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now, we can see that there is atleast one point in (1,2) where curve is cutting x axis..

this means (1+cos8x)(ax2+bx+c) = 0 for some x$\epsilon$(1,2)....., but,

(1+cos8x)$\neq$0 , since 1+cos8x>=1 $\forall$ x , therefore

(ax2+bx+c)=0 for some x$\epsilon$(1,2)...

hence (ax2+bx+c) has atleast one root in (0,2)..

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