Q51 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Mathematics

The number of integral vaules of $x$ satisfying $5 x-1 < (x+1)^2 < 7 x-3$ is

  • A
    0
  • Correct answer
    B
    1
  • C
    2
  • D
    3

Solution

We have, $5 x-1 < (x+1)^2$ $$ \begin{aligned} & \Rightarrow \quad 5 x-1 < x^2+2 x+1 \\ & \Rightarrow \quad x^2-3 x+2>0 \\ & \Rightarrow \quad(x-1)(x-2)>0 \\ & \Rightarrow \quad x \in(-\infty, 1) \cup(2, \infty) \\ & \end{aligned} $$ Again, $$ \begin{array}{r} \quad(x+1)^2 < 7 x-3 \\ \Rightarrow \quad x^2+2 x+1 < 7 x-3 \end{array} $$ $$ \begin{aligned} & \Rightarrow \quad x^2-5 x+4 < 0 \\ & \Rightarrow \quad(x-1)(x-4) < 0 \\ & \Rightarrow \quad x \in(1,4) \\ & \therefore \quad x=3 \quad[\because x \text { is integer }] \\ & \end{aligned} $$

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

Relevant Questions