Q43 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

If $f(x)$ is a real valued function defined by $f(x)=\frac{a x^{10}+b x^8+c x^6+d x^4+e x^2+12 x+15}{x}$ $(x \neq 0)$ and $f(4)=-4$, then $f(-4)=$

  • Correct answer
    A
    28
  • B
    39
  • C
    4
  • D
    24

Solution

$f(x)=\frac{a x^{10}+b x^8+c x^6+d x^4+e x^2+12 x+15}{x}$ and $f(4)=-4$ $\because f(x)=a x^9+b x^7+c x^5+d x^3+e x+12+\frac{15}{x}$ $\therefore f(-x)=-a x^9-b x^7-c x^5-d x^3-e x+12-\frac{15}{x}$ $\therefore f(x)=f(-x)=24$ $\begin{aligned} & f(4)+f(-4)=24 \\ & f(-4)=24-f(4) \\ & =24-(-4)=28\end{aligned}$

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

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