Question:

Given, the function f(x)=ax+ax2 f(x) = \frac{a^x + a^{-x}}{2} (a>2 a > 2 ), then f(x+y)+f(xy) f(x+y) + f(x-y) is equal to

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To simplify expressions involving exponential terms like ax+y a^{x+y} and axy a^{x-y} , use exponent rules to group terms efficiently.
Updated On: Mar 29, 2025
  • f(x)f(y) f(x) - f(y)
  • f(y) f(y)
  • 2f(x)f(y) 2f(x)f(y)
  • f(x)f(y) f(x)f(y)
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The Correct Option is C

Solution and Explanation

We are given the function: f(x)=ax+ax2. f(x) = \frac{a^x + a^{-x}}{2}. Now, let's calculate f(x+y) f(x+y) and f(xy) f(x-y) : f(x+y)=ax+y+a(x+y)2,f(xy)=axy+a(xy)2. f(x+y) = \frac{a^{x+y} + a^{-(x+y)}}{2}, \quad f(x-y) = \frac{a^{x-y} + a^{-(x-y)}}{2}. By adding these two expressions together: f(x+y)+f(xy)=ax+y+a(x+y)+axy+a(xy)2. f(x+y) + f(x-y) = \frac{a^{x+y} + a^{-(x+y)} + a^{x-y} + a^{-(x-y)}}{2}. Next, simplify this using the properties of exponents: f(x+y)+f(xy)=ax(ay+ay)+ax(ay+ay)2. f(x+y) + f(x-y) = \frac{a^x(a^y + a^{-y}) + a^{-x}(a^y + a^{-y})}{2}. Factor the expression: f(x+y)+f(xy)=(ax+ax)(ay+ay)2. f(x+y) + f(x-y) = \frac{(a^x + a^{-x})(a^y + a^{-y})}{2}. Now, recognizing that f(x)=ax+ax2 f(x) = \frac{a^x + a^{-x}}{2} and f(y)=ay+ay2 f(y) = \frac{a^y + a^{-y}}{2} , we substitute these back in: f(x+y)+f(xy)=2f(x)f(y). f(x+y) + f(x-y) = 2f(x)f(y). Final Answer: 2f(x)f(y) \boxed{2f(x)f(y)}
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