Question

Let $f\left(x\right)=a{x}^2+bx+c$ are integers. Suppose $f\left(1\right)=0,,40<f\left(6\right)<50,60<f\left(7\right)<70$ , and $1000t<f\left(50\right)<1000(t+1)$ for some integer $t$. Then the value of its

• A. $2$
• B. $3$
• C. $4$
• D. $5$ or more

Answer 1

The correct option is C $4$

$f\left(x\right)=a{x}^2+bx+c,$ $f\left(1\right)=0$
given, $f\left(1\right)=0$
$a+b+c=0$
and $40<f\left(6\right)<50$
$40<36a+6b+c<50$
$40+35a+5b<50$
$8<7a+b<10$
$7a+b=integer=9$ ........(1)
and $60<f\left(7\right)<70$
$60<49a+7b+c<70$
$60<48a+6b<70$
$10<8a+b<11.6$
$8a+b=11$ ........(2)
Solving equations (1) and (2)
$a=2,b=-5,$ so $c=3$
$f\left(x\right)=2x^2-5x+3$
$f\left(50\right)=4753$
$t=4$