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Rolle's Theorem is applicable in the interval [-5, 5] for the function1. \(\rm f (x) = x^4\)2. \(\rm f (x) = 5x^4\)3. \(\rm f (x) = 2x^3 + 3\)4. \(\rm f(x) = \pi |x|\) |
Answer» Correct Answer - Option 2 : \(\rm f (x) = 5x^4\) Concept: If real valued function f (x) (i) is continuous in [a, b] (ii) is differentiable on (a, b) (iii) f (a) = f(b) Then there exists at least one real value c in the interval (a, b) such that f'(c) = 0 Calculation: If we take\(\rm f (x) = 5x^4\) (i) f (x) is continuous in (-5, 5) (ii) f (x) is differentiable in (-5, 5) (iii) f (-5) = f (5) So,\(\rm f (x) = 5x^4\)satisfies all the conditions of Rolle's Theorem, therefore a point c, such that f (c') = 0 20\(\rm c^3\)= 0 = c = 0\(\rm \epsilon \)(-5, 5) Hence Option 2 is correct. |
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