find solution of the following question of pgt mathematics question paper

Question : Find solution of the following question of pgt mathematics question paperSolution :From continuity \[f\left(1\right)=-\frac{1}{\left|1\right|}=-1\] \[limit_{h\rightarrow0}f\left(1+h\right)=limit_{h\rightarrow0}-\frac{1}{\left|1+h\right|}=-1\] \[limit_{h\rightarrow0}f\left(1-h\right)=limit_{h\rightarrow0}a\left(1-h\right)^2-b =a-b\] f is continuous at x =1 if \[a-b=-1\] Now from differentiability \[limit_{h\rightarrow0}\frac{f\left(1+h\right)-f(1)}{h}=limit_{h\rightarrow0}\frac{f\left(1-h\right)-f(1)}{-h}\] \[limit_{h\rightarrow0}\frac{-\frac{1}{\left|1+h\right|}+1}{h}=limit_{h\rightarrow0}\frac{a\left(1-h\right)^2-b+1}{-h}\] \[limit_{h\rightarrow0}\frac{-\frac{1}{\left(1+h\right)}+1}{h}=limit_{h\rightarrow0}\frac{a\left(1-h\right)^2-b+1}{-h}\] \[limit_{h\rightarrow0}\frac{-1+1+h}{h\left(1+h\right)}\ =limit_{h\rightarrow0}\frac{a\left(1-h\right)^2-b+1}{-h}\] \[1=limit_{h\rightarrow0}\frac{a\left(1+h^2-2h\right)-b+1}{-h}\] \[1=limit_{h\rightarrow0}\frac{a\left(h^2-2h\right)+a-b+1}{-h}\] \[1=limit_{h\rightarrow0}\frac{a\left(h^2-2h\right)}{-h}+limit_{h\rightarrow0}\frac{a-b+1}{-h}\] Possible only…

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