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 )

Preparation Products

Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2021

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

₹ 13999/- ₹ 9999/-
Buy Now
Test Series JEE Main April 2022

Take chapter-wise, subject-wise and Complete syllabus mock tests and get an in-depth analysis of your test..

₹ 6999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions