Question

If the inequality $\frac{m{x}^2+3x+4}{x^2+2x+2}<5$ is satisfied for all $x\in R,$ then which one of following is correct ?

• A. $1<m<5$
• B. $-1<m<5$
• C. $1<m<6$
• D. $m<\frac{71}{24}$

Answer 1

The correct option is D $m<\frac{71}{24}$

$\frac{m{x}^2+3x+4}{x^2+2x+2}<5$

$\therefore m{x}^2+3x+4<5(x^2+2x+2)$

$\therefore m{x}^2+3x+4<5x^2+10x+10$

$\therefore x^2(m-5)+3x+0x+4-10<0$

$\therefore x^2(m-5)-7x-6<0$

Since above Quadratic Equation is always less than 0

i) $m-5<0$

ii) $0<0$

$\therefore m<5;b^2-4ac<0$ __(i)

$\therefore 49-4(-6)(m-5)<0$

$\therefore 49<-24(m-5)$

$\therefore 49<-24m+120$

$\therefore 24m<120-43$

$\therefore m<\frac{71}{24}$ __(ii)

Hence from (i) and (ii) we obtain

$m<\frac{71}{24}$