Let f(x) = min {1, 1 + x sin x } for 0 leq x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We are given $f(x) = \min \{1, 1 + x \sin x \}$.This means $f(x) = 1$ when $1 + x \sin x \geq 1$,i.e.,$x \sin x \geq 0$,and $f(x) = 1 + x \sin x$

Vedclass · Accepted Answer

$(1, 0)$

Vedclass · Answer

(B) We are given $f(x) = \min \{1, 1 + x \sin x \}$.This means $f(x) = 1$ when $1 + x \sin x \geq 1$,i.e.,$x \sin x \geq 0$,and $f(x) = 1 + x \sin x$ when $x \sin x In the interval $[0, 2\pi]$,$x \sin x \geq 0$ for $x \in [0, \pi]$ and $x \sin x Thus,$f(x) = 1$ for $x \in [0, \pi]$ and $f(x) = 1 + x \sin x$ for $x \in (\pi, 2\pi]$.At $x = \pi$,$f(\pi) = 1$. The left-hand limit is $\lim_{x 	o \pi^-} f(x) = 1$ and the right-hand limit is $\lim_{x 	o \pi^+} f(x) = 1 + \pi \sin(\pi) = 1$.Since the limits are equal,$f(x)$ is continuous at $x = \pi$,so $n = 0$.Now,check for differentiability at $x = \pi$:$f'(x) = 0$ for $x For $x > \pi$,$f'(x) = \sin x + x \cos x$. As $x 	o \pi^+$,$f'(x) 	o \sin(\pi) + \pi \cos(\pi) = 0 + \pi(-1) = -\pi$.Since the left-hand derivative $(0)$ is not equal to the right-hand derivative $(-\pi)$,the function is not differentiable at $x = \pi$.Thus,$m = 1$.The ordered pair $(m, n) = (1, 0)$.

Let $f(x) = \min \{1, 1 + x \sin x \}$ for $0 \leq x \leq 2\pi$. If $m$ is the number of points where $f$ is not differentiable and $n$ is the number of points where $f$ is not continuous,then the ordered pair $(m, n)$ is equal to

Similar Questions

If $f(x)= \begin{cases} \frac{x-[x]}{x-2}, & x>2 \\ b, & x=2 \\ \frac{|x^2-x-2|}{a(2+x-x^2)}, & -1 < x \leq 2 \\ 2a-b, & x \leq -1 \end{cases}$ is continuous on $R$,then $\lim _{x \rightarrow 0} \frac{\sin ^2 ax+x \tan bx}{x^2}=$

If $f(x)=\begin{cases} \frac{x-3}{|x-3|}+a & , x<3 \\ a+b & , x=3 \\ \frac{|x-3|}{x-3}+b & , x>3 \end{cases}$ is continuous at $x=3$,then the value of $a-b$ is

Let $f : [-1,3] \to R$ be defined as $f(x) = \begin{cases} |x| + [x], & -1 \leq x < 1 \\ x + |x|, & 1 \leq x < 2 \\ x + |x|, & 2 \leq x \leq 3 \end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$. Then,$f$ is discontinuous at:

Let $f(x) = \begin{cases} 2 - |x^2 + 5x + 6|, & x \neq -2 \\ a^2 + 1, & x = -2 \end{cases}$. Then the range of $a$ such that $f(x)$ has a maximum at $x = -2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module