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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
Let $\quad \mathrm{f}: \mathbb{R}-\{0\} \rightarrow \mathbb{R}$ be a function satisfying $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$ for all $x, y, f(y) \neq 0$. If $f^{\prime}(1)=2024$ then
  • ✓
    $\mathrm{xf}^{\prime}(\mathrm{x})-2024 \mathrm{f}(\mathrm{x})=0$
  • B
    $x f^{\prime}(x)-2024 f(x)=0$
  • C
    $\mathrm{xf}^{\prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})=2024$
  • D
    $x f^{\prime}(x)-2023 f(x)=0$
✓

Answer

Correct option: A.
$\mathrm{xf}^{\prime}(\mathrm{x})-2024 \mathrm{f}(\mathrm{x})=0$
a
$f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$

$\mathrm{f}^{\prime}(1)=2024$

${f}(1)=1$

Partially differentiating w. r. t. $x$

$f^{\prime}\left(\frac{x}{y}\right) \cdot \frac{1}{y}=\frac{1}{f(y)} f^{\prime}(x)$

$ y \rightarrow x $

$ f^{\prime}(1) \cdot \frac{1}{x}=\frac{f^{\prime}(x)}{f(x)} $

$ 2024 f(x)=x f^{\prime}(x) \Rightarrow x f^{\prime}(x)-2024 f(x)=0$

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