Question

# For any real parameter a, let L(a)=Ltx→0(1+x)3x−(1+ax)1xx. Then which of the following statement(s) is/are true?

A
L(a) is continuous aϵR
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B

L(a) is continuous everywhere except a = 0, 1

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C
L(2)=e2
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D
L(12)=e

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Solution

## The correct option is C L(2)=e2Let y1=(1+x)axandy2=(1+ax)1x Then Ltx→0y1=Ltx→0y2=ea .....(i) Now by L’Hospitals rule L(a)=Ltx→0y′1−y′21 ....(ii) Hence we need to take derivatives of the functions and then apply limit We have, lny1=axln(1+x). Taking derivative, 1y1y′1=ax(1+x)−ax2ln(1+x) ...(iii) Similarly, lny2=1xln(1+ax) ⇒1y2y′2=ax(1+ax)−1x2ln(1+ax) ....(iv) Applying (iii) and (iv) in (ii) L(a)=Ltx→0y1=[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)=eaLtx→0a(a−1)x(1+x)(1+ax)−1x2(ax−ax22+……−ax+a2x22) =ea[a(a−1)−Ltx→01x2(a2−a)x22] =ea[a(a−1)−Ltx→0−a(a−1)2]=a(a−1)2ea Hence, L(a) is continuous ∀aϵR L(2)=e2 L18=−18√e

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