The smallest positive integral value of f ( x )=( x 2+ x +7/ x +2), x ∈ R is equal to

Question

The smallest positive integral value of f ( x )=( x 2+ x +7/ x +2), x ∈ R is equal to (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

y x+2 y=x2+x+7 x2+x(1-y)+7-2 y=0 text as x ∈ R D ≥ 0 (1-y)2-4(7-2 y) ≥ 0 y2+6 y-27 ≥ 0 (y+9)(y-3) ≥ 0 y ∈(-∞,-9] ∪[3 . ∞) ∴ text Smallest positive value of f(x) text is 3 .

Q. The smallest positive integral value of $f (x) = \frac{x ^{2} + x + 7}{x + 2}, x \in R$ is equal to

1

2

3

4

$y x + 2 y = x^{2} + x + 7$
$x^{2} + x (1 - y) + 7 - 2 y = 0 as x \in R$
$D \geq 0$
$(1 - y)^{2} - 4 (7 - 2 y) \geq 0$
$y^{2} + 6 y - 27 \geq 0$
$(y + 9) (y - 3) \geq 0$
$y \in (- \infty, - 9] \cup [3.\infty)$
$∴ Smallest positive value of f (x) is 3$ .

Q. The smallest positive integral value of f(x)=x+2x2+x+7​,x∈R is equal to

1

2

3

4

yx+2y=x2+x+7 x2+x(1−y)+7−2y=0 as x∈R D≥0 (1−y)2−4(7−2y)≥0 y2+6y−27≥0 (y+9)(y−3)≥0 y∈(−∞,−9]∪[3.∞) ∴ Smallest positive value of f(x) is 3 .

Q. The smallest positive integral value of $f (x) = \frac{x ^{2} + x + 7}{x + 2}, x \in R$ is equal to

$y x + 2 y = x^{2} + x + 7$
$x^{2} + x (1 - y) + 7 - 2 y = 0 as x \in R$
$D \geq 0$
$(1 - y)^{2} - 4 (7 - 2 y) \geq 0$
$y^{2} + 6 y - 27 \geq 0$
$(y + 9) (y - 3) \geq 0$
$y \in (- \infty, - 9] \cup [3.\infty)$
$∴ Smallest positive value of f (x) is 3$ .