If x is real, then the maximum value of 3-6 x-8 x2 is

Question

If x is real, then the maximum value of 3-6 x-8 x2 is (A) (17/8) (B) (33/8) (C) (21/8) (D) None of these

Tardigrade Admin · Accepted Answer

Let y=3-6 x-8 x2 then 8 x2+6 x+y-3=0. Since x is real, ∴ 62-4 ⋅ 8(y-3) ≥ 0, or 36-32 y+96 ≥ 0 or 32 y ≤ 132 ∴ y ≤ (132/32) or y ≤ (33/8) Hence, maximum value of y=(33/8).

Q. If $x$ is real, then the maximum value of $3 - 6 x - 8 x^{2}$ is

$\frac{17}{8}$

$\frac{33}{8}$

$\frac{21}{8}$

None of these

Let $y = 3 - 6 x - 8 x^{2}$
then $8 x^{2} + 6 x + y - 3 = 0$ .
Since $x$ is real,
$∴ 6^{2} - 4 \cdot 8 (y - 3) \geq 0,$
or $36 - 32 y + 96 \geq 0$
or $32 y \leq 132$
$∴ y \leq \frac{132}{32}$
or $y \leq \frac{33}{8}$
Hence, maximum value of $y = \frac{33}{8}$ .

Q. If x is real, then the maximum value of 3−6x−8x2 is

817​

833​

821​