Hard INTEGER +4 / -1 PYQ · JEE Mains 2022

Let c, k $\in$ R. If $f(x) = (c + 1){x^2} + (1 - {c^2})x + 2k$ and $f(x + y) = f(x) + f(y) - xy$, for all x, y $\in$ R, then the value of $|2(f(1) + f(2) + f(3) + \,\,......\,\, + \,\,f(20))|$ is equal to ____________.

Answer (integer) 3395

Solution

<p>f(x) is polynomial</p> <p>Put y = 1/x in given functional equation we get</p> <p>$f\left( {x + {1 \over x}} \right) = f(x) + f\left( {{1 \over x}} \right) - 1$</p> <p>$$ \Rightarrow (c + 1){\left( {x + {1 \over x}} \right)^2} + (1 - {c^2})\left( {x + {1 \over x}} \right) + 2K$$</p> <p>$$ = (c + 1){x^2} + (1 - {c^2})x + 2K + (c + 1){1 \over {{x^2}}} + (1 - {c^2}){1 \over x} + 2K - 1$$</p> <p>$\Rightarrow 2(c + 1) = 2K - 1$ ..... (1)</p> <p>and put $x = y = 0$ we get</p> <p>$f(0) = 2 + f(0) - 0 \Rightarrow f(0) = 0 \Rightarrow k = 0$</p> <p>$\therefore$ $k = 0$ and $2c = - 3 \Rightarrow c = - 3/2$</p> <p>$f(x) = - {{{x^2}} \over 2} - {{5x} \over 4} = {1 \over 4}(5x + 2{x^2})$</p> <p>$$\left| {2\sum\limits_{i = 1}^{20} {f(i)} } \right| = \left| {{{ - 2} \over 4}\left( {{{5.20.21} \over 2} + {{2.20.21.41} \over 6}} \right)} \right|$$</p> <p>$= \left| {{{ - 1} \over 2}(2730 + 5740)} \right|$</p> <p>$= \left| { - {{6790} \over 2}} \right| = 3395$.</p>

About this question

Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations

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