Let $A$ be the set of all functions $f:mathbb{R}tomathbb{R}$ that satisfy the following two properties.












2












$begingroup$


Let $A$ be the set of all functions $f:mathbb{R}tomathbb{R}$ that satisfy the following two properties:



(1) $f$ has derivative of all orders, and



(2) for all $x,y in mathbb{R}$, $f(x+y)-f(y-x)=2xf'(y)$.



Which of the following sentences is true?



(a) Any $fin A$ is a polynomial of degree less than or equal to 1



(b) Any $f in A$ is a polynomial of degree less than or equal to 2



(c) $exists f in A$ which is not polynomial



(d) $exists f in A$ which is a polynomial of degree 4



It is a problem from TIFR GS-2018. I have found polynomials upto degree 2 hold this properties but have to exclude the possibility of (c) and (d). I also found out if $f'(x)$ has at most one real root and clearly graph of $f$ is symetric about that root. How can I proceed further?






calculus real-analysis







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $A$ be the set of all functions $f:mathbb{R}tomathbb{R}$ that satisfy the following two properties:



    (1) $f$ has derivative of all orders, and



    (2) for all $x,y in mathbb{R}$, $f(x+y)-f(y-x)=2xf'(y)$.



    Which of the following sentences is true?



    (a) Any $fin A$ is a polynomial of degree less than or equal to 1



    (b) Any $f in A$ is a polynomial of degree less than or equal to 2



    (c) $exists f in A$ which is not polynomial



    (d) $exists f in A$ which is a polynomial of degree 4



    It is a problem from TIFR GS-2018. I have found polynomials upto degree 2 hold this properties but have to exclude the possibility of (c) and (d). I also found out if $f'(x)$ has at most one real root and clearly graph of $f$ is symetric about that root. How can I proceed further?






    calculus real-analysis







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      0



      $begingroup$


      Let $A$ be the set of all functions $f:mathbb{R}tomathbb{R}$ that satisfy the following two properties:



      (1) $f$ has derivative of all orders, and



      (2) for all $x,y in mathbb{R}$, $f(x+y)-f(y-x)=2xf'(y)$.



      Which of the following sentences is true?



      (a) Any $fin A$ is a polynomial of degree less than or equal to 1



      (b) Any $f in A$ is a polynomial of degree less than or equal to 2



      (c) $exists f in A$ which is not polynomial



      (d) $exists f in A$ which is a polynomial of degree 4



      It is a problem from TIFR GS-2018. I have found polynomials upto degree 2 hold this properties but have to exclude the possibility of (c) and (d). I also found out if $f'(x)$ has at most one real root and clearly graph of $f$ is symetric about that root. How can I proceed further?






      calculus real-analysis







      share|cite|improve this question









      $endgroup$




      Let $A$ be the set of all functions $f:mathbb{R}tomathbb{R}$ that satisfy the following two properties:



      (1) $f$ has derivative of all orders, and



      (2) for all $x,y in mathbb{R}$, $f(x+y)-f(y-x)=2xf'(y)$.



      Which of the following sentences is true?



      (a) Any $fin A$ is a polynomial of degree less than or equal to 1



      (b) Any $f in A$ is a polynomial of degree less than or equal to 2



      (c) $exists f in A$ which is not polynomial



      (d) $exists f in A$ which is a polynomial of degree 4



      It is a problem from TIFR GS-2018. I have found polynomials upto degree 2 hold this properties but have to exclude the possibility of (c) and (d). I also found out if $f'(x)$ has at most one real root and clearly graph of $f$ is symetric about that root. How can I proceed further?






      calculus real-analysis




      calculus real-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 26 '18 at 3:47









      OfflawOfflaw

      2689




      2689






















          1 Answer
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          $begingroup$

          We have
          $$f(x+y)-f(y-x)=2xf'(y) $$
          Apply $frac{mathrm d}{mathrm dx}$:
          $$f'(x+y)+f'(y-x)=2f'(y) $$
          and once more:
          $$f''(x+y)-f''(y-x)=0. $$
          With $xleftarrow frac t2$, $tleftarrow frac t2$, this becomes
          $$f''(t)=f''(0). $$
          Thus $f$ is a polynomal and of degree $le 2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. But, there may be a typing error, it should $y leftarrow frac{t}{2}$
            $endgroup$
            – Offlaw
            Nov 26 '18 at 5:21











          Your Answer





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          1 Answer
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          1 Answer
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          active

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          active

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          3












          $begingroup$

          We have
          $$f(x+y)-f(y-x)=2xf'(y) $$
          Apply $frac{mathrm d}{mathrm dx}$:
          $$f'(x+y)+f'(y-x)=2f'(y) $$
          and once more:
          $$f''(x+y)-f''(y-x)=0. $$
          With $xleftarrow frac t2$, $tleftarrow frac t2$, this becomes
          $$f''(t)=f''(0). $$
          Thus $f$ is a polynomal and of degree $le 2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. But, there may be a typing error, it should $y leftarrow frac{t}{2}$
            $endgroup$
            – Offlaw
            Nov 26 '18 at 5:21
















          3












          $begingroup$

          We have
          $$f(x+y)-f(y-x)=2xf'(y) $$
          Apply $frac{mathrm d}{mathrm dx}$:
          $$f'(x+y)+f'(y-x)=2f'(y) $$
          and once more:
          $$f''(x+y)-f''(y-x)=0. $$
          With $xleftarrow frac t2$, $tleftarrow frac t2$, this becomes
          $$f''(t)=f''(0). $$
          Thus $f$ is a polynomal and of degree $le 2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. But, there may be a typing error, it should $y leftarrow frac{t}{2}$
            $endgroup$
            – Offlaw
            Nov 26 '18 at 5:21














          3












          3








          3





          $begingroup$

          We have
          $$f(x+y)-f(y-x)=2xf'(y) $$
          Apply $frac{mathrm d}{mathrm dx}$:
          $$f'(x+y)+f'(y-x)=2f'(y) $$
          and once more:
          $$f''(x+y)-f''(y-x)=0. $$
          With $xleftarrow frac t2$, $tleftarrow frac t2$, this becomes
          $$f''(t)=f''(0). $$
          Thus $f$ is a polynomal and of degree $le 2$.






          share|cite|improve this answer









          $endgroup$



          We have
          $$f(x+y)-f(y-x)=2xf'(y) $$
          Apply $frac{mathrm d}{mathrm dx}$:
          $$f'(x+y)+f'(y-x)=2f'(y) $$
          and once more:
          $$f''(x+y)-f''(y-x)=0. $$
          With $xleftarrow frac t2$, $tleftarrow frac t2$, this becomes
          $$f''(t)=f''(0). $$
          Thus $f$ is a polynomal and of degree $le 2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 4:08









          Hagen von EitzenHagen von Eitzen

          277k22269496




          277k22269496












          • $begingroup$
            Thanks. But, there may be a typing error, it should $y leftarrow frac{t}{2}$
            $endgroup$
            – Offlaw
            Nov 26 '18 at 5:21


















          • $begingroup$
            Thanks. But, there may be a typing error, it should $y leftarrow frac{t}{2}$
            $endgroup$
            – Offlaw
            Nov 26 '18 at 5:21
















          $begingroup$
          Thanks. But, there may be a typing error, it should $y leftarrow frac{t}{2}$
          $endgroup$
          – Offlaw
          Nov 26 '18 at 5:21




          $begingroup$
          Thanks. But, there may be a typing error, it should $y leftarrow frac{t}{2}$
          $endgroup$
          – Offlaw
          Nov 26 '18 at 5:21


















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