A function f: R → R satisfies the equation f(x + y) = f(x)f(y) for all x, y R ; f(x) ≠ 0 . Suppose that the function f(x) is differentiable at x = 0 a - Mathematics MCQ Mathematics Mock Test - 10 2026 | ExamTest.live

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A function f : R \to R satisfies the equation f (x + y) = f (x) f (y) for all x, y \in R; f (x) \neq = 0 . Suppose that the function f (x) is differentiable at x = 0 and f^{'} (0) = 2 . If f^{'} (x) = λ f (x), then the value of λ is:

Solution & Step-by-step Explanation

Given f (x + y) = f (x) f (y) .Put x = y = 0 : f (0) = f (0)^{2} . Since f (x) \neq = 0, f (0) = 1 .Definition of derivative at x : f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h} f^{'} (x) = lim_{h \to 0} \frac{f ( x ) f ( h ) - f ( x )}{h} f^{'} (x) = f (x) lim_{h \to 0} \frac{f ( h ) - 1}{h} Since f (0) = 1, the limit is the definition of f^{'} (0) : f^{'} (x) = f (x) \cdot f^{'} (0) Given f^{'} (0) = 2 : f^{'} (x) = 2 f (x) Comparing with f^{'} (x) = λ f (x), we find λ = 2 .

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A function f : R \to R satisfies the equation f (x + y) = f (x) f (y) for all x, y \in R; f (x) \neq = 0 . Suppose that the function f (x) is differentiable at x = 0 and f^{'} (0) = 2 . If f^{'} (x) = λ f (x), then the value of λ is:

1

2

1/2

Any real number

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