If 40 5 m + 3 − [ 40 5 m + 3 ] = λ ( m ∈ N , m ≥ 3 ) and [ ] denote the G.I.F., then λ can take

40 5 m + 3 = 10 1 ( 5 + 5 2 + 5 3 + … + 5 m − 1 + 2 ) ⇒ λ = 5 1 , 10 7

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If $\frac{5 ^{m} + 3}{40} - [\frac{5 ^{m} + 3}{40}] = λ (m \in N, m \geq 3)$ and [ ] denote the G.I.F., then $λ$ can take

Answer:

Solution:

$\frac{5 ^{m} + 3}{40} = \frac{1}{10} (5 + 5^{2} + 5^{3} + \dots + 5^{m - 1} + 2) \Rightarrow λ = \frac{1}{5}, \frac{7}{10}$

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Domain-Range

⏱ 2mℹ️ Source: QB

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