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JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy
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Application +4 | -1
If a function F is such that F(0)=2,β€…β€ŠF(1)=3,β€…β€ŠF(n+2)=2F(n)βˆ’F(n+1)β€…β€Šforβ€…β€Šnβ‰ 0F\left( 0 \right) = 2,\;F\left( 1 \right) = 3,\;F\left( {n + 2} \right) = 2F\left( n \right) - F\left( {n + 1} \right)\;for\;n \ne 0, then F(5)F\left( 5 \right) is equal to
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Correct option: D
13
Option D (13) is correct because by iteratively applying the given recurrence relation, we find that F(5) equals 13.

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