1hm...no tukka mrunal [3][3][3].......give the solution.....[1][1][1]
f(x)=x^{2}+1 ,x<3
f(x)=5,x=3.
f(x)=1-x^{2} ,x>3.
g(x)=logx,0<x≤e
g(x)=3 x>e. and h(x)=(fog)x-(gof)x.......
then,h(1)+h'(1) is equal to......
a.log2 b)log(1/2) c)log(2/e) d)log(2e)...
here log is to the base 'e'.
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9 Answers
1f(h(x)=[ln(x)]^2+1 -ln(x^2+1)
for othr two itz will b constnt
so nw
h(1) i gt as 1-ln2
h'(1)=-1
so i gt dat answer :P :P
1hmm....der should be some mistake in my way of doing.....not getting -1..............
1dhundhkar nikhal diya.......simple mistake yaar......never thought that would come between solving..........[2][2][2]....but nw i got -1.......so [1][1][1]..