Calculus: Limits, continuity and differentiability. Maxima and minima. Mean value theorem. Integration
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2. The function f(x) = x3 - 6x2 + 9x + 25 has
a maxima at x= 1 and a minima at x = 3
a maxima at x = 3 and a minima at x = 1
no maxima, but a minima at x = 1
a maxima at x = 1, but no minima
f '(x) = 3(x2-4x+3) f ''(x)=6(x)-12 at x=1, f "(1) at x=3, f ''(3) = +ve
3. The value of a =
0 - 1 + 100 - 10 + 1
4. The interval in which the Lagrange's theorem is applicable for the function f(x) = 1/x is
5. If f(x) = | x | , then for interval [-1, 1] ,f(x)
satisfied all the conditions of Rolle's Theorem
satisfied all the conditions of Mean Value Theorem
does not satisfied the -conditions of Mean Value Theorem
None of these
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