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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)

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

Posted by

Vakul

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