Set, Relations and FunctionsMTP Nov 20Question 1975 of 217
All Questions

Let f:R→R\displaystyle f: R \to R be such that f(x)=2x\displaystyle f(x) = 2^x, then f(x+y)\displaystyle f(x+y) equals

Options

Af(x)+f(y)\displaystyle f(x) + f(y)
Bf(x)⋅f(y)\displaystyle f(x) \cdot f(y)
Cf(x)÷f(y)\displaystyle f(x) \div f(y)
DNone of these
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Correct Answer

✅ Option a — f(x)+f(y)\displaystyle f(x) + f(y)

All Options:

  • Af(x)+f(y)\displaystyle f(x) + f(y)
  • Bf(x)⋅f(y)\displaystyle f(x) \cdot f(y)
  • Cf(x)÷f(y)\displaystyle f(x) \div f(y)
  • DNone of these

Detailed Solution & Explanation

Given the function:
f(x)=2xf(x) = 2^x
We want to evaluate f(x+y)\displaystyle f(x+y):
f(x+y)=2x+yf(x+y) = 2^{x+y}
Using the laws of exponents, we know that:
2x+y=2x⋅2y2^{x+y} = 2^x \cdot 2^y
Since f(x)=2x\displaystyle f(x) = 2^x and f(y)=2y\displaystyle f(y) = 2^y, we can substitute these values back into the expression:
2x⋅2y=f(x)⋅f(y)2^x \cdot 2^y = f(x) \cdot f(y)
Therefore, we have:
f(x+y)=f(x)⋅f(y)f(x+y) = f(x) \cdot f(y)

**Discrepancy & Typographical Error:**
Mathematically, the expression equals f(x)⋅f(y)\displaystyle f(x) \cdot f(y), which corresponds to **Option B**. However, the textbook answer key for this specific mock test paper (MTP Nov 20) has a typographical error listing **Option A** (f(x)+f(y)\displaystyle f(x) + f(y)) as the correct answer. The mathematically correct result is indeed f(x)⋅f(y)\displaystyle f(x) \cdot f(y) (Option B), which is correctly keyed in similar questions elsewhere (e.g. MTP March 22).

Hence, **Option A** is the correct answer (according to the textbook key's typo), but the mathematically proven correct value is f(x)⋅f(y)\displaystyle f(x) \cdot f(y) (**Option B**).

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