Let f : R rightarrow R and g : R rightarrow R | vedclass

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

Question

(A) Given $f(x) = \log_{e}(x^{2}+1) - e^{-x} + 1$. Calculating the derivative,$f'(x) = \frac{2x}{x^{2}+1} + e^{-x}$. Since $x^{2}+1 \geq 2|x|$,we have

Vedclass · Accepted Answer

$(2,3)$

Vedclass · Answer

(A) Given $f(x) = \log_{e}(x^{2}+1) - e^{-x} + 1$. Calculating the derivative,$f'(x) = \frac{2x}{x^{2}+1} + e^{-x}$. Since $x^{2}+1 \geq 2|x|$,we have $\frac{2x}{x^{2}+1} \geq -1$. Also $e^{-x} > 0$. Actually,$f'(x) = \frac{2x}{x^{2}+1} + e^{-x}$. For $x \geq 0$,$f'(x) > 0$. For $x  0$,then $f'(-t) = \frac{-2t}{t^{2}+1} + e^{t}$. Since $e^{t} > 1$ and $\frac{2t}{t^{2}+1} \leq 1$,$f'(x) > 0$ for all $x \in R$. Thus,$f$ is strictly increasing. Given $g(x) = \frac{1-2e^{2x}}{e^{x}} = e^{-x} - 2e^{x}$. Calculating the derivative,$g'(x) = -e^{-x} - 2e^{x} = -(e^{-x} + 2e^{x}) Thus,$g$ is strictly decreasing. Given the inequality $f(g(\frac{(\alpha-1)^{2}}{3})) > f(g(\alpha-\frac{5}{3}))$. Since $f$ is strictly increasing,this implies $g(\frac{(\alpha-1)^{2}}{3}) > g(\alpha-\frac{5}{3})$. Since $g$ is strictly decreasing,this implies $\frac{(\alpha-1)^{2}}{3} Multiplying by $3$,we get $(\alpha-1)^{2} $\alpha^{2} - 2\alpha + 1 $\alpha^{2} - 5\alpha + 6 $(\alpha-2)(\alpha-3) Therefore,$\alpha \in (2, 3)$.

Let $f : R \rightarrow R$ and $g : R \rightarrow R$ be two functions defined by $f(x)=\log _{e}(x^{2}+1)-e^{-x}+1$ and $g(x)=\frac{1-2e^{2x}}{e^{x}}$. Then,for which of the following range of $\alpha$,the inequality $f(g(\frac{(\alpha-1)^{2}}{3})) > f(g(\alpha-\frac{5}{3}))$ holds?

Similar Questions

If the graph of a non-constant function $f(x)$ is symmetric about the point $(3,4)$,then the value of $\sum\limits_{r = 0}^6 {f(r) + f(3)}$ is equal to

If $N$ denotes the set of all positive integers and if $f: N \rightarrow N$ is defined by $f(n) = \text{the sum of positive divisors of } n$, then $f(2^k \cdot 3)$, where $k$ is a positive integer, is

$f(x) = \frac{x}{\ln x}$ and $g(x) = \frac{\ln x}{x}$. Identify the $CORRECT$ statement.

If $R \subset A \times B$ and $S \subset B \times C$ be two relations,then $(S \circ R)^{-1} = $

The probability that a relation $R$ from $\{x, y\}$ to $\{x, y\}$ is both symmetric and transitive is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module