MHTCET 16 May 2024 Shift 1 Paper

Concept:To find the range of a rational function, set $y = f(x)$, form a quadratic equation in $x$, and use the condition $D \ge 0$ for real $x$.Explanation:Let $$y = \frac{x^2 + x + 2}{x^2 + x + 1}.$$ Since $x^2+x+1 > 0$ for all real $x$, we can cross-multiply: $$y(x^2+x+1) = x^2+x+2.$$ Rearranging gives: $$(y-1)x^2 + (y-1)x + (y-2) = 0.$$ For real $x$, the discriminant must be non-negative: $$(y-1)^2 - 4(y-1)(y-2) \ge 0.$$ Simplifying: $$(y-1)(3y-7) \le 0.$$ Hence: $$1 \le y \le \frac{7}{3}.$$ But check $y=1$: $$x^2+x+2 = x^2+x+1,$$ which gives $2=1$, impossible. So $y \neq 1$ for any real $x$. Also, the maximum value occurs when $x=-\frac12$, giving $y = \frac{7}{3}$. Therefore, the range is: $$\mathrm{R}_f = \left(1, \frac{7}{3}\right].$$Answer:$\left(1, \frac{7}{3}\right]$, i.e. option C.