Let f(x) = ax^2 + bx + c,where a, b, c are in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(x) = ax^2 + bx + c$ with $a, b, c \in \mathbb{Z}$.Since $f(1) = 0$,we have $a + b + c = 0$,which implies $c = -a - b$.Substituting $c$ in

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given $f(x) = ax^2 + bx + c$ with $a, b, c \in \mathbb{Z}$.Since $f(1) = 0$,we have $a + b + c = 0$,which implies $c = -a - b$.Substituting $c$ into $f(x)$,we get $f(x) = ax^2 + bx - a - b = a(x^2 - 1) + b(x - 1) = (x - 1)(ax + a + b)$.Given $40 Since $a, b$ are integers,$7a + b = 9$.Given $60 Since $a, b$ are integers,$8a + b = 11$.Subtracting the two equations: $(8a + b) - (7a + b) = 11 - 9 \implies a = 2$.Substituting $a = 2$ into $7a + b = 9$,we get $14 + b = 9 \implies b = -5$.Then $c = -a - b = -2 - (-5) = 3$.Thus,$f(x) = 2x^2 - 5x + 3$.Calculating $f(50) = 2(50)^2 - 5(50) + 3 = 5000 - 250 + 3 = 4753$.Given $1000t < 4753 < 1000(t + 1)$,we find $t = 4$.

Let $f(x) = ax^2 + bx + c$,where $a, b, c$ are integers. Suppose $f(1) = 0$,$40 < f(6) < 50$,$60 < f(7) < 70$,and $1000t < f(50) < 1000(t+1)$ for some integer $t$. Then,the value of $t$ is

Similar Questions

If $\alpha$ and $\beta$ are roots of the equation $x^2 - x + 1 = 0$,then $\alpha^{2009} + \beta^{2009} = \dots$

If $mn=3$ and $\frac{1}{m}+\frac{1}{n}=\frac{4}{3}$,then the value of $0.1+0.1^{\frac{1}{m}}+0.1^{\frac{1}{n}}$ is

If $(x-1)$ is a factor of the polynomial $x^{5}-4 x^{3}+2 x^{2}-3 x+k=0$,then the value of $k$ is:

The solution set of the equation $\sqrt{x + 3 - 4\sqrt{x - 1}} + \sqrt{x + 8 - 6\sqrt{x - 1}} = 1$ is

The difference between the roots of the equation $x^2 + 2x + 4 = 0$ is ... (where $i = \sqrt{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module