A calculus problem by Atul Kumar Ashish
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 , then find the value of h ( 1 0 ) if h ( 5 ) = 1 1 .
The answer is 11.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
If f ′ ′ ( x ) = − f ( x ) , then f ′ ′ ( x ) + f ( x ) = 0 ⇒ f ( x ) = A c o s ( x ) + B s i n ( x ) and g ( x ) = f ′ ( x ) = − A s i n ( x ) + B c o s ( x ) . If h ( x ) = [ f ( x ) ] 2 + [ g ( x ) ] 2 , then:
h ( x ) = [ A c o s ( x ) + B s i n ( x ) ] 2 + [ − A s i n ( x ) + B c o s ( x ) ] 2 ;
or h ( x ) = [ A 2 c o s 2 ( x ) + 2 A B c o s ( x ) s i n ( x ) + B 2 s i n 2 ( x ) ] + [ A 2 s i n 2 ( x ) − 2 A B c o s ( x ) s i n ( x ) + B 2 c o s 2 ( x ) ] ;
or h ( x ) = ( A 2 + B 2 ) [ c o s 2 ( x ) + s i n 2 ( x ) ] ;
or h ( x ) = A 2 + B 2 = c o n s t a n t .
If h ( 5 ) = 1 1 , then h ( 1 0 ) = 1 1 as well.
h ( x ) = [ f ( x ) ] 2 + [ g ( x ) ] 2 h ′ ( x ) = 2 f ( x ) f ′ ( x ) + 2 g ( x ) g ′ ( x )
From given information f ′ ( x ) = g ( x ) ⟹ f ′ ′ ( x ) = g ′ ( x ) = − f ( x )
Putting this
h ′ ( x ) = 2 f ( x ) g ( x ) − 2 g ( x ) f ( x ) = 0 .So, h ( x ) is a constant function.
Therefore, h ( 1 0 ) = h ( 5 ) = 1 1