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AIEEE 2006

JEE Main 2006 Mathematics question on Limits Continuity and Differentiability

If f′′(x)=−f(x)f''(x) = -f(x), where f(x)f(x) is a continuous double differentiable function and g(x)=f′(x)g(x) = f'(x).

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 and F(5)=5F(5) = 5,

then F(10)F(10) is
  1. A.
    0
  2. B.
    5
  3. C.
    10
  4. D.
    25
Show answer and solution

Correct option: B

Solution

F'(x) = 2f(x/2)·f'(x/2)·(1/2) + 2g(x/2)·g'(x/2)·(1/2)

= f(x/2)·g(x/2) + g(x/2)·f''(x/2)·(1/2)... let me redo carefully.

F'(x) = 2f(x/2)·(1/2)f'(x/2) + 2g(x/2)·(1/2)g'(x/2)

= f(x/2)·g(x/2) + g(x/2)·g'(x/2)

Now g'(x/2) = f''(x/2) = -f(x/2), so:

F'(x) = f(x/2)·g(x/2) + g(x/2)·(-f(x/2)) = 0

Since F'(x) = 0 for all x, F is a constant function. Therefore F(10) = F(5) = 5.

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