Question

The set of all values of the parameters a for which the points of minimum of the function y = 1 + a²x – x³ satisfy the inequality $\frac { x ^ { 2 } + x + 2 } { x ^ { 2 } + 5 x + 6 }$£ 0 is –

• A. An empty set
• B. $( - 3 \sqrt { 3 } , - 2 \sqrt { 3 } )$
• C. $( 2 \sqrt { 3 } , 3 \sqrt { 3 } )$
• D. ) $( - 3 \sqrt { 3 } , - 2 \sqrt { 3 } )$Č $( 2 \sqrt { 3 } , 3 \sqrt { 3 } )$

Answer 1

Answer: (C)

$( 2 \sqrt { 3 } , 3 \sqrt { 3 } )$

Answer 2

$\frac { d y } { d x }$ = a² – 3x² = 0 Ū x = ± a/ $\sqrt { 3 }$ .

Since = –6x so y is minimum for x = –a/$\sqrt { 3 }$.

Since x² + x + 2 > 0 for all x so for $\frac { x ^ { 2 } + x + 2 } { x ^ { 2 } + 5 x + 6 }$ £ 0,

we must have x² + 5x + 6 < 0.

If x = –a/$\sqrt { 3 }$, we have a²/3 – 5a/$\sqrt { 3 }$ + 6 < 0 i.e. a² – 5$\sqrt { 3 }$ a + 18 < 0

Ū (a – 2$\sqrt { 3 }$) (a – 3$\sqrt { 3 }$) < 0 i.e. a Ī (2$\sqrt { 3 }$, 3$\sqrt { 3 }$).