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fog is differentiable at x=-1 and gof is differentiable at x=1 for is differentiable at x=-1 and gof is not differentiable at x=1fog is differentiable at x=1 and gof is differentiable at x=-1 none of these

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We have <br> `fog(x)={{:(,1+x,-2 le x lt -1),(,-x-1,-1le x lt 0),(,x-1,0 le x le 2):}` <br> `and gof(x)={{:(,2+x,-1 le x lt 0),(,2-x,0 le x lt 1),(,x,1 le x lt 2),(,4-x,2 le x le 3):}` <br> We observe that <br> `("LHD of fog(x) at x=-1")=((d)/(dx)(1+x))_(atx=-1)=1` <br> `("RHD of fog(x) at x=-1")=((d)/(dx)(-x-1))_(atx=-1)=-1` <br> Clearly, `("LHD of fog(x) at x=-1")ne ("RHD of fog(x) at x=-1")` <br> So, fog is not differentiable at x=-1 <br> Clearly, fog is a polynomial function on [0,2]. <br> So, it is differentiable at `x =1 in [0,2]` <br> Also, `("LHD of gof(x) at x=1")=((d)/(dx)(2-x))_("at x=1")=-1` <br> `("RHD of gof(x) at x=1")=((d)/(dx)(x))_("at x=1")=1` <br> Clearly, these two are not same, So, gof is not differentiable at x=1. **Theorem :-Every first degree equation in x,y represents a straight line.**

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