Let f: R rightarrow R be given byf(x) = begin | vedclass

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

Question

(B) First,we analyze the function $f(x)$ in different intervals.For $x < 0$,$f(x) = (x+1)^5 - 2x$. As $x \rightarrow -\infty$,$f(x) \rightarrow -\inft

Vedclass · Accepted Answer

$1, 2, 4$

Vedclass · Answer

(B) First,we analyze the function $f(x)$ in different intervals.For $x $f^{\prime}(x) = 5(x+1)^4 - 2$. Setting $f^{\prime}(x) = 0$ gives $(x+1)^4 = 2/5$,so $x = -1 \pm (2/5)^{1/4}$. Since $f^{\prime}(x)$ changes sign in $(-\infty, 0)$,$f(x)$ is not monotonic on $(-\infty, 0)$. Thus,statement $(3)$ is incorrect.For $x \geq 3$,$f(x)$ is continuous,$f(3) = 1/3$,and $\lim_{x \rightarrow \infty} f(x) = \infty$. The range is $[1/3, \infty)$.Combining the ranges of all intervals,we find that the range of $f(x)$ is $R$,so $f$ is onto. Statement $(2)$ is correct.Now,consider $f^{\prime}(x)$ for $x$ near $1$:For $0 \leq x For $1 \leq x For $x$ slightly greater than $1$,$f^{\prime}(x) = 2x^2 - 8x + 7$. The derivative of $f^{\prime}(x)$ is $f^{\prime\prime}(x) = 4x - 8$. At $x=1$,$f^{\prime\prime}(1^+) = -4$.Since the left-hand derivative of $f^{\prime}$ at $x=1$ is $2$ and the right-hand derivative is $-4$,$f^{\prime}$ is not differentiable at $x=1$. Statement $(4)$ is correct.Also,since $f^{\prime}(x)$ increases to $1$ at $x=1$ and decreases for $x > 1$,$f^{\prime}$ has a local maximum at $x=1$. Statement $(1)$ is correct.Therefore,statements $(1), (2),$ and $(4)$ are correct.

Similar Questions

Let $f: R \rightarrow R$ be defined as $f(x)=\begin{cases} \frac{a-b \cos 2 x}{x^2} & ; x<0 \\ x^2+c x+2 & ; 0 \leq x \leq 1 \\ 2 x+1 & ; x>1 \end{cases}$. If $f$ is continuous everywhere in $R$ and $m$ is the number of points where $f$ is $NOT$ differentiable,then $m+a+b+c$ equals:

If $f(x) = \begin{cases} \int_{0}^{x} (5 + |1-t|) \, dt, & x > 2 \\ 5x + 1, & x \leq 2 \end{cases}$,then:

If $f(x) = \begin{cases} ax^2 - bx + 2, & x < 3 \\ bx^2 - 3, & x \geq 3 \end{cases}$ is differentiable at every $x \in R$,then the area (in sq units) of the triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ with the coordinate axes is

The function $f(x) = \begin{cases} \frac{\pi}{4} + \tan^{-1} x, & |x| \leq 1 \\ \frac{1}{2}(|x|-1), & |x| > 1 \end{cases}$ is:

If $f(x) = \begin{cases} 3x^2 + 12x - 1, & -1 \le x \le 2 \\ 37 - x, & 2 < x \le 3 \end{cases}$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module