A function f satisfies the relation f x - y - f x f y - x - y - N- Further if f 1-3 and - r -1 n f a - r -81-23 n -1 then a -A- 1B- 2C- 3D- 4

The correct option is C 3f(n) = f(1+1+1+ . . . n times) = ={f(1)^n}=3^n Hence ∑_r=1^nf(a+r) = ∑_r=1^n3^a+r=3^a. 3(3^n 1)/3 1 =81/2(3^n 1) ∴ 3^a+1=81 ⇒ a=3 ...

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Standard XII

Mathematics

Integrating Factor

A function f ...

Question

A function f satisfies the relation f(x + y) = f(x) f(y) $\forall$ x, y $ϵ$ N. Further if f(1) = 3 and $\sum_{r = 1}^{n} f (a + r) = \frac{81}{2} (3^{n} - 1)$ then a=

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Solution

The correct option is C 3
f(n) = f(1+1+1+ . . . n times) = $= {f (1)^{n}} = 3^{n}$
Hence $\sum_{r = 1}^{n} f (a + r) = \sum_{r = 1}^{n} 3^{a + r} = 3^{a} . \frac{3 (3^{n} - 1)}{3 - 1}$
$= \frac{81}{2} (3^{n} - 1) ∴ 3^{a + 1} = 81 \Rightarrow a = 3$

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A function f satisfies the relation f(x + y) = f(x) f(y) ∀ x, y ϵ N. Further if f(1) = 3 and ∑nr=1f(a+r)=812(3n−1) then a=

The correct option is C 3f(n) = f(1+1+1+ . . . n times) = ={f(1)n}=3n Hence ∑nr=1f(a+r)=∑nr=13a+r=3a.3(3n−1)3−1 =812(3n−1)∴ 3a+1=81⇒a=3

A function f satisfies the relation f(x + y) = f(x) f(y) $\forall$ x, y $ϵ$ N. Further if f(1) = 3 and $\sum_{r = 1}^{n} f (a + r) = \frac{81}{2} (3^{n} - 1)$ then a=

The correct option is C 3
f(n) = f(1+1+1+ . . . n times) = $= {f (1)^{n}} = 3^{n}$
Hence $\sum_{r = 1}^{n} f (a + r) = \sum_{r = 1}^{n} 3^{a + r} = 3^{a} . \frac{3 (3^{n} - 1)}{3 - 1}$
$= \frac{81}{2} (3^{n} - 1) ∴ 3^{a + 1} = 81 \Rightarrow a = 3$