JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 3)
The number of ordered pairs (r, k) for which
6.35Cr = (k2 - 3). 36Cr + 1, where k is an integer, is :
6.35Cr = (k2 - 3). 36Cr + 1, where k is an integer, is :
6
3
2
4
Explanation
6.35Cr
= (k2 - 3). 36Cr + 1
$$ \Rightarrow $$ 6.35Cr = (k2 - 3).$${{36} \over {r + 1}}$$35Cr
$$ \Rightarrow $$ k2 - 3 = $${{r + 1} \over 6}$$
Possible values of r for integral values of k, are
r = 5, 35
number of ordered pairs are 4
(5, 2), (5, –2), (35, 3), (35, 3)
$$ \Rightarrow $$ 6.35Cr = (k2 - 3).$${{36} \over {r + 1}}$$35Cr
$$ \Rightarrow $$ k2 - 3 = $${{r + 1} \over 6}$$
Possible values of r for integral values of k, are
r = 5, 35
number of ordered pairs are 4
(5, 2), (5, –2), (35, 3), (35, 3)
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