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Question

For any real parameter a, let L(a)=Ltx0(1+x)3x(1+ax)1xx. Then which of the following statement(s) is/are true?
  1. L(a) is continuous everywhere except a = 0, 1

  2. L(a) is continuous aϵR
  3. L(2)=e2
  4. L(12)=e

     


Solution

The correct option is C L(2)=e2
Let y1=(1+x)axandy2=(1+ax)1x
Then Ltx0y1=Ltx0y2=ea   .....(i)
Now by L’Hospitals rule
L(a)=Ltx0y1y21   ....(ii)
Hence we need to take derivatives of the functions and then apply limit
We have, lny1=axln(1+x). Taking derivative,
1y1y1=ax(1+x)ax2ln(1+x)  ...(iii)
Similarly, lny2=1xln(1+ax)
1y2y2=ax(1+ax)1x2ln(1+ax) ....(iv)
Applying (iii) and (iv) in (ii)
L(a)=Ltx0y1=[ax(1+x)ax2ln(1+x)]y2[ax(1+ax)1x2ln(1+ax)]
Using (i) as both limits on y1 and y2 are finite and equal, we can factor it out. Hence,
L(a)=eaLtx0a(a1)x(1+x)(1+ax)1x2(axax22+ax+a2x22)
=ea[a(a1)Ltx01x2(a2a)x22]
=ea[a(a1)Ltx0a(a1)2]=a(a1)2ea
Hence, L(a) is continuous aϵR
L(2)=e2
L18=18e

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