If (x^2+mx+1)/(x^2+x+1)<3 for all real x, then

Correct answer: (C) -1 0 So, -3 0 and 2x^2+3-mx+2>0 for real x ⇒m+3^2-4×4×4<0 and 3-m^2-4×2×2<0 ⇒m+3+8m+3-8<0 and m-3+4m-3-4<0 ⇒m+11m-5<0 and m+1m-7<0 ⇒-11<m<5 and -1<m<7 Hence, -1<m<5

Q26 / 87 · NTA Abhyas (JEE Main) 2020 · Mathematics · Functions

If $|\frac{x^{2} + m x + 1}{x^{2} + x + 1}| < 3$ for all real $x,$ then

A
$m < - 1$
B
$- 1 < m < 6$
Correct answer
C
$- 1 < m < 5$
D
$m > 6$

Solution

We have

x^{2} + x + 1 = {(x + \frac{1}{2})}^{2} + \frac{3}{4} > 0

So,

- 3 < \frac{x^{2} + m x + 1}{x^{2} + x + 1} < 3

\Rightarrow - 3 (x^{2} + x + 1) < x^{2} + m x + 1 < 3 (x^{2} + x + 1)

\Rightarrow 4 x^{2} + (m + 3) x + 4 > 0

and

2 x^{2} + (3 - m) x + 2 > 0

for real x

\Rightarrow {(m + 3)}^{2} - 4 \times 4 \times 4 < 0

and

{(3 - m)}^{2} - 4 \times 2 \times 2 < 0

\Rightarrow (m + 3 + 8) (m + 3 - 8) < 0

and

(m - 3 + 4) (m - 3 - 4) < 0

\Rightarrow (m + 11) (m - 5) < 0

and

(m + 1) (m - 7) < 0

\Rightarrow - 11 < m < 5

and

- 1 < m < 7

Hence,

- 1 < m < 5

Hence, the correct answer is option: 'C'.

If x2+mx+1x2+x+1<3 for all real x, then

Solution

More Functions Questions

If $|\frac{x^{2} + m x + 1}{x^{2} + x + 1}| < 3$ for all real $x,$ then