=0,ln1=CC=0
ln2xf(x)=02xf(x)=1f(x)=2x
Ψ(x)=(f(f(x)))=2g(x)=212|x1|=2.2|x1|
a+baΨ(x)dx=a+ba2g(x)dx
a+baΨ(x)dx has same value '' aR as Ψ(x) is periodic with period 2.
b=2

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