Q.18. The number of critical points of the function in the interval is equal to:

Q: What is the correct answer for this exam Mathematics Q18?

The correct answer is C. Sol. First, check continuity at $x=0$. $\lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{\sin x}{x} = 1$. And $f(0) = 1$. So $f(x)$ is continuous at $x=0$. Critical points occur where $f'(x)=0$ or $f'(x)$ is undefined. For $x \ne 0$, $f(x) = \frac{\sin x}{x}$. $f'(x) = \frac{x \cos x - \sin x}{x^2}$. Critical points for $x \ne 0$ are when $f'(x)=0$, which means $x \cos x - \sin x = 0$. $x \cos x = \sin x$

Q: Q.18. The number of critical points of the function in the interval is equal to:

The correct answer is C. Sol. First, check continuity at $x=0$. $\lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{\sin x}{x} = 1$. And $f(0) = 1$. So $f(x)$ is continuous at $x=0$. Critical points occur where $f'(x)=0$ or $f'(x)$ is undefined. For $x \ne 0$, $f(x) = \frac{\sin x}{x}$. $f'(x) = \frac{x \cos x - \sin x}{x^2}$. Critical points for $x \ne 0$ are when $f'(x)=0$, which means $x \cos x - \sin x = 0$. $x \cos x = \sin x$

Question

Q.18. The number of critical points of the function  in the interval  is equal to:

FirstInTest · Accepted Answer

Sol. First, check continuity at $x=0$.
$\lim_{x	o 0} f(x) = \lim_{x	o 0} \frac{\sin x}{x} = 1$.
And $f(0) = 1$. So $f(x)$ is continuous at $x=0$.
Critical points occur where $f'(x)=0$ or $f'(x)$ is undefined.
For $x 
e 0$, $f(x) = \frac{\sin x}{x}$.
$f'(x) = \frac{x \cos x - \sin x}{x^2}$.
Critical points for $x 
e 0$ are when $f'(x)=0$, which means $x \cos x - \sin x = 0$.
$x \cos x = \sin x$.
If $\cos x = 0$, then $\sin x = \pm 1$, so $x \cos x = 0$ but $\sin x 
e 0$. So $\cos x 
e 0$.
Divide by $\cos x$: $x = 	an x$.
We need to find the number of solutions to $x = 	an x$ in the interval $(-2\pi, 2\pi)$, excluding $x=0$ (since $f'(0)$ needs separate consideration).
Graphically, the solutions are the intersections of $y=x$ and $y=	an x$.
In $(0, 2\pi)$, $y=	an x$ has vertical asymptotes at $x=\frac{\pi}{2}$, $x=\frac{3\pi}{2}$.
$y=x$ intersects $y=	an x$ once in $(\frac{\pi}{2}, \frac{3\pi}{2})$ and once in $(\frac{3\pi}{2}, 2\pi)$. (2 solutions)
In $(-2\pi, 0)$, $y=	an x$ has vertical asymptotes at $x=-\frac{\pi}{2}$, $x=-\frac{3\pi}{2}$.
$y=x$ intersects $y=	an x$ once in $(-\frac{3\pi}{2}, -\frac{\pi}{2})$ and once in $(-2\pi, -\frac{3\pi}{2})$. (2 solutions)
So there are 4 solutions for $x 
e 0$ where $f'(x)=0$.
Now consider $x=0$. $f'(0)$ is undefined because the derivative of $\frac{\sin x}{x}$ at $x=0$ is not defined in the usual way (it's a removable singularity for $f'(x)$ if we define $f'(0)=0$). However, for critical points, we check where the derivative is 0 or undefined. The function $f(x)$ is continuous at $x=0$. The derivative $f'(x) = \frac{x\cos x - \sin x}{x^2}$. As $x 	o 0$, using L'Hopital's rule on the numerator and denominator separately:
$\lim_{x	o 0} \frac{\cos x - x\sin x - \cos x}{2x} = \lim_{x	o 0} \frac{-x\sin x}{2x} = \lim_{x	o 0} -\frac{\sin x}{2} = 0$.
So $f'(0)=0$. Thus $x=0$ is also a critical point.
Total critical points = 4 (from $x=	an x, x 
e 0$) + 1 (from $x=0$) = 5.

Let's verify the graph of $y=x$ and $y=	an x$. The line $y=x$ passes through the origin. The branches of $	an x$ are symmetric about the origin. The first positive intersection is between $\pi/2$ and $3\pi/2$. The second positive intersection is between $3\pi/2$ and $2\pi$. Similarly for negative values. So there are 2 positive solutions and 2 negative solutions. And $x=0$ is also a solution to $x=	an x$. So $x=0$ is a critical point where $f'(x)=0$. Total 5 critical points.

Answer

1

Answer

3

Answer

5

Answer

7

Related Questionsin Mathematics

Browse Engineering Questions by Subject

Practice Other Engineering Exams

Practice More Engineering Questions