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Question
Let f(x) be a polynomial function of degree 6 such that `d/dx (f(x))` = (x – 1)3 (x – 3)2, then
Assertion (A): f(x) has a minimum at x = 1.
Reason (R): When `d/dx (f(x)) < 0, ∀ x ∈ (a - h, a)` and `d/dx (f(x)) > 0, ∀ x ∈ (a, a + h)`; where 'h' is an infinitesimally small positive quantity, then f(x) has a minimum at x = a, provided f(x) is continuous at x = a.
Options
Both (A) and (R) are true and (R) is the correct explanation of (A).
Both (A) and (R) are true but (R) is not the correct explanation of (A).
(A) is true but (R) is false.
(A) is false but (R) is true.
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Solution
Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation:
`d/dx (f(x))` = (x – 1)3 (x – 3)2
Assertion : f(x) has a minimum at x = 1 is true as
`d/dx (f(x)) < 0, ∀ x ∈ (1 - h, 1)` and `d/dx (f(x)) > 0, ∀ x ∈ (1, 1 + h)`; where 'h' is an infinitesimally small positive quantity, which is in accordance with the Reason statement.
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