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JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1
ff is a real-valued differentiable and " aa " is a real constant. Further f′f^{\prime} is continuous. If

F(x)=∣f(x+a)f(x+2a)f(x+3a)f(a)f(2a)f(3a)f′(a)f′(2a)f′(3a)∣F(x)=\left|\begin{array}{ccc} f(x+a) & f(x+2 a) & f(x+3 a) \\ f(a) & f(2 a) & f(3 a) \\ f^{\prime}(a) & f^{\prime}(2 a) & f^{\prime}(3 a) \end{array}\right|

then lim⁡x→0F(x)/x\lim \limits_{{x \rightarrow 0}} F(x) / x is
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Correct option: D
0
Option D (0) is correct because the limit evaluates to 0 after applying L'Hôpital's rule and using the properties of determinants.

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