Question

Let \([x]\) represent the greatest integer function. If \(\lim_{x \to 0^+} \frac{\cos[x] - \cos(kx - [x])}{x^2} = 5\), then \(k =\)

• A. \(\sqrt{10}\)
• B. \(\sqrt{11}\)
• C. 3
• D. \(\frac{9}{2}\)

Hint:

For limits involving \(\cos x\) as \(x \to 0\), use \(\cos x \approx 1 - \frac{x^2}{2}\). For \(x \to 0^+\), \([x] = 0\).

Answer 1

The Correct Option is A

As \(x \to 0^+\), \(0<x<1\), so \([x] = 0\). The limit becomes: \[ \lim_{x \to 0^+} \frac{\cos 0 - \cos(kx - 0)}{x^2} = \lim_{x \to 0^+} \frac{1 - \cos(kx)}{x^2} \] Use the approximation for small \(u\): \(\cos u \approx 1 - \frac{u^2}{2}\). Let \(u = kx\): \[ 1 - \cos(kx) \approx 1 - \left(1 - \frac{(kx)^2}{2}\right) = \frac{k^2 x^2}{2} \] \[ \frac{1 - \cos(kx)}{x^2} \approx \frac{\frac{k^2 x^2}{2}}{x^2} = \frac{k^2}{2} \] \[ \lim_{x \to 0^+} \frac{1 - \cos(kx)}{x^2} = \frac{k^2}{2} \] Given the limit equals 5: \[ \frac{k^2}{2} = 5 \implies k^2 = 10 \implies k = \sqrt{10} \] Option (1) is correct. Options (2), (3), and (4) do not satisfy \(k^2 = 10\).