Let a , n ∈ N such that a ≥ n 3 then 3 a + 1 − 3 a is always

Let f ( x ) = x 1/3 ⇒ f ′ ( x ) = 3 x 2/3 1 , applying LMVT in [ a , a + 1 ] , we get one c ∈ ( a , a + 1 ) f ′ ( c ) = a + 1 − a f ( a + 1 ) − f ( a ) ⇒ 3 a + 1 − 3 a = 3 c 2/3 1 < 3 a 2/3 1 ≤ 3 n 2 1 ⇒ 3 a + 1 − 3 a < 3 n 2 1 ∀ a ≥ n 3

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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMonotonicity-LMVTMedium2 minQB

MathematicsMediumsingle choice

Let $a, n \in N$ such that $a \geq n^{3}$ then $3 a + 1 - 3 a$ is always

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Let $f (x) = x^{1/3} \Rightarrow f^{'} (x) = \frac{1}{3 x ^{2/3}}$ , applying LMVT in $[a, a + 1]$ , we get one $c \in (a, a + 1)$ $f^{'} (c) = \frac{f ( a + 1 ) - f ( a )}{a + 1 - a} \Rightarrow 3 a + 1 - 3 a = \frac{1}{3 c ^{2/3}} < \frac{1}{3 a ^{2/3}} \leq \frac{1}{3 n ^{2}} \Rightarrow 3 a + 1 - 3 a < \frac{1}{3 n ^{2}} \forall a \geq n^{3}$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Monotonicity-LMVT

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