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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Limits

NDA 2023 · Mathematics · Limits

Let f(x) a^ x-1 b^ x-1 and g(x)=x-1 . What is _ x arrow 1...

Let f(x)ax−1bx−1f(x) \quad \frac{a^{x-1} b^{x-1}}{} and g(x)=x−1g(x)=x-1. What is lim⁡x→1f(x)−1g(x)\lim _{x \rightarrow 1} \frac{f(x)-1}{g(x)} equal to?

Options

  1. Aln⁡(ab)4\frac{\ln (a b)}{4}
  2. Bln⁡(ab)2\frac{\ln (a b)}{2}
  3. Cln⁡(ab)\ln (a b)
  4. D2ln⁡(ab)2 \ln (a b)

Correct answer

B. ln⁡(ab)2\frac{\ln (a b)}{2}

Step-by-step solution

Since, \lim _{x \rightarrow 1} \frac{f(x)-1}{g(x)}=lim⁡x→1ax−1+bx−12−1x−1=lim⁡x→1ax−1+bx−1−22(x−1)=lim⁡x→1(ax−1−12(x−1)+bx−1−12(x−1))=12ln⁡a+12ln⁡(b)=12ln⁡(ab)\begin{aligned} & =\lim _{x \rightarrow 1} \frac{\frac{a^{x-1}+b^{x-1}}{2}-1}{x-1}=\lim _{x \rightarrow 1} \frac{a^{x-1}+b^{x-1}-2}{2(x-1)} \\ & =\lim _{x \rightarrow 1}\left(\frac{a^{x-1}-1}{2(x-1)}+\frac{b^{x-1}-1}{2(x-1)}\right)=\frac{1}{2} \ln a+\frac{1}{2} \ln (b) \\ & =\frac{1}{2} \ln (a b) \end{aligned}
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