Let f : N -> R be a function such that f(x + y) = 2f(x) f(y) for natural number x and y. If f(1) = 2, then the value of a for which
holds is:
Let f : N -> R be a function such that f(x + y) = 2f(x) f(y) for natural number x and y. If f(1) = 2, then the value of a for which holds is:
Option 1 -
2
Option 2 -
3
Option 3 -
4
Option 4 -
6
-
1 Answer
-
Correct Option - 3
Detailed Solution:=4
Similar Questions for you
f (x) = 2e²? / (e²? +e? ) and f (1-x) = 2e²? / (e²? +e¹? )
∴ f (x) + f (1-x) = 1/2
i.e. f (x) + f (1-x) = 2
∴ f (1/100) + f (2/100) + . + f (99/100)
Σ? f (x/100) + f (1-x/100) + f (1/2)
= 49 x 2 + 1 = 99
lim? (x→7) (18- [1-x])/ ( [x-3a])
exist & a∈I.
= lim? (x→7) (17- [-x])/ ( [x]-3a)
exist
RHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 25/ (7-3a) [a ≠ 7/3]
LHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 24/ (6-3a) [a ≠ 2]
LHL = RHL
25/ (7-3a) = 8/ (2-a)
∴ a = -6
A function f (x) is continuous at x=1, so lim (x→1? ) f (x) = lim (x→1? ) f (x) = f (1).
Assuming a piecewise function like f (x) = { -x, x<1; ax+b, x≥1 } (structure inferred from derivative).
Continuity at x=1: f (1) = 1. a (1)+b = 1 => a+b=1.
The function is differentiable at x=1. The derivative of f (x) at x=1 from the left is -1. The derivative from the right is a.
So, a = -1. (The image has 2a = -1, which would imply a function like -x and ax²+b). Let's assume f' (x) = 2a for x>1.
2a = -1 => a = -1/2.
From a+b=1, b = 1 - a = 1 - (-1/2) = 3/2.
So, a = -1/2 and b = 3/2.
Given the function f (x) = cosec? ¹ (x) / √ {x - [x]} where [x] is the greatest integer function.
The domain of cosec? ¹ (x) is (-∞, -1] U [1, ∞).
For the denominator to be defined, x - [x] ≠ 0, which means {x} ≠ 0 (the fractional part of x is not zero). This implies that x cannot be an integer (x ∉ I).
Combining these conditions, the domain is all non-integer numbers except for those in the interval (-1, 1).
y = log10 x ×
Now,
So, (x, y) = (106, 9)
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