JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

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Analyzing +4 | -1

0 < a < b < \frac{\pi }{2}\,\,{\text{and}}\,\,f\left( {a,b} \right) = \frac{{\tan b - \tan a}}{{b - a}},

then

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Correct option: B

f(a, b)>1

Option B (f(a, b)>1) is correct because by applying Lagrange's mean value theorem to f(x) = tan x, we find that f(a, b) = sec²(c) > 1.

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