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Question

(c)

We have,

$f(x)=a x^2+b x+c, a, b, c, \in Z$

$	ext { Also } f(1)=0,40 < f(6) < 50,60 < f(7) < 70$

$	herefore a+b+c=0,40 &

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(c)

We have,

$f(x)=a x^2+b x+c, a, b, c, \in Z$

$	ext { Also } f(1)=0,40 < f(6) < 50,60 < f(7) < 70$

$	herefore a+b+c=0,40 < 36 a+6 b+c < 50$

$60 < 49 a+7 b+c < 70$

$	herefore \quad 40 < 36 a+6 b-a-b < 50$

and $\quad 60 < 49 a+7 b-a-b < 70$

$\Rightarrow \quad 40 < 35 a+5 b < 50$

and $\quad 60 < 48 a+6 b < 70$

$\Rightarrow 8 < 7 a+b < 10$ and $10 < 8 a+b < \frac{70}{6}$

Now, $a$ and $b$ are integer

$	herefore \quad 7 a+b=9 	ext { and } 8 a+b=11$

On solving these equation, we get

$a=2, b=-5$

Put the value of $a, b$ in $Eq (i)$, we get $c=3$

$	herefore f(x)=2 x^2-5 x+3$

$f(50)=2(50)^2-5(50)+3$

$=5000-250+3=4753$

$	ext { Now, } 1000 t < f(50) < 1000(t+1)$

$	herefore 1000 t < 4753 < 1000(t+1)$

$\Rightarrow t < 4753 < t+1$

$	herefore t=4, t 	ext { is integer. }$

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

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