DifferentiationHard
Question
If f is twice differentiable such that f¢¢(x) = - f(x) and f′(x) = g(x). If h(x) is a twice differentiable function such that h′(x) = (f(x))2 + (g(x))2 . If h(0) = 2, h(1) = 4, then the equation y = h(x) represents
Options
A.a curve of degree 2
B.a curve passing through the origin
C.a straight line with slope 2
D.a straight line with y intercept equal to 2.
Solution
f″(x) = - f(x) .... (i)
f′(x) = g(x) .... (ii)
h′(x) = (f(x))2 + (g(x))2 .... (iii)
h(0) = 2, h(1) = 4
Differentiating equation (ii) w.r.t. x
f″(x) = g′(x) = - f(x)
Differentiating equation (iii) w.r.t. x
h″(x) = 2f(x) . f′(x) + 2 g(x) . g′(x)
= 2f(x) . f′(x) - 2f′(x) . f(x) = 0 {∵ g′(x) = - f(x)}
h′(x) is constant
⇒ h(x) is linear function
∴ h(0) = 2 ⇒ h(x) not passing through (0, 0)
Let y = h(x) = ax + b
at x = 0
y = 2 = b ⇒ y = ax + 2
at x = 1
a + 2 = 4
a = 2
⇒ curve is y = 2x + 2
f′(x) = g(x) .... (ii)
h′(x) = (f(x))2 + (g(x))2 .... (iii)
h(0) = 2, h(1) = 4
Differentiating equation (ii) w.r.t. x
f″(x) = g′(x) = - f(x)
Differentiating equation (iii) w.r.t. x
h″(x) = 2f(x) . f′(x) + 2 g(x) . g′(x)
= 2f(x) . f′(x) - 2f′(x) . f(x) = 0 {∵ g′(x) = - f(x)}
h′(x) is constant
⇒ h(x) is linear function
∴ h(0) = 2 ⇒ h(x) not passing through (0, 0)
Let y = h(x) = ax + b
at x = 0
y = 2 = b ⇒ y = ax + 2
at x = 1
a + 2 = 4
a = 2
⇒ curve is y = 2x + 2
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