Question

Let $f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy)) \tan(xy)}{y^3}$. Then the number of solutions of the equation $f(x) = \sin x$, $x \in \mathbb{R}$ is :

• A. 0
• B. 2
• C. 3
• D. 1

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We first evaluate the limit to find the expression for $f(x)$ and then solve the trigonometric equation. 
: Key Formula or Approach: 
Standard limits: $\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$ and $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$. 
Step 2: Detailed Explanation: 
$f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy)) \tan(xy)}{y^3}$. 
Let $xy = \theta$. As $y \to 0$, $\theta \to 0$. 
$f(x) = \lim_{y \to 0} \left[ \frac{1 - \cos(xy)}{(xy)^2} \cdot \frac{\tan(xy)}{xy} \cdot \frac{x^3 y^3}{y^3} \right]$. 
$f(x) = \frac{1}{2} \cdot 1 \cdot x^3 = \frac{x^3}{2}$. 
Equation to solve: $\frac{x^3}{2} = \sin x \implies x^3 = 2 \sin x$. 
Let $g(x) = x^3 - 2\sin x$. 
$g(0) = 0$ (One solution is $x=0$). 
$g'(x) = 3x^2 - 2\cos x$. $g'(0) = -2 < 0$. 
As $x \to \infty$, $g(x) \to \infty$. By IVT, there is a root in $(0, \infty)$. 
Since $g(x)$ is an odd function, if $x_0$ is a root, $-x_0$ is also a root. 
Checking number of intersections of $y = x^3$ and $y = 2\sin x$: 
At $x=0$, slopes are 0 and 2. $2\sin x$ is initially above. 
Since $x^3$ eventually exceeds $2\sin x$ (which is bounded by 2), they must intersect once more in $x>0$. 
Total solutions: $x=0$, one positive, and one negative. Total = 3. 
Step 3: Final Answer: 
The number of solutions is 3.