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


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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
	$let f(x)=p and g(x)= q (dp/dx)=q and (dq/dx)=−p (d^2 p/dx^2 )=(dq/dx) (d^2 p/dx^2 )=−p so (d^2 p/dx^2 )+p=0 let p=Ae^(αx) is a solution (dp/dx)=Aαe^(αx) (d^2 p/dx^2 )=Aα^2 e^(αx) Aα^2 e^(αx) +Ae^(αx) =0 Ae^(αx) (α^2 +1)=0 α^2 +1=0 α=±i so p=c_1 e^(ix) +c_2 e^(−ix) (dp/dx)=c_1 ie^(ix) −c_2 ie^(−ix) f(x)=c_1 e^(ix) +c_2 e^(−ix) f(x)=c_1 (cosx+isinx)+c_2 (cosx−isinx) =cosx(c_1 +c_2 )+sinx(ic_1 −ic_2 ) =Rsinθcosx+Rcosθsinx {c_1 +c_2 =Rsinθ} =Rsin(x+θ) {ic_1 −ic_2 =Rcosθ} so f(x)=Rsin(x+θ) g(x)=Rcos(x+θ) f(5)=Rsin(5+θ)=2 f′(x)=Rcos(x+θ) f′(5)=Rcos(5+θ)=2 R^2 sin^2 (5+θ)+R^2 cos^2 (5+θ)=8 R=2(√2) f^2 (10)+g^2 (10) =R^2 sin^2 (10+θ)+R^2 cos^2 (10+θ) =R^2 =8 ANS is C=8$
	$l e t f (x) = p a n d g (x) = q$ $\frac{d p}{d x} = q a n d \frac{d q}{d x} = - p$ $\frac{d^{2} p}{{d x}^{2}} = \frac{d q}{d x}$ $\frac{d^{2} p}{{d x}^{2}} = - p s o \frac{d^{2} p}{{d x}^{2}} + p = 0$ $l e t p = {A e}^{α x} i s a s o l u t i o n$ $\frac{d p}{d x} = A α e^{α x} \frac{d^{2} p}{{d x}^{2}} = A α^{2} e^{α x}$ $A α^{2} e^{α x} + {A e}^{α x} = 0$ ${A e}^{α x} (α^{2} + 1) = 0$ $α^{2} + 1 = 0 α = \pm i$ $s o p = c_{1} e^{i x} + c_{2} e^{- i x}$ $\frac{d p}{d x} = c_{1} {i e}^{i x} - c_{2} {i e}^{- i x}$ $f (x) = c_{1} e^{i x} + c_{2} e^{- i x}$ $f (x) = c_{1} (c o s x + i s i n x) + c_{2} (c o s x - i s i n x)$ $= c o s x (c_{1} + c_{2}) + s i n x ({i c}_{1} - {i c}_{2})$ $= R s i n θ c o s x + R c o s θ s i n x {c_{1} + c_{2} = R s i n θ}$ $= R s i n (x + θ) {{i c}_{1} - {i c}_{2} = R c o s θ}$ $s o f (x) = R s i n (x + θ)$ $g (x) = R c o s (x + θ)$ $f (5) = R s i n (5 + θ) = 2$ $f^{'} (x) = R c o s (x + θ)$ $f^{'} (5) = R c o s (5 + θ) = 2$ $R^{2} {s i n}^{2} (5 + θ) + R^{2} {c o s}^{2} (5 + θ) = 8$ $R = 2 \sqrt{2}$ $f^{2} (10) + g^{2} (10)$ $= R^{2} {s i n}^{2} (10 + θ) + R^{2} {c o s}^{2} (10 + θ)$ $= R^{2} = 8$ $A N S i s C = 8$
	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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