Q53 / 160 · AP EAMCET 2024 (18 May Shift 1) · Mathematics

$f(x+h)=0$ represents the transformed equation of the equation $f(x)=x^4+2 x^3-19 x^2-8 x+60=0$. If this transformation removes the term containing $x^3$ from $f(x)=0$, then $h=$

  • Correct answer
    A
    $-\frac{1}{2}$
  • B
    $1$
  • C
    $2$
  • D
    $-1$

Solution

Given, $f(x)=x^4+2 x^3-19 x^2-8 x+60=0$ Now, $f(x+h)=0$ $\begin{aligned} & \Rightarrow(x+h)^4+2(x+h)^3-19(x+h)^2-8(x+h)+60=0 \\ & \Rightarrow x^4+4 x^3 h+6 x^2 h^2+4 x h^3+h^4+2 x^3+2 h^3\end{aligned}$ $\begin{aligned} & +6 x h(x+h)-19(x+h)^2-8(x+h)+60\end{aligned}=0$ It does not contain $x^3$ So, $4 h+2=0 \Rightarrow h=\frac{-2}{4}=\frac{-1}{2}$

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

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