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)

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 )

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