﻿ Let f be a twice differentiable function such that f’’(x) = –f(x) and f’(x) = g(x). If h’(x) = [f(x)]2 + [g(x)]2, h(1) = 8 and  h(0) = 2, then h(2) is equal to : Kaysons Education

# Let f be A Twice Differentiable Function Such That f’’(x) = –f(x) And f’(x) = G(x). If h’(x) = [f(x)]2 + [g(x)]2, h(1) = 8 And  H(0) = 2, Then h(2) Is Equal To

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## Question

### Solution

Correct option is

None of these

Thus h’(x) = k, a constant, for all x Ïµ R. Hence h(x) = kx + m, so that formh(0) = 2, we get m = 2 and from h(1) = 8, we get k = 6. Therefore, h(2) = 14.

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