f-x-g-x-and-g-x-f-x-for-all-real-x-andf-5-2-f-5-then-f-2-10-g-2-10-is-a-2-b-4-c-8-d-none-

Question Number 40052 by LXZ last updated on 15/Jul/18

f^{'} (x) = g (x) a n d g^{'} (x) = - f (x) f o r

a l l r e a l x a n d f (5) = 2 = f^{'} (5) t h e n

f^{2} (10) + g^{2} (10) i s

(a) 2 (b) 4 (c) 8 (d) n o n e

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18

l e t f (x) = p a n d g (x) = q

\frac{d p}{d x} = q a n d \frac{d q}{d x} = - p

\frac{d^{2} p}{{d x}^{2}} = \frac{d q}{d x}

\frac{d^{2} p}{{d x}^{2}} = - p s o \frac{d^{2} p}{{d x}^{2}} + p = 0

l e t p = {A e}^{α x} i s a s o l u t i o n

\frac{d p}{d x} = A α e^{α x} \frac{d^{2} p}{{d x}^{2}} = A α^{2} e^{α x}

A α^{2} e^{α x} + {A e}^{α x} = 0

{A e}^{α x} (α^{2} + 1) = 0

α^{2} + 1 = 0 α = \pm i

s o p = c_{1} e^{i x} + c_{2} e^{- i x}

\frac{d p}{d x} = c_{1} {i e}^{i x} - c_{2} {i e}^{- i x}

f (x) = c_{1} e^{i x} + c_{2} e^{- i x}

f (x) = c_{1} (c o s x + i s i n x) + c_{2} (c o s x - i s i n x)

= c o s x (c_{1} + c_{2}) + s i n x ({i c}_{1} - {i c}_{2})

= R s i n θ c o s x + R c o s θ s i n x {c_{1} + c_{2} = R s i n θ}

= R s i n (x + θ) {{i c}_{1} - {i c}_{2} = R c o s θ}

s o f (x) = R s i n (x + θ)

g (x) = R c o s (x + θ)

f (5) = R s i n (5 + θ) = 2

f^{'} (x) = R c o s (x + θ)

f^{'} (5) = R c o s (5 + θ) = 2

R^{2} {s i n}^{2} (5 + θ) + R^{2} {c o s}^{2} (5 + θ) = 8

R = 2 \sqrt{2}

f^{2} (10) + g^{2} (10)

= R^{2} {s i n}^{2} (10 + θ) + R^{2} {c o s}^{2} (10 + θ)

= R^{2} = 8

A N S i s C = 8

Commented by LXZ last updated on 15/Jul/18

t h a n k s s i r

Answered by ajfour last updated on 16/Jul/18

f^{″} (x) = g^{'} (x) = - f (x)

\Rightarrow f (x) = A \sin x + B \cos x

g (x) = f^{'} (x) = A \cos x - B \sin x

f^{2} (10) + g^{2} (10) = A^{2} + B^{2} = f^{2} (5) + g^{2} (5)

= 4 + 4 = 8 .

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