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If f(x)= int(0)^(x)(f(t))^(2) dt, f:R ra...

If `f(x)= int_(0)^(x)(f(t))^(2) dt, f:R rarr R ` be differentiable function and `f(g(x))` is differentiable at `x=a`, then

A

(a)g(x) must be differentiable at x = a

B

(b)g(x) is discontinuous, then f(a) = 0

C

(c)`f(a) ne 0`, then g(x) must be differentiable

D

(d)None of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we start with the given function: \[ f(x) = \int_0^x (f(t))^2 \, dt \] ### Step 1: Differentiate \( f(x) \) Using the Fundamental Theorem of Calculus, we differentiate \( f(x) \): \[ f'(x) = (f(x))^2 \] ### Step 2: Analyze the differentiability of \( f(g(x)) \) We are given that \( f(g(x)) \) is differentiable at \( x = a \). By applying the chain rule, we have: \[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \] ### Step 3: Substitute \( f'(g(x)) \) From Step 1, we know that: \[ f'(g(x)) = (f(g(x)))^2 \] Thus, we can rewrite the derivative as: \[ \frac{d}{dx} f(g(x)) = (f(g(x)))^2 \cdot g'(x) \] ### Step 4: Evaluate at \( x = a \) Since \( f(g(x)) \) is differentiable at \( x = a \), we can evaluate the expression at \( x = a \): \[ \frac{d}{dx} f(g(x)) \bigg|_{x=a} = (f(g(a)))^2 \cdot g'(a) \] ### Step 5: Conclusion about differentiability For \( f(g(x)) \) to be differentiable at \( x = a \), both \( f(g(a)) \) and \( g'(a) \) must exist. This implies that \( g(x) \) must be differentiable at \( x = a \). ### Final Answer Thus, we conclude that \( g(x) \) must be differentiable at \( x = a \). ---
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