Q103 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
A function $f: R \rightarrow R$ is such that $y f(x+y)+\cos m x y=1+y f(x)$. If $m=2$, then $f^1(x)=$
- A$-2 \sin 2 x y$
- B$4 x$
- C$\frac{2 \sin 2 x y}{y}$
- Correct answerD$2 x^2$
Solution
$\begin{aligned}
& \text y(f(x+y)-f(x))=1-\cos m x y \\
& \Rightarrow f(x+y)-f(x)=\frac{1-\cos 2 x y}{y} \\
& \Rightarrow \frac{f(x+y)-f(x)}{y}=\frac{2 \sin ^2 x y}{y^2}
\end{aligned}$
Let $y \rightarrow 0$ $\Rightarrow \frac{f(x+y)-f(x)}{y}=2\left(\frac{\sin x y}{x y}\right) \times x^2$ $\begin{aligned} & \Rightarrow \text { LHS }=\lim _{y \rightarrow 0} \frac{f(x+y)-f(x)}{y}=f^{\prime}(x) \\ & \Rightarrow \text { RHS }=\lim _{y \rightarrow 0}\left(2 x^2\left(\frac{\sin x y}{x y}\right)^2\right)=2 x^2 .\end{aligned}$
Let $y \rightarrow 0$ $\Rightarrow \frac{f(x+y)-f(x)}{y}=2\left(\frac{\sin x y}{x y}\right) \times x^2$ $\begin{aligned} & \Rightarrow \text { LHS }=\lim _{y \rightarrow 0} \frac{f(x+y)-f(x)}{y}=f^{\prime}(x) \\ & \Rightarrow \text { RHS }=\lim _{y \rightarrow 0}\left(2 x^2\left(\frac{\sin x y}{x y}\right)^2\right)=2 x^2 .\end{aligned}$
Hence, the correct answer is option: 'D'.