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f'' = -f, g = f', h = f^2 + g^2: given h at one point, find h at another

The same question, asked 5 times: 2 May 2023 S2*, 3 May 2023 S1*, 11 May 2023 S2*, 16 Apr 2026 S2*, 19 May 2026 S1. Every wording below, oldest first.

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  1. Question 1

    2 May 2023 S2

    143
    If F(x)=(f ⁣(x2))2+(g ⁣(x2))2F(x) = \left(f\!\left(\frac{x}{2}\right)\right)^2+\left(g\!\left(\frac{x}{2}\right)\right)^2, where f′′(x)=−f(x)f''(x) = -f(x) and g(x)=f′(x)g(x) = f'(x), and F(5)=5F(5) = 5, then F(10)F(10) is equal to

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  2. Question 2

    3 May 2023 S1

    111
    Let f be a twice differentiable function such that f′′(x)=−f(x)f''(x)=-f(x), f′(x)=g(x)f'(x)=g(x) and h(x)=[f(x)]2+[g(x)]2h(x)=[f(x)]^2+[g(x)]^2. If h(5)=1h(5)=1, then h(10)h(10) is

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  3. Question 3

    11 May 2023 S2

    144
    Let f be twice differentiable function such that f′′(x)=−f(x)f''(x)=-f(x), f′(x)=g(x)f'(x)=g(x) and h(x)=(f(x))2+(g(x))2h(x)=(f(x))^2+(g(x))^2. If h(5)=1h(5)=1, then the value of h(10)h(10) is

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  4. Question 4

    16 Apr 2026 S2

    71
    Let f(x) be a twice differentiable function such that f′′(x)=−f(x)f''(x) = -f(x), f′(x)=g(x)f'(x) = g(x) and h(x)={f(x)}2+{g(x)}2h(x) = \{f(x)\}^2 + \{g(x)\}^2. If h(5)=11h(5) = 11, then h(10)h(10) is equal to ...

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  5. Question 5

    19 May 2026 S1

    75
    Let f(x)f(x) be a twice differentiable function such that f′′(x)=−f(x)f''(x) = -f(x), and f′(x)=g(x)f'(x) = g(x). Let h(x)={f(x)}2+{g(x)}2h(x) = \{f(x)\}^2 + \{g(x)\}^2. If h(2)=7h(2) = 7, then h(8)h(8) is equal to:

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