Question

If the given points $(a,-5,a-3),(a+1,a-2,a),(-1,-85,16)$ are collinear, then which of the following is/are correct ?

• A. $\underset{x\to 2}{lim}(a{x}^2+22x+1{)}^{1/(2a+11x)}=e^2$
• B. $\underset{x\to 2}{lim}(a{x}^2+22x+1{)}^{1/(2a+11x)}=e^{-2}$
• C. $\underset{x\to a}{lim}\frac{x^2+8x-33}{x^2+x-110}=\frac{2}{3}$
• D. The number of real solutions of the equation $x^2+ax+2021=\sin x$ is zero

Answer 1

Answer: (B), (C), (D)

The number of real solutions of the equation $x^2+ax+2021=\sin x$ is zero

For collinear points:
$\Rightarrow \frac{x_3-x_1}{x_2-x_1}=\frac{y_3-y_1}{y_2-y_1}=\frac{z_3-z_1}{z_2-z_1}$
$\Rightarrow \frac{-1-a}{1}=\frac{-85+5}{a+3}=\frac{16-a+3}{3}$
$\Rightarrow \frac{-1-a}{1}=\frac{19-a}{3}$
$\Rightarrow -3-3a=19-a$
$\Rightarrow 2a=-22$
$\Rightarrow a=-11$
satisfy all three conditions.
Now,
$\to \underset{x\to 2}{lim}(a{x}^2+22x+1{)}^{1/(2a+11x)}$
$=e^{\underset{x\to 2}{lim}\frac{a{x}^2+22x}{2a+11x}}$
$=e^{\underset{x\to 2}{lim}\frac{11x(-x+2)}{11(x-2)}}=e^{-2}$

$\to \underset{x\to a}{lim}\frac{x^2+8x-33}{x^2+x-110}$
$=\underset{x\to a}{lim}\frac{(x+11)(x-3)}{(x+11)(x-10)}$
$=\frac{-14}{-21}=\frac{2}{3}$

$\to x^2+ax+2021=x^2-11x+2021$
Let $f\left(x\right)=x^2-11x+2021$
$f^{\prime }\left(x\right)=2x-11$
$\therefore f^{\prime }\left(x\right)=0\Rightarrow x=\frac{11}{2}$
$\therefore$ minimum of $f\left(x\right)$ is equal to $f\left(\frac{11}{2}\right)=2021-\frac{121}{4}=\frac{7963}{4}>1$
But $g\left(x\right)=\sin x\le 1$
Hence, $f\left(x\right)=g\left(x\right)$ has no real solutions.