If f(x) = \(\left\{ \begin{aligned} & \frac{1-\cos Kx}{x\sin x} ,\ \ \text{if x}\neq0 \\ &\ \ \ \ \ \ \ \frac{1}{2}\ \ \ \ \ \ \ \ \ ,\ \ \ \text{if x=0} \end{aligned} \right.\) is continuous at x = 0, then the value of K is
If \(f(x) = \begin{cases} \frac{1 - \cos(Kx)}{x \sin x}, & \text{if } x \neq 0 \\ \frac{1}{2}, & \text{if } x = 0 \end{cases}\) is continuous at x = 0, then we need to find the value of K.
For f(x) to be continuous at x = 0, we must have \(\lim_{x \to 0} f(x) = f(0)\).
So, \(\lim_{x \to 0} \frac{1 - \cos(Kx)}{x \sin x} = \frac{1}{2}\).
We can use the trigonometric identity \(1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})\):
\(\lim_{x \to 0} \frac{2 \sin^2(\frac{Kx}{2})}{x \sin x} = \frac{1}{2}\)
\(\lim_{x \to 0} \frac{2 \sin^2(\frac{Kx}{2})}{x^2} \cdot \frac{x}{\sin x} = \frac{1}{2}\)
We know that \(\lim_{x \to 0} \frac{\sin x}{x} = 1\), so \(\lim_{x \to 0} \frac{x}{\sin x} = 1\).
\(\lim_{x \to 0} \frac{2 \sin^2(\frac{Kx}{2})}{x^2} \cdot 1 = \frac{1}{2}\)
\(2 \lim_{x \to 0} \frac{\sin^2(\frac{Kx}{2})}{x^2} = \frac{1}{2}\)
\(\lim_{x \to 0} \frac{\sin^2(\frac{Kx}{2})}{x^2} = \frac{1}{4}\)
\(\lim_{x \to 0} (\frac{\sin(\frac{Kx}{2})}{x})^2 = \frac{1}{4}\)
We can rewrite this as:
\(\lim_{x \to 0} (\frac{\sin(\frac{Kx}{2})}{\frac{Kx}{2}} \cdot \frac{K}{2})^2 = \frac{1}{4}\)
Since \(\lim_{x \to 0} \frac{\sin(\frac{Kx}{2})}{\frac{Kx}{2}} = 1\), we have:
\((1 \cdot \frac{K}{2})^2 = \frac{1}{4}\)
\(\frac{K^2}{4} = \frac{1}{4}\)
\(K^2 = 1\)
\(K = \pm 1\)
Therefore, the correct option is (D) \(\pm 1\).
For $ f(x) $ to be continuous at $ x = 0 $, we must have $ \lim_{x \to 0} f(x) = f(0) $.
We are given $ f(0) = \frac{1}{2} $.
We need to find $ \lim_{x \to 0} \frac{1 - \cos Kx}{x \sin x} $.
We can rewrite $ 1 - \cos Kx = 2 \sin^2 \frac{Kx}{2} $.
Then:
$$ \lim_{x \to 0} \frac{1 - \cos Kx}{x \sin x} = \lim_{x \to 0} \frac{2 \sin^2 \frac{Kx}{2}}{x \sin x}. $$
Using the small-angle approximation $ \sin u \approx u $ as $ u \to 0 $, we simplify:
$$ \lim_{x \to 0} \frac{2 \sin^2 \frac{Kx}{2}}{x \sin x} = \lim_{x \to 0} \frac{2 \left( \frac{Kx}{2} \right)^2}{x^2}. $$
Simplify further:
$$ \lim_{x \to 0} \frac{2 \frac{K^2 x^2}{4}}{x^2} = \frac{K^2}{2}. $$
We want $ \frac{K^2}{2} = \frac{1}{2} $, so:
$$ K^2 = 1, $$
which implies $ K = \pm 1 $.
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He applied a resistance of \( 520 \, \Omega \) in \( R \). When \( K_1 \) is closed and \( K_2 \) is open, the deflection observed in the galvanometer is 20 div. When \( K_1 \) is also closed and a resistance of \( 90 \, \Omega \) is removed in \( S \), the deflection becomes 13 div. The resistance of galvanometer is nearly: