The number of solutions to sin x = frac{6}{x} | vedclass

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(C) To find the number of solutions to $\sin x = \frac{6}{x}$ in the interval $0 \leq x \leq 12 \pi$,we look for the points of intersection between th

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) To find the number of solutions to $\sin x = \frac{6}{x}$ in the interval $0 \leq x \leq 12 \pi$,we look for the points of intersection between the graphs of $y = \sin x$ and $y = \frac{6}{x}$.$1$. The function $y = \sin x$ oscillates between $-1$ and $1$ with a period of $2 \pi$.$2$. The function $y = \frac{6}{x}$ is a hyperbola that decreases as $x$ increases.$3$. For $x > 0$,the intersection points occur where $\sin x = \frac{6}{x}$. Since $|\sin x| \leq 1$,we must have $|\frac{6}{x}| \leq 1$,which implies $x \geq 6$.$4$. In the interval $[0, 12 \pi]$,the function $y = \frac{6}{x}$ intersects the peaks of the sine wave. By observing the graph,the curve $y = \frac{6}{x}$ intersects the sine wave twice in each period of $2 \pi$ starting from the first positive cycle where the value of $\sin x$ can reach $\frac{6}{x}$.$5$. Counting the intersection points from the graph,there are $10$ such points.Thus,the correct option is $C$.

List-$I$	List-$II$
$(A) \ r_1 r_2 \sqrt{\frac{4R-r_1-r_2}{r_1+r_2}}$	$1. \ b$
$(B) \ \frac{r_2(r_3+r_1)}{\sqrt{r_1r_2+r_2r_3+r_3r_1}}$	$2. \ a^2, b^2, c^2 \text{ are in } AP$
$(C) \ \frac{a}{c} = \frac{\sin(A-B)}{\sin(B-C)}$	$3. \ \Delta$
$(D) \ bc \cos^2 \frac{A}{2}$	$4. \ R r_1 r_2 r_3$
	$5. \ s(s-a)$

The number of solutions to $\sin x = \frac{6}{x}$ with $0 \leq x \leq 12 \pi$ is

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