If ⨍ '(x)=11+x2 for all x and ⨍ (0)=0, then :

f1(x)=f(2)-f(0)2-0=f(2)2 f(2)=21+x2 0<x<2 ⇒ 0<x2<4 1<1+x2<5 0.2<11+x2<1 0.4<21+x2<2 0.4<f(2)<2

If ⨍ '(x)=11+x2 for all x and ⨍ (0)=0, then :

f1(x)=f(2)-f(0)2-0=f(2)2 f(2)=21+x2 0<x<2 ⇒ 0<x2<4 1<1+x2<5 0.2<11+x2<1 0.4<21+x2<2 0.4<f(2)<2

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If $⨍^{'} (x)= \frac{1}{1 + x^{2}}$ for all x and $⨍ (0)=0,$ then :

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$⨍ (2)>2$

$0.4 < ⨍ (2)<2$

$⨍ (2)<0.4$

$⨍ (2)=2$

answer is C.

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$f^{1} (x) = \frac{f (2) - f (0)}{2 - 0} = \frac{f (2)}{2} f (2) = \frac{2}{1 + x^{2}} 0 < x < 2 \Rightarrow 0 < x^{2} < 4 1 < 1 + x^{2} < 5 0.2 < \frac{1}{1 + x^{2}} < 1 0.4 < \frac{2}{1 + x^{2}} < 2 0.4 < f (2) < 2$

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If ⨍ '(x)=11+x2 for all x and ⨍ (0)=0, then :

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If $⨍^{'} (x)= \frac{1}{1 + x^{2}}$ for all x and $⨍ (0)=0,$ then :