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Q.

Let ⨍  be twice differentiable function such that

⨍ 11x=-⨍ x and ⨍ 1x=gx,

 hx=⨍ x2+ gx2. If h5=11 then h10= 

 

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a

11

b

22

c

6

d

30

answer is B.

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Detailed Solution

f11x=-fx f1x=gx hx=fx2+gx2 h1x=2fx f1x+2g(x) g1(x)          =2f(x) g(x)+2g(x) -f(x)          =0 h(x) =constant since h(5)=11     h(10)=11

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