get app

Go To Your Study Dashboard

Let $f(x)=e^{x}-x$ and $g(x)=x^{2}-x$ , $\vee x\epsilon R$ .

Then the set of all $x\epsilon R$ , where the function $h(x)=(fog)(x)$

is increasing , is :

Option 1)
$[-1,\frac{-1}{2}]\cup [\frac{1}{2},\infty )$

Option 2)
$[0,\frac{1}{2}]\cup [1,\infty )$

Option 3)
$[0,\infty )$

Option 4)
$[\frac{-1}{2},0]\cup [1,\infty )$

Answers (1)

V Vakul

$f(x)=e^{x}-x$

$g(x)=x^{2}-x$

$h(x)=fog(x)=e^{g(x)}-g(x)$

$h(x)=e^{x^{2}-x}-x^{2}+x$

$h'(x)=e^{x^{2}-x}(2x-1)-2x+1=(2x-1)e^{x^{2}-1}$

$\because f(x)\: is\: \: increasing\: \: h'(x)\geq 0$

Case 1 : $(2x-1)\geq 0 \: \:and\: \: x^{2}-x\geq 0$

$x\geq \frac{1}{2}\: \: \: and\: \: \: x\epsilon(-\infty,0]\cup [1,\infty)$

Case 2 : $(2x-1)\leq 0 \: \:and\: \: x^{2}-x\leq 0$

$x\leq \frac{1}{2}\: \: \: and\: \: \: x\epsilon[0,1]$

=> $x\epsilon [0,\frac{1}{2}]$

$\therefore$ $x\epsilon [0,\frac{1}{2}]\cup [1,\infty )$

correct option (2)

Option 1)

$[-1,\frac{-1}{2}]\cup [\frac{1}{2},\infty )$

Option 2)

$[0,\frac{1}{2}]\cup [1,\infty )$

Option 3)

$[0,\infty )$

Option 4)

$[\frac{-1}{2},0]\cup [1,\infty )$

Similar Questions

Latest Asked Questions

Most Viewed Questions

Preparation Products

JEE Main Rank Booster 2022

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, 24x7 Doubt Chat Support,.

₹ 13999/- ₹ 9999/-

Buy Now

Knockout JEE Main 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes(Classes will start from Last week of September), Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 24999/-

Buy Now

Test Series JEE Main 2022

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 4999/-

Buy Now

Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 2/-

Buy Now

Knockout JEE Main 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes(Classes will start from Last week of September), Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 24999/-

Buy Now

Boost your Preparation for JEE Main 2021 with Personlized Coaching

Start Now

Exams

Articles

Questions