Q82 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

By shifting the origin to the point $(h, 5)$ by the translation of coordinate axes, if the equation $y=x^3-9 x^2+c x-d$ transforms to $\mathrm{Y}=X^3$, then $\left(d-\frac{c}{h}\right)=$

  • A
    0
  • Correct answer
    B
    13
  • C
    11
  • D
    25

Solution

Shifting origin to $(h, 5)$ $\mathrm{x}=\mathrm{X}-\mathrm{h}, \mathrm{y}=\mathrm{Y}-5$ Equation $y=x^3-9 x^2+c x-c$ becomes $\begin{aligned} & \mathrm{Y}-5=(\mathrm{X}-h)^3-9(\mathrm{X}-h)^2+\mathrm{c}(\mathrm{X}-h)-d \\ & \mathrm{Y}-5=\mathrm{X}^3-(3 h+9) \mathrm{X}^2+\left(3 h^2+18 h+\mathrm{c}\right) \mathrm{X}-h^3 \\ & -9 h^2-\mathrm{c} h-d \\ & \mathrm{Y}=\mathrm{X}^3-(3 h+9) \mathrm{X}^2+\left(3 h^2+18 h+\mathrm{c}\right) \mathrm{X}-h^3-9 h^2- \\ & \mathrm{c} / \mathrm{h}-d+5\end{aligned}$ Comparing with $\mathrm{Y}=\mathrm{X}^3 ; 3 h=-9 \Rightarrow h=-3$ $\begin{aligned} & 3 h^2+18 h+c=0 \Rightarrow \mathrm{c}=27 \\ & -h^9-9 h^2-c h-d+5=0 \Rightarrow a l=4\end{aligned}$ $S O\left(d-\frac{c}{h}\right)=4+9=13$

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

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