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Functions question

2022 · 29 Jun · Shift 1 · Q39
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  5. /2022 · 29 Jun · Shift 1 · Q39

Functions question

2022 · 29 Jun · Shift 1 · Q39

JEE MainMathematicsFunctionsNumerical+4 / −1
Let c, k ∈\in∈ R. If f(x)=(c+1)x2+(1−c2)x+2kf(x) = (c + 1){x^2} + (1 - {c^2})x + 2kf(x)=(c+1)x2+(1−c2)x+2k and f(x+y)=f(x)+f(y)−xyf(x + y) = f(x) + f(y) - xyf(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))∣|2(f(1) + f(2) + f(3) + \,\,......\,\, + \,\,f(20))|∣2(f(1)+f(2)+f(3)+......+f(20))∣ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 3395

  1. We are given f(x)=(c+1)x2+(1−c2)x+2kf(x)=(c+1)x^2+(1-c^2)x+2kf(x)=(c+1)x2+(1−c2)x+2k and also f(x+y)=f(x)+f(y)−xy  for all x,y∈R.f(x+y)=f(x)+f(y)-xy \,\text{ for all }x,y\in\mathbb R.f(x+y)=f(x)+f(y)−xy for all x,y∈R.

We must find ∣2∑n=120f(n)∣.\left|2\sum_{n=1}^{20} f(n)\right|.​2∑n=120​f(n)​.

  1. Use the functional relation.

First compute both sides using the quadratic form.

Since f(t)=(c+1)t2+(1−c2)t+2k,f(t)=(c+1)t^2+(1-c^2)t+2k,f(t)=(c+1)t2+(1−c2)t+2k, we have f(x+y)=(c+1)(x+y)2+(1−c2)(x+y)+2k.f(x+y)=(c+1)(x+y)^2+(1-c^2)(x+y)+2k.f(x+y)=(c+1)(x+y)2+(1−c2)(x+y)+2k. Expanding, f(x+y)=(c+1)x2+2(c+1)xy+(c+1)y2+(1−c2)x+(1−c2)y+2k.f(x+y)=(c+1)x^2+2(c+1)xy+(c+1)y^2+(1-c^2)x+(1-c^2)y+2k.f(x+y)=(c+1)x2+2(c+1)xy+(c+1)y2+(1−c2)x+(1−c2)y+2k.

Now compute f(x)+f(y)−xy.f(x)+f(y)-xy.f(x)+f(y)−xy. That is, \begin{align*} f(x)+f(y)-xy &= \left[(c+1)x^2+(1-c^2)x+2k\right] \ &\quad + \left[(c+1)y^2+(1-c^2)y+2k\right]-xy \ &=(c+1)x^2+(c+1)y^2+(1-c^2)x+(1-c^2)y+4k-xy. \end{align*}

Since these are equal for all x,yx,yx,y, compare coefficients.

  1. Compare coefficients.

From the coefficient of xyxyxy: 2(c+1)=−12(c+1)=-12(c+1)=−1 so c+1=−12  ⟹  c=−32.c+1=-\frac12 \implies c=-\frac32.c+1=−21​⟹c=−23​.

From the constant terms: 2k=4k  ⟹  2k=0  ⟹  k=0.2k=4k \implies 2k=0 \implies k=0.2k=4k⟹2k=0⟹k=0.

So the function becomes

= -\frac12 x^2-\frac54 x.$$ Thus, $$f(x)=-\frac12 x^2-\frac54 x.$$ --- 4. Compute the sum $\sum_{n=1}^{20} f(n)$. \begin{align*} \sum_{n=1}^{20} f(n) &= \sum_{n=1}^{20}\left(-\frac12 n^2-\frac54 n\right) \\ &= -\frac12 \sum_{n=1}^{20} n^2 - \frac54 \sum_{n=1}^{20} n. \end{align*} Use standard formulas: $$\sum_{n=1}^{20} n = \frac{20\cdot 21}{2}=210,$$ $$\sum_{n=1}^{20} n^2 = \frac{20\cdot 21\cdot 41}{6}=2870.$$ Therefore, \begin{align*} \sum_{n=1}^{20} f(n) &= -\frac12(2870)-\frac54(210) \\ &= -1435-\frac{1050}{4} \\ &= -1435-262.5 \\ &= -1697.5. \end{align*} Hence, $$2\sum_{n=1}^{20} f(n)=2(-1697.5)=-3395.$$ Therefore, $$\left|2\sum_{n=1}^{20} f(n)\right|=|-3395|=3395.$$ --- 5. Final answer: $$\boxed{3395}$$ This matches the stored correct answer.
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