Given h′(x)=[f(x)]2+[g(x)]2
Differentiating both side w.r.t x, we get
h′′(x)=2f(x)f′(x)+2g(x)g′(x)=2f(x)g(x)+2g(x)g′′(x)[∵f′(x)=g(x)]=2f(x)g(x)−2g(x)f(x)=0[∵f′′(x)=−f(x)]
Thus h′(x)=k, a constant for all x∈R.
Hence h(x)=ax+b, so that form h(0)=4, we get b=4
and from h(1)=6 we get a=2
Therefore h(4)=12