Let f(x) and g(x) be two continuous functions | vedclass

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

Question

(C) Given that $f(x_1) > f(x_2)$ for all $x_1 > x_2$,$f$ is a strictly increasing function.Given that $g(x_1)  x_2$,$g$ is a s

Vedclass · Accepted Answer

$(1, 4)$

Vedclass · Answer

(C) Given that $f(x_1) > f(x_2)$ for all $x_1 > x_2$,$f$ is a strictly increasing function.Given that $g(x_1)  x_2$,$g$ is a strictly decreasing function.Since $f$ is strictly increasing,the inequality $f(g(\alpha^2 - 2\alpha)) > f(g(3\alpha - 4))$ implies $g(\alpha^2 - 2\alpha) > g(3\alpha - 4)$.Since $g$ is strictly decreasing,the inequality $g(\alpha^2 - 2\alpha) > g(3\alpha - 4)$ implies $\alpha^2 - 2\alpha Rearranging the terms,we get $\alpha^2 - 5\alpha + 4 Factoring the quadratic expression,we have $(\alpha - 1)(\alpha - 4) This inequality holds when $\alpha$ lies between the roots $1$ and $4$.Thus,the solution set is $\alpha \in (1, 4)$.

Let $f(x)$ and $g(x)$ be two continuous functions defined from $R \rightarrow R$,such that $f(x_1) > f(x_2)$ and $g(x_1) < g(x_2)$ for all $x_1 > x_2$. Then the solution set of $f(g(\alpha^2 - 2\alpha)) > f(g(3\alpha - 4))$ is

Similar Questions

Let $f(x) = a^x$ $(a > 0)$ be written as $f(x) = f_1(x) + f_2(x)$,where $f_1(x)$ is an even function and $f_2(x)$ is an odd function. Then $f_1(x + y) + f_1(x - y)$ equals

Let $[x]$ denote the integral part of $x \in R$. Let $g(x) = x - [x]$. Let $f(x)$ be any continuous function with $f(0) = f(1)$. Then the function $h(x) = f(g(x))$:

If $R \subset A \times B$ and $S \subset B \times C$,then the relation $(SoR)^{-1} = $

If $f(x) = \cos(\log x)$,then the value of $f(x^2) \cdot f(y^2) - \frac{1}{2} \left[ f\left(\frac{x^2}{y^2}\right) + f(x^2 y^2) \right]$ is:

The number of solutions of the equation $2^x = x^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module