Q52 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics

If $x$ is real, then the range of $\frac{x^2+2 x+1}{x^2+2 x+7}$ is

  • Correct answer
    A
    $[0,1)$
  • B
    $(-\infty, 0) \cup(1, \infty)$
  • C
    $(0,1)$
  • D
    $R$

Solution

$$ \begin{array}{lc} \text {} & \text { Let } \frac{x^2+2 x+1}{x^2+2 x+7}=y, \\ \because & y \neq 1 \\ \Rightarrow & (y-1) x^2+2(y-1) x+(7 y-1)=0 \\ \because & x \in R \\ \text { so, } & D \geq 0 \\ \Rightarrow & 4(y-1)^2-4(y-1)(7 y-1) \geq 0 \\ \Rightarrow & (y-1)[y-1-7 y+1] \geq 0 \\ \Rightarrow & y(y-1) \leq 0 \\ \Rightarrow & y \in[0,1] \end{array} $$ From Eqs. (i) and (ii), we are getting $$ y \in[0,1) $$

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

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