Question

Let $f\left(x\right)=x^3+3x^2+9x+6\sin x$ then the roots of the equation
$\frac{1}{x-f\left(1\right)}+\frac{2}{x-f\left(2\right)}+\frac{3}{x-f\left(3\right)}=0$ has

• A. No real roots
• B. One real root
• C. Two real roots
• D. More than 2 real roots

Answer 1

Answer: (C)

Two real roots

Given $f\left(x\right)=x^3+3x^2+9x+6\sin x$

$f\left(0\right)=0$

$f^{\prime }\left(x\right)=3x^2+6x+9+6\cos x$

$=3{(x+1)}^2+6(1+\cos x)$

$f^{\prime }\left(x\right)>0$

$\therefore f\left(x\right)$ is monotonically increasing graph

$\therefore f\left(3\right)>f\left(2\right)>f\left(1\right)>f\left(0\right)$

$\therefore f\left(3\right)>f\left(2\right)>f\left(1\right)>0$ ($\because$ if $x_1>x_2\Rightarrow f\left(x_1\right)>f\left(x_2\right)$)

Let $f\left(1\right)=a;f\left(2\right)=9;f\left(3\right)=c$

$g\left(x\right)=\frac{1}{x-f\left(1\right)}+\frac{2}{x-f\left(2\right)}+\frac{3}{x-f\left(3\right)}=0$

$(x-b)(x-c)+2(x-a)(x-c)+3(x-b)(x-a)=0$

$(x\ne a,b,c)$

$3x^2-(b+c+2a+2c+3b+3a)x+(bc+2ac+3ab)=0$

$3x^2-(5a+4b+3c)x+(3ab+2ac+bc)=0$

$△={(5a+4b+3c)}^2-4\left(3\right)(3ab+bc+2ac)$

$=25a^2+16b^2+a{c}^2+4ab+6ac+12bc$

$\therefore △>0(\because a>b>c>0)$

$\therefore g\left(x\right)$ has two distinct real roots.