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# Calculus Problem - I.S.I 2019 Subjective Problem -4

## Understand the problem

Let $f:\mathbb{R}\to\mathbb{R}$ be a twice differentiable function such that$\frac{1}{2y}\int_{x-y}^{x+y}f(t)\, dt=f(x)\qquad\forall~x\in\mathbb{R}~\&~y>0$Show that there exist $a,b\in\mathbb{R}$ such that $f(x)=ax+b$ for all $x\in\mathbb{R}$.
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I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 4
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### Calculus

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.7" open="off"]8.5 out of 10

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Problems In CALCULUS OF ONE VARIABLE-i.a maron

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Do you really need a hint? Try it first!

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Do you know Taylor series expansion good then see..... $f(x+h) = f(x) + f'(x)h+f''(x)\frac{h^2}{2} + f'''(x)\frac{h^3}{3!}+\cdots$ $f(x-h) = f(x) - f'(x)h+f''(x)\frac{h^2}{2} - f'''(x)\frac{h^3}{3!}+\cdots$

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adding previous expression we get  $\frac{f(x+h) - 2f(x) + f(x-h)}{h^2} = f''(x) + 2\frac{f''''(x)}{4!}h^2+\cdots$  taking the limit of the above equation as h goes to zero gives  $\Rightarrow f''(x) = \lim_{h\to0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2} \,.$

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Since $f$ is continuous, there exists $F$ such that $F' = f.$ The given identity becomes  $F(x+y)-F(x-y) = 2yf(x).$ Fix $x \in \mathbb{R}$ and differentiate the above identity with respect to $y$ and get $f(x+y)+f(x-y)=2f(x).$

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now we already get ,  If $f$ is twice differentiable at $x$ then $\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x).$ by rearranging our given problem we get  that $f''(x)=0$ Therefore , $f(x)=ax+b$ for some $a,b\in\mathbb{R}.$

## Connected Program at Cheenta

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

## Similar Problem

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