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JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1
Let f:R→Rf:R \to R be a differentiable function satisfying f′(3)+f′(2)=0.f'(3) + f'(2) = 0. then lim⁡x→0(1+f(3+x)−f(3)1+f(2−x)−f(2))1x\lim\limits_{x \to 0} {\left( {\frac{{1 + f(3 + x) - f(3)}}{{1 + f(2 - x) - f(2)}}} \right)^{\frac{1}{x}}} is equal to
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Correct option: D
1
Option D (1) is correct because applying L'Hopital's rule and using the given condition f′(3)+f′(2)=0f'(3) + f'(2) = 0 simplifies the limit to 1.

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