Question

Let $f\left(x\right)=a{x}^2+bx+c,$ where $a,b,c$ 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 t is

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

Answer 1

Answer: (C)

$4$

$f\left(x\right)=a{x}^2+bx+ca,b,c$ are integers

$f\left(1\right)=0\Rightarrow a+b+c=0$

$4-<f\left(6\right)<50\Rightarrow 40<36a+6b+c<50$

$40<35a+5b<50$

$8<7a+bc<10\Rightarrow 7a+b=a$ ($\because a,b$ are integers)

$60<f\left(7\right)<70\Rightarrow 60<49a+7b+c<70$

$60<48a+6b<70$

$10<8a+b<\frac{35}{3}$

$\therefore 8a+b=11$ ($\because a,b$ are integers)

$(8a+b)-(7a+b)=11-9$

$a=2\Rightarrow b=-5$

$a+b+c=0\Rightarrow c=3$

$f\left(x\right)=2x^2-5x+3$

$f\left(50\right)=2{\left(50\right)}^2-5\left(50\right)+3$

$=5000-250+3$

$=4753$

Given $1000t<f\left(50\right)<1000(t+1)$

$\therefore t=4$