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Functions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSFunctions1 Mark
MCQ
If $\text{e}^{\text{f(x)}}=\frac{10+\text{x}}{10-\text{x}},\text{ x}\in(-10,10)$ and $\text{f(x)}=\text{kf}\Big(\frac{200\text{x}}{100+\text{x}^2}\Big),$ then k =
  • ✓
    0.5
  • B
    0.6
  • C
    0.7
  • D
    0.8
✓

Answer

Correct option: A.
0.5
$\text{e}^{\text{f(x)}}=\frac{10+\text{x}}{10-\text{x}}$
$\Rightarrow\text{ f(x)}=\log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)\ ...(\text{i})$
$\Rightarrow\ \text{f(x)}=\text{kf}\Big(\frac{200\text{x}}{100+\text{x}^2}\Big)$
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\Bigg(\frac{10+\frac{200\text{x}}{100+\text{x}^2}}{10-\frac{200\text{x}}{100+\text{x}^2}}\Bigg)$ {from (1)}
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\Big(\frac{1000+10\text{x}^2+200\text{x}}{1000+10\text{x}^2-200\text{x}}\Big)$
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\bigg(\frac{(\text{x}+10)^2}{(\text{x}-10)^2}\bigg)$
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=2\text{k}\log_\text{e}\frac{(\text{x}+10)}{(\text{x}+10)}$
$\Rightarrow\ 1=2\text{k}$
$\Rightarrow\ \text{k}=\frac{1}{2}=0.5$

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