If the equation x 3-12 x + a =0 has exactly one real root, then range of a is equal to

Question

If the equation x 3-12 x + a =0 has exactly one real root, then range of a is equal to (A) (-∞,-16) ∪(16, ∞) (B)  -16,16  (C) (-16,16) (D) (-∞,-16] ∪[16, ∞)

Tardigrade Admin · Accepted Answer

Let y=f(x)=x3-12 x and y=-a For f(x)=-a to have exactly one real root, we must have -a >16 text or -a<-16 ⇒ a ∈(-∞,-16) ∪(16, ∞)

Q. If the equation $x^{3} - 12 x + a = 0$ has exactly one real root, then range of a is equal to

$(- \infty, - 16) \cup (16, \infty)$

${- 16, 16}$

$(- 16, 16)$

$(- \infty, - 16] \cup [16, \infty)$

Let $y = f (x) = x^{3} - 12 x$ and $y = - a$
For $f (x) = - a$ to have exactly one real root, we must have
$- a > 16 or - a < - 16$
$\Rightarrow a \in (- \infty, - 16) \cup (16, \infty)$

Q. If the equation x3−12x+a=0 has exactly one real root, then range of a is equal to

(−∞,−16)∪(16,∞)

{−16,16}

(−16,16)

(−∞,−16]∪[16,∞)