Q120 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics

If $f$ is defined on $\mathbb{R}$ such that $f(\mathrm{x}) f(-\mathrm{x})=9$, then $\int_{-23}^{23} \frac{d x}{3+f(x)}$

  • A
    $\frac {51}{3}$
  • B
    $\frac {49}{3}$
  • C
    $\frac {46}{3}$
  • Correct answer
    D
    $\frac {46}{6}$

Solution

$\begin{aligned} & f(x) \cdot f(-x)=9 \\ & I=\int_{-23}^{23} \frac{d x}{3+f(x)}... (i) \\ & I=\int_{-23}^{23} \frac{d x}{3+f(-x)} \quad(f(a+b-x) \text { property }) \\ & I=\int_{-23}^{23} \frac{d x}{3+\frac{9}{f(x)}}=\int_{-23}^{23} \frac{f(x) d x}{3(f(x)+3)}... (ii) \\ & \text { (i) }+ \text { (ii), } \\ & 2 I=\int_{-23}^{23}\left(\frac{d x}{3+f(x)}+\frac{f(x) d x}{3(3+f(x))}\right) \\ & =\int_{-23}^{23} \frac{(3+f(x)) d x}{3(3+f(x))} \\ & 2 I=\frac{1}{3} \int_{-23}^{23} 1 d x \Rightarrow I=\frac{1}{6}[x]_{-23}^{23}=\frac{46}{6} \\ & \end{aligned}$

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

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