Let f, g and h be real-valued functions defin | vedclass

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

Question

(D) We are given the functions $f(x)=e^{x^2}+e^{-x^2}$,$g(x)=x e^{x^2}+e^{-x^2}$,and $h(x)=x^2 e^{x^2}+e^{-x^2}$ on the interval $[0,1]$.For $x \in [0

Vedclass · Accepted Answer

$a=b=c$

Vedclass · Answer

(D) We are given the functions $f(x)=e^{x^2}+e^{-x^2}$,$g(x)=x e^{x^2}+e^{-x^2}$,and $h(x)=x^2 e^{x^2}+e^{-x^2}$ on the interval $[0,1]$.For $x \in [0,1]$,we have $0 \leq x^2 \leq x \leq 1$.Since $e^{x^2} > 0$ and $e^{-x^2} > 0$,we compare the functions:$f(x) - g(x) = e^{x^2} + e^{-x^2} - x e^{x^2} - e^{-x^2} = e^{x^2}(1-x) \geq 0$ for $x \in [0,1]$. Thus $f(x) \geq g(x)$.$g(x) - h(x) = x e^{x^2} + e^{-x^2} - x^2 e^{x^2} - e^{-x^2} = e^{x^2}(x-x^2) = x e^{x^2}(1-x) \geq 0$ for $x \in [0,1]$. Thus $g(x) \geq h(x)$.Therefore,$f(x) \geq g(x) \geq h(x)$ for all $x \in [0,1]$.At $x=1$,$f(1) = e^1 + e^{-1} = e + \frac{1}{e}$,$g(1) = 1 \cdot e^1 + e^{-1} = e + \frac{1}{e}$,and $h(1) = 1^2 \cdot e^1 + e^{-1} = e + \frac{1}{e}$.Since $f(x)$,$g(x)$,and $h(x)$ are increasing functions on $[0,1]$ (as their derivatives are non-negative),their maximum values occur at $x=1$.Thus,$a = f(1) = e + \frac{1}{e}$,$b = g(1) = e + \frac{1}{e}$,and $c = h(1) = e + \frac{1}{e}$.Hence,$a=b=c$.

Let $f, g$ and $h$ be real-valued functions defined on the interval $[0,1]$ by $f(x)=e^{x^2}+e^{-x^2}$,$g(x)=x e^{x^2}+e^{-x^2}$ and $h(x)=x^2 e^{x^2}+e^{-x^2}$. If $a, b$ and $c$ denote,respectively,the absolute maximum of $f, g$ and $h$ on $[0,1]$,then

Similar Questions

The maximum area of a triangle whose one vertex is at $(0,0)$ and the other two vertices lie on the curve $y = -2x^2 + 54$ at points $(x, y)$ and $(-x, y)$ where $y > 0$ is:

If $f(x)$ is a function such that $f^{\prime}(x)=(x-1)^{2}(4-x),$ then

$20 \ m$ of wire is available to fence a flower bed in the form of a circular sector. If the flower bed should have the greatest possible surface area,then the radius of the circle is (in $m$)

On the interval $[0, 1]$, the function $f(x) = x^{25}(1 - x)^{75}$ attains its maximum value at the point $x = ....$

Let $f : (-\infty, \infty) \to (-\infty, \infty)$ be defined by $f(x) = x^3 + 1$.
Statement-$1$: The function has a local extremum at $x = 0$.
Statement-$2$: The function $f(x)$ is continuous and differentiable on $(-\infty, \infty)$ and $f'(0) = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module