Set, Relation and FunctionEasy
Question
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 where h(2) = 2. Then the value of h(1) is
Options
A.1
B.2
C.1/2
D.5
Solution
f″(x) = - f(x)
Multiplying both sides by f′(x), we get
f′(x).f″(x) = - f(x).f′(x).
Integrating
⇒ (f(x))2 + (f′(x))2 = k ⇒ (f(x))2 + (g(x))2 = k
⇒ h(x) = k
∵ h(2) = 2 ⇒ h(x) = 2 ∀ x ⇒ h(1) = 2
Multiplying both sides by f′(x), we get
f′(x).f″(x) = - f(x).f′(x).
Integrating
⇒ (f(x))2 + (f′(x))2 = k ⇒ (f(x))2 + (g(x))2 = k
⇒ h(x) = k
∵ h(2) = 2 ⇒ h(x) = 2 ∀ x ⇒ h(1) = 2
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