Question

Let \([x]\) denotes the greatest integer less than or equal to x and \(f(x) = [tan^2x]\), then which of the following is true ?

• A. \(\underset{x\rightarrow 0}{lim}f(x)\) does not exist
• B. \(f(x)\) is continuous at \(x = 0\)
• C. \(f(x)\) is not differentiable at \(x = 0\)
• D. \(f^'(0) = 1\)

Hint:

Near 0, tan squared x lies between 0 and 1, so the floor is 0.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For \(f(x) = [\tan^2 x]\), find the values of \(\tan^2x\) near \(x = 0\).

Step 2: Detailed Explanation:
When \(|x| < \frac{\pi}{4}\), we have \(0 \le \tan^2x < 1\), so \([\tan^2x] = 0\). At \(x = 0\), \(f(0) = 0\). So \(f(x) = 0\) in a whole neighbourhood of 0.

Step 3: Conclusions:
\(\lim_{x \to 0} f(x) = 0 = f(0)\), so \(f\) is continuous at 0. Since \(f\) is constant near 0, it is also differentiable there with \(f'(0) = 0\).

Step 4: Check the options:
(A) is false because the limit exists and equals 0. (B) is true. (C) is false because \(f\) is differentiable at 0. (D) is false because \(f'(0) = 0\), not 1.

Final Answer:
The function is continuous at 0, option (B). \[ \boxed{\text{(B)}} \]