If the point of minima of the function, f ( x )=1+ a 2 x - x 3 satisfy the inequality (x2+x+2/x2+5 x+6)

Question

If the point of minima of the function, f ( x )=1+ a 2 x - x 3 satisfy the inequality (x2+x+2/x2+5 x+6)<0, then 'a' must lie in the interval: (A) (-3 √3, 3 √3) (B) (-2 √3,-3 √3) (C) (2 √3, 3 √3) (D) (-3 √3,-2 √3) Y (2 √3, 3 √3)

Tardigrade Admin · Accepted Answer

f prime( x )=0 ⇒ x =( a /√3) or -( a /√3) f prime prime( x )=-6 x Case I if a>0 ⇒ x=-(a/√3) is minima Case II if a <0 ⇒ x =( a /√3) is minima put x=(a/√3) and then x=-(a/√3) in the given inequality to get the result

Q. If the point of minima of the function, $f (x) = 1 + a^{2} x - x^{3}$ satisfy the inequality $\frac{x ^{2} + x + 2}{x ^{2} + 5 x + 6} < 0$ , then 'a' must lie in the interval:

$(- 33, 33)$

$(- 23, - 33)$

$(23, 33)$

$(- 33, - 23)$ Y $(23, 33)$

$f^{'} (x) = 0 \Rightarrow x = \frac{a}{3}$
or $- \frac{a}{3} f^{''} (x) = - 6 x$
Case I if $a > 0 \Rightarrow x = - \frac{a}{3}$ is minima
Case II if $a < 0 \Rightarrow x = \frac{a}{3}$ is minima
put $x = \frac{a}{3}$ and then $x = - \frac{a}{3}$ in the given inequality to get the result

Q. If the point of minima of the function, f(x)=1+a2x−x3 satisfy the inequality x2+5x+6x2+x+2​<0, then 'a' must lie in the interval:

(−33​,33​)

(−23​,−33​)

(23​,33​)

(−33​,−23​) Y (23​,33​)