Question

Number of integers lying in the range of
is
f(x)=
- 12x - 4
x2 - 2x - 2​

Answer 1

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Answer 2

✪ᴀɴsᴡᴇʀ✪

• There are 13 integers in the range of f(x).

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

Here,

$\:\:\:\:\: f(x) = \frac{x^2 + 12x + 4} {x^2 + 2x + 2}$

Put $f(x) = y$ for simplicity.

$\:\:\:\:\: y = \frac{x^2 + 12x + 4} {x^2 + 2x + 2}$

$✈︎\:\: y(x^2 + 2x + 2) = x^2 + 12x + 4$

$✈︎\:\: (y-1)x^2 + (2y-12)x$

$\:\:\:\:\:\:\:\:\:+ (2y-4)= 0$

$✈︎\:\: (y-1)x^2 + 2(y-6)x$

$\:\:\:\:\:\:\:\:\:+ 2(y-2)= 0$

Since $x \in R \implies D ≥ 0$

$✈︎\:\: [2(y-6)]^2$

$\:\:\:\:\:\:\:\:\:- 4(y-1).2(y-2) ≥ 0$

$✈︎\:\: 4(y-6)^2$

$\:\:\:\:\:\:\:\:\:- 8(y-1)(y-2) ≥ 0$

$✈︎\:\: (y^2 -12y + 36)$

$\:\:\:\:\:\:\:\:\: - 2(y^2 - 3y +2) ≥ 0$

$✈︎\:\: y^2 -12y + 36$

$\:\:\:\:\:\:\:\:\:- 2y^2 + 6y - 4 ≥ 0$

$✈︎\:\: - y^2 -6y + 32 ≥ 0$

$✈︎\:\: y^2 + 6y - 32 ≤ 0$

$✈︎\:\: y^2 + 6y + 9 - 41 ≤ 0$

$✈︎\:\:(y+3)^2 - (\sqrt{41})^2 ≤ 0$

$✈︎\:\: [y+3 - \sqrt{41}][y+3 + \sqrt{41}]≤ 0$

$✈︎\:\: -3 - \sqrt{41} ≤ y ≤ -3 + \sqrt{41}$

$✈︎\:\: -3 - 6.4 ≤ y ≤ -3 + 6.4$

$✈︎\:\:\boxed{ - 9.4 ≤ y ≤ 3.4}$

So, integers in range of f(x) are

$- 9, - 8, - 7, - 6, - 5, - 4,$

$- 3, - 2, - 1, 0, 1, 2, 3.$

Hence there are 13 integers in the range of f(x).