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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

If f (x) = ((2 + x)/(1 + x)) ^ 1 + x , then f ^ ′ (0) is equal to

If ff\left(x\right) = {\left(\dfrac{2 + x}{1 + x}\right)}^{1 + x}, then f′(0){f}^{′} \left(0\right) is equal to

Options

  1. A2log⁡22\log 2
  2. Blog⁡2\log 2
  3. C3log⁡2−13\log 2 - 1
  4. D2log⁡2−12\log 2 - 1

Correct answer

D. 2log⁡2−12\log 2 - 1

Step-by-step solution

By taking logarithm on both sides,
logf(x)=(1+x)[log⁡(2+x)−log⁡(1+x)]logf ( x ) = ( 1 + x ) \left[\log ( 2 + x ) - \log ( 1 + x )\right]
On differentiating, we get,

1f(x)⋅f′(x)=(1+x)(2+x)+log⁡(2+x)−(1+x)(1+x)−log⁡(1+x)\dfrac{1}{f\left(x\right)} \cdot {f}^{'}\left(x\right) = \dfrac{\left(1 + x\right)}{\left(2 + x\right)} + \log ( 2 + x ) - \dfrac{\left(1 + x\right)}{\left(1 + x\right)} - \log ( 1 + x )

Now, put x=0x = 0

f′(0)=f(0){1+log⁡22−1}=f(0)[log⁡2−12]{f}^{'} ( 0) = f\left(0\right)\left\{\dfrac{1 + \log 2}{2} - 1\right\} = f\left(0\right)\left[\log 2 - \dfrac{1}{2}\right]

= 2\log 2 - 1\left(∵ f\left(0\right) = 2\right)
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