A) \[\frac{-2({{x}^{2}}-x-1).{{\sin }^{2}}\left( \frac{2x-1}{{{x}^{2}}+1} \right)}{{{({{x}^{2}}+1)}^{2}}}\]
B) \[\frac{-2({{x}^{2}}-x-1).{{\cos }^{2}}\left( \frac{2x-1}{{{x}^{2}}+1} \right)}{{{({{x}^{2}}+1)}^{2}}}\]
C) \[\frac{({{x}^{2}}-x-1).{{\sin }^{2}}\left( \frac{x-1}{{{x}^{2}}+1} \right)}{{{({{x}^{2}}+1)}^{2}}}\]
D) None of these
Correct Answer: A
Solution :
[a] \[\because y=f\left( \frac{2x-1}{{{x}^{2}}+1} \right)\And f'\left( x \right)=si{{n}^{2}}x\] \[\therefore \frac{dy}{dx}=f'\left( \frac{2x-1}{{{x}^{2}}+1} \right).\frac{d}{dx}=\left( \frac{2x-1}{{{x}^{2}}+1} \right)\] \[={{\sin }^{2}}\left( \frac{2x-1}{{{x}^{2}}+1} \right).\left\{ \frac{\left( {{x}^{2}}+1 \right).2.-\left( 2x-1 \right).2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \right\}\] \[={{\sin }^{2}}\left( \frac{2x-1}{{{x}^{2}}+1} \right).\left\{ \frac{2{{x}^{2}}+2-4{{x}^{2}}+2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \right\}\] \[={{\sin }^{2}}\left( \frac{2x-1}{{{x}^{2}}+1} \right).\left\{ \frac{-2{{x}^{2}}+2x+2}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \right\}\] \[={{\sin }^{2}}\left( \frac{2x-1}{{{x}^{2}}+1} \right).\left\{ \frac{-2\left( {{x}^{2}}-x-1 \right)}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \right\}\] Hence, option [a] is correct.
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