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JEE Main 201910 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2019 · 10 January · Evening Shift · Mathematics · Determinants

Let a _ 1 , a _ 2 , a _ 3 , a _ 10 be in G . P . with a _ i > 0 for...

Let a1,a2,a3…,a10{a}_{1}, {a}_{2}, {a}_{3} \ldots , {a}_{10} be in G.P.G . P .with ai>0{a}_{i} > 0 for i=1,2,…,10andSi = 1, 2, \ldots , 10 and S be the set of pairs \left(r, k\right), r, k \in N(the set of natural numbers) for which<br /><br />\left|\begin{cases}\log_{e} {a}_{1}^{r}{a}_{2}^{k} & \log_{e} {a}_{2}^{r}{a}_{3}^{k} & \log_{e} {a}_{3}^{r}{a}_{4}^{k} \\ \log_{e} {a}_{4}^{r}{a}_{5}^{k} & \log_{e} {a}_{5}^{r}{a}_{6}^{k} & \log_{e} {a}_{6}^{r}{a}_{7}^{k} \\ \log_{e} {a}_{7}^{r}{a}_{8}^{k} & \log_{e} {a}_{8}^{r}{a}_{9}^{k} & \log_{e} {a}_{9}^{r}{a}_{10}^{k}\end{cases}\right| = 0

Then the number of elements in S,S, is:

Options

  1. AInfinitely many
  2. B44
  3. C1010
  4. D22

Correct answer

A. Infinitely many

Step-by-step solution

{C}_{3} \to {C}_{3} - {C}_{2}, {C}_{2} \to {C}_{2} - {C}_{1} (LetLet\alphaiscommonratioofis common ratio ofGP )<br/><br/> \left|\begin{cases}\log_{e}{a}_{1}^{r}{a}_{2}^{k} & \log_{e}\left({\alpha}^{r + k}\right) & \log_{e}\left({\alpha}^{r + k}\right) \\ \log_{e}{a}_{4}^{r}{a}_{5}^{k} & \log_{e}\left({\alpha}^{r + k}\right) & \log_{e} \left({\alpha}^{r + k}\right) \\ \log_{e}{a}_{7}^{r}{a}_{8}^{k} & \log_{e} \left({\alpha}^{r + k}\right) & \log_{e} \left({\alpha}^{r + k}\right)\end{cases}\right| = 0 which is always true because C2 and  {C}_{2} \text{\text{ and } } {C}_{3} are identical.
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