Q53 / 160 · AP EAMCET 2023 (15 May Shift 1) · Mathematics

If -1 is a twice repeated root of the equation $a\left(x^3+x^2\right)+$ $\mathrm{bx}+\mathrm{c}=0$, then $\mathrm{a}: \mathrm{b}: \mathrm{c}=$

  • A
    $:-1: 1$
  • Correct answer
    B
    $-1: 1: 1$
  • C
    $1: 1:-1$
  • D
    $1: 1: 1$

Solution

Since $x=1$ is a twice repeated root of $f(x)=a x^3+a x^2$ $+b x+c=0$. Hence $f(-1)=0$ and $f^{\prime}(-1)=0$ Now $f(-1)=0$ $$ \Rightarrow-a+a+b+c=0 \Rightarrow c=b $$ Now $f^{\prime}(x)=3 a x^2+2 a x+b$ $$ \begin{aligned} & \Rightarrow f^{\prime}(-1)=3 a-2 a+b \\ & \Rightarrow a+b=0 \\ & \Rightarrow a=-b \\ & \therefore a: b: c=(-b):(b):(b) \\ & =-1: 1: 1 \end{aligned} $$

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

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