• # question_answer The function $f(x)={{(x-3)}^{2}}$ satisfies all the conditions of mean value theorem in $\{3,\,\,4\}$. A point on $y={{(x-3)}^{2}}$, where the tangent is parallel to the chord joining $(3,\,\,0)$ and $(4,\,\,1)$ is A) $\left( \frac{7}{2},\,\,\frac{1}{2} \right)$                  B) $\left( \frac{7}{2},\,\,\frac{1}{4} \right)$C) $(1,\,\,4)$                   D) $(4,\,\,1)$

Let the point be$({{x}_{1}},\,\,{{y}_{1}})$. Therefore$={{({{x}_{1}}-3)}^{2}}$                                ... (i) $\therefore$Now slope of the tangent at $({{x}_{1}},\,\,{{y}_{1}})$ is $2({{x}_{1}}-3),$ but it is equal to $1$. Therefore,$2({{x}_{1}}-3)=1\Rightarrow {{x}_{1}}=\frac{7}{2}$             ${{y}_{1}}={{\left( \frac{7}{2}-3 \right)}^{2}}=\frac{1}{4}$. Hence the point is$\left( \frac{7}{2},\,\,\frac{1}{4} \right)$.