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If \(f(x)=\begin{cases}\frac{\sin^{2}ax}{x^{2}},&x\ne0\\ 1,&x=0\end{cases}\) is continuous at \(x=0\), then the value of a is:
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Q: If \(f(x)=\begin{cases}\frac{\sin^{2}ax}{x^{2}},&x\ne0\\ 1,&x=0\end{cases}\) is continuous at \(x=0\), then the value of a is:

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Detailed Solution

Step 1: Understanding Continuity

For a function to be continuous at a point, the limit of the function as it approaches that point must be equal to the function's value at that point. In this case, we need to ensure that $$\lim_{x \to 0} f(x) = f(0)$$.

Step 2: Evaluating the Limit

We need to find the limit of \(f(x)\) as \(x\) approaches 0. Since \(f(x) = \frac{\sin^{2}ax}{x^{2}}\) for \(x \ne 0\), we evaluate the following limit: $$\lim_{x \to 0} \frac{\sin^{2}ax}{x^{2}}$$

Step 3: Manipulating the Limit

We can rewrite the expression as: $$\lim_{x \to 0} \frac{\sin^{2}ax}{x^{2}} = \lim_{x \to 0} \left(\frac{\sin ax}{x}\right)^{2}$$ To solve this, we can multiply and divide by \(a\) inside the sine function: $$\lim_{x \to 0} \left(\frac{\sin ax}{ax} \cdot a\right)^{2} = \lim_{x \to 0} a^{2} \left(\frac{\sin ax}{ax}\right)^{2}$$

Step 4: Applying the Standard Limit

We know that $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$. Therefore, $$\lim_{x \to 0} \frac{\sin ax}{ax} = 1$$. So, the limit becomes: $$a^{2} \left(\lim_{x \to 0} \frac{\sin ax}{ax}\right)^{2} = a^{2} (1)^{2} = a^{2}$$

Step 5: Applying the Continuity Condition

For \(f(x)\) to be continuous at \(x = 0\), we must have: $$\lim_{x \to 0} f(x) = f(0)$$ $$a^{2} = 1$$

Step 6: Solving for a

Solving the equation \(a^{2} = 1\), we get: $$a = \pm 1$$

Final Answer: \(\pm1\)

AI Suggestion: Option C
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