Question:
Consider a function ƒ satisfying ƒ(x + y) = ƒ(x)ƒ(y) where x,y are positive integers , and ƒ(1) = 2 . If ƒ(a+1) + ƒ(a+2) + ... + ƒ(a+n) = 16(2n - 1) then a is equal to
Consider a function ƒ satisfying ƒ(x + y) = ƒ(x)ƒ(y) where x,y are positive integers , and ƒ(1) = 2 . If ƒ(a+1) + ƒ(a+2) + ... + ƒ(a+n) = 16(2n - 1) then a is equal to
Updated On: Aug 20, 2024
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The Correct Option is C
Solution and Explanation
ƒ(a+1) + ƒ(a+2) + ... + ƒ(a+n) = 16(2n - 1)
⇒ ƒ(a)ƒ(1) + ƒ(a)(2) + ... + ƒ(a)ƒ(n) = 16(2n - 1)
⇒ ƒ(a)ƒ(1) + ƒ(2) + ... + ƒ(n) = 16(2n - 1)
By substituting n=1,
⇒ ƒ(a)ƒ(1) = 16( 21 - 1) = 16
⇒ ƒ(a)×2 = 16 ⇒ ƒ(a) = 8
Therefore ,
ƒ(a)(ƒ(1) + ƒ(2) + ... + ƒ(n)) = 16(2n - 1)
⇒ƒ(1) + ƒ(2) + ... + ƒ(n) = 2(2n - 1)
If n=2 , then ƒ(1) + ƒ(2) = 2(22 - 1) = 6
⇒ ƒ(2) = 6 - ƒ(1) = 6 - 2 = 4
If n=3 , then ƒ(1) + ƒ(2) + ƒ(3) = 2(23 - 1) = 14
⇒ ƒ(3) = 6 - ƒ(1) - ƒ(2) = 14 - 2 - 4 = 8 = ƒ(a)
Therefore, a=3.
⇒ ƒ(a)ƒ(1) + ƒ(a)(2) + ... + ƒ(a)ƒ(n) = 16(2n - 1)
⇒ ƒ(a)ƒ(1) + ƒ(2) + ... + ƒ(n) = 16(2n - 1)
By substituting n=1,
⇒ ƒ(a)ƒ(1) = 16( 21 - 1) = 16
⇒ ƒ(a)×2 = 16 ⇒ ƒ(a) = 8
Therefore ,
ƒ(a)(ƒ(1) + ƒ(2) + ... + ƒ(n)) = 16(2n - 1)
⇒ƒ(1) + ƒ(2) + ... + ƒ(n) = 2(2n - 1)
If n=2 , then ƒ(1) + ƒ(2) = 2(22 - 1) = 6
⇒ ƒ(2) = 6 - ƒ(1) = 6 - 2 = 4
If n=3 , then ƒ(1) + ƒ(2) + ƒ(3) = 2(23 - 1) = 14
⇒ ƒ(3) = 6 - ƒ(1) - ƒ(2) = 14 - 2 - 4 = 8 = ƒ(a)
Therefore, a=3.
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