\hspace{-15}$(1)\;\;::\;If $\mathbf{f:\mathbb{Z}\rightarrow \mathbb{Z}\;,}$ and $\mathbf{f(x)-f(x-1)=x^3}$ and $\mathbf{f(2)=-1}$\\\\ Then $\mathbf{f(x)=}$\\\\ $\mathbf{Ans:\Leftrightarrow f(x)=\left(\frac{n.(n+1)}{2}\right)^2-10}$\\\\\\ (2)\;\;::\;If $\mathbf{f:\mathbb{Q}\rightarrow \mathbb{Q}\;,}$ and $\displaystyle\mathbf{\frac{1}{3}.f\left(x+\frac{1}{10}\right)=\frac{1}{3}.f\left(x\right)+\frac{x}{7}}$ and $\mathbf{f(0)=\displaystyle\frac{1}{2}}$\\\\ Then $\mathbf{f(1)=}$\\\\ $\mathbf{Ans:\Leftrightarrow f(x)=\frac{17}{7}}$
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1 Answers
211)Just telescope, or there are routine induction
2) plugging 0 , gives f(1/10) , plugging 1/10 gives f(2/10) continue till u get f(10/10) i.e f(1)..
