JEE Mains · Maths · STD 12 - 1. relation and function
Let \(\mathrm{A}=\{1,2,3, \ldots, 10\}\) and R be a relation on A such that \(\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}=2 \mathrm{~b}+1\}\). Let \(\left(\mathrm{a}_1, \mathrm{a}_2\right)\), \(\left(a_2, a_3\right),\left(a_3, a_4\right), \ldots .,\left(a_k, a_{k+1}\right)\) be a sequence of \(k\) elements of R such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k , for which such a sequence exists, is equal to :
- A \(6\)
- B \(7\)
- C \(5\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{a}=2 \mathrm{~b}+1 \\ & 2 \mathrm{~b}=\mathrm{a}-1 \\ & \mathrm{R}=\{(3,1),(5,2), \ldots,(99,49)\} \end{aligned}\) Let \((2 m+1, m),(2 \lambda-1, \lambda)\) are such ordered pairs. According to the condition…
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