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),`
`h(a)=8" and "h(0)=2," then "h(2)=`

Options

A. 1
B. 2
C. 3
D. none of these

Answer

D

Explanation

We have, `h'(x)=[f(x)]^(2)+[g(x)]^(2)`
`implies" "h''(x)=2f(x)f'(x)+2g(x)g'(x)`
`implies" "h''(x)=2f(x)g(x)+2g(x)f''(x)" "[{:(becauseg(x)-f'(x)),( :.g'(x)=f'(x)):}]`
`implies" "h''(x)=2f(x)g(x)+2g(x)(-f(x))" "[becausef''(x)=-f(x)]`
`implies" "h''(x)=0`
`implies" "h'(x)=c,a" constant for all "x inR.`
`implies" "h(x)=cx+c_(1)`
`implies" "h(0)=c_(1)" and "h(1)=c+c_(1)`
`implies" "2=c_(1)" and "8=c+c_(1)`
`implies" "c_(1)=2" and "c=6`
`:." "h(x)=6x+2impliesh(2)=6xx2+2=14`