Đáp án:
$D$
Giải thích các bước giải:
$y=\dfrac{x+1}{\sqrt{mx^2+1}}\\ \text{ĐKXĐ}: mx^2+1 > 0 \ \forall x\\ \circledast m=0; mx^2+1=1 >0\\ \circledast m \ne 0; mx^2+1 > 0 \ \forall x\\ \Leftrightarrow m > -\dfrac{1}{x^2} \ \forall x\\ \Leftrightarrow m > max_{-\tfrac{1}{x^2}} \ \forall x\\ \Leftrightarrow m \ge 0 \left(\text{Do }-\dfrac{1}{x^2} <0 \ \forall x \ne 0\right)\\ \text{Kết hợp 2 trường hợp} \Rightarrow \text{ĐKXĐ}:m\ge 0$
$\circledast m=0 ,y=x+1$ không có tiệm cận ngang
$\circledast m>0 ,y=\dfrac{x+1}{\sqrt{mx^2+1}}=\dfrac{x+1}{|x|\sqrt{m+\dfrac{1}{x^2}}}\\ \displaystyle \lim_{x \to -\infty} y\\ =\displaystyle \lim_{x \to -\infty}\dfrac{x+1}{|x|\sqrt{m+\dfrac{1}{x^2}}}\\ =\displaystyle \lim_{x \to -\infty} y\\ =\dfrac{x+1}{-x\sqrt{m+\dfrac{1}{x^2}}}\\ =\displaystyle \lim_{x \to -\infty} \dfrac{1+\dfrac{1}{x}}{-1.\sqrt{m+\dfrac{1}{x^2}}}\\ =-\dfrac{1}{\sqrt{m}}\\ \displaystyle \lim_{x \to +\infty} y\\= \displaystyle \lim_{x \to +\infty}\dfrac{x+1}{|x|\sqrt{m+\dfrac{1}{x^2}}}\\=\displaystyle \lim_{x \to +\infty} y=\dfrac{x+1}{x\sqrt{m+\dfrac{1}{x^2}}}\\ =\displaystyle \lim_{x \to +\infty} \dfrac{1+\dfrac{1}{x}}{1.\sqrt{m+\dfrac{1}{x^2}}}\\ =\dfrac{1}{\sqrt{m}}$
$\Rightarrow$ Hàm số luôn có 2 TCN với $m>0.$
