A) \[\frac{\pi }{3}\]
B) \[-\frac{\pi }{6}\]
C) \[\frac{2\pi }{3}\]
D) \[\frac{5\pi }{6}\]
Correct Answer: A , B , C , D
Solution :
| (Bonus) |
| \[f(x)={{\left\{ 2\sin \left( \frac{{{\tan }^{-1}}x+{{\cot }^{-1}}x}{2} \right)\cos \left( \frac{{{\tan }^{-1}}x-{{\cot }^{-1}}x}{2} \right) \right\}}^{2}}-1\]\[={{\left\{ 2\frac{1}{\sqrt{2}}\cos \left( \frac{\pi }{4}-{{\cos }^{-1}}x \right) \right\}}^{2}}-1\] |
| \[={{\left( \sqrt{2}\left( \frac{1}{\sqrt{2}}\cos (co{{t}^{-1}}x)+\frac{1}{\sqrt{2}}\sin (co{{t}^{-1}}x) \right) \right)}^{2}}-1\] |
| \[=1+2\sin (co{{t}^{-1}}x)cos(co{{t}^{-1}}x)-1\] |
| \[f(x)=sin(2co{{t}^{-1}}x)=sin(2ta{{n}^{-1}}x)=sin2\theta \] |
| As \[\left| x \right|<|{{\tan }^{-1}}x<-\frac{\pi }{4}or{{\tan }^{-1}}x>\frac{\pi }{4}\] |
| \[\therefore 2<-\frac{\pi }{2}\,\text{or}\,2\theta >\frac{\pi }{2}\] |
| Hence \[{{\sin }^{-1}}(f(x))=si{{n}^{-1}}(sin(2ta{{n}^{-1}}x))=\pi -2ta{{n}^{-1}}x\] |
| or \[-\pi -2{{\tan }^{-1}}x\] |
| \[\therefore \frac{dy}{dx}=\frac{1}{2}\cdot \frac{-2}{1+{{x}^{2}}}\Rightarrow y={{\cot }^{-1}}x+c\] |
| or \[y(\sqrt{3})=\frac{\pi }{6}\] we get c = 0 |
| \[\therefore y={{\cot }^{-1}}x\,\,\text{if}\,\,x>1\] |
| But if \[x<-1\] |
| \[y={{\cot }^{-1}}x+d\] |
| Where constant 'd' cannot be found to less data given. |
Solution :
| (Bonus) |
| \[f(x)={{\left\{ 2\sin \left( \frac{{{\tan }^{-1}}x+{{\cot }^{-1}}x}{2} \right)\cos \left( \frac{{{\tan }^{-1}}x-{{\cot }^{-1}}x}{2} \right) \right\}}^{2}}-1\]\[={{\left\{ 2\frac{1}{\sqrt{2}}\cos \left( \frac{\pi }{4}-{{\cos }^{-1}}x \right) \right\}}^{2}}-1\] |
| \[={{\left( \sqrt{2}\left( \frac{1}{\sqrt{2}}\cos (co{{t}^{-1}}x)+\frac{1}{\sqrt{2}}\sin (co{{t}^{-1}}x) \right) \right)}^{2}}-1\] |
| \[=1+2\sin (co{{t}^{-1}}x)cos(co{{t}^{-1}}x)-1\] |
| \[f(x)=sin(2co{{t}^{-1}}x)=sin(2ta{{n}^{-1}}x)=sin2\theta \] |
| As \[\left| x \right|<|{{\tan }^{-1}}x<-\frac{\pi }{4}or{{\tan }^{-1}}x>\frac{\pi }{4}\] |
| \[\therefore 2<-\frac{\pi }{2}\,\text{or}\,2\theta >\frac{\pi }{2}\] |
| Hence \[{{\sin }^{-1}}(f(x))=si{{n}^{-1}}(sin(2ta{{n}^{-1}}x))=\pi -2ta{{n}^{-1}}x\] |
| or \[-\pi -2{{\tan }^{-1}}x\] |
| \[\therefore \frac{dy}{dx}=\frac{1}{2}\cdot \frac{-2}{1+{{x}^{2}}}\Rightarrow y={{\cot }^{-1}}x+c\] |
| or \[y(\sqrt{3})=\frac{\pi }{6}\] we get c = 0 |
| \[\therefore y={{\cot }^{-1}}x\,\,\text{if}\,\,x>1\] |
| But if \[x<-1\] |
| \[y={{\cot }^{-1}}x+d\] |
| Where constant 'd' cannot be found to less data given. |
Solution :
| (Bonus) |
| \[f(x)={{\left\{ 2\sin \left( \frac{{{\tan }^{-1}}x+{{\cot }^{-1}}x}{2} \right)\cos \left( \frac{{{\tan }^{-1}}x-{{\cot }^{-1}}x}{2} \right) \right\}}^{2}}-1\]\[={{\left\{ 2\frac{1}{\sqrt{2}}\cos \left( \frac{\pi }{4}-{{\cos }^{-1}}x \right) \right\}}^{2}}-1\] |
| \[={{\left( \sqrt{2}\left( \frac{1}{\sqrt{2}}\cos (co{{t}^{-1}}x)+\frac{1}{\sqrt{2}}\sin (co{{t}^{-1}}x) \right) \right)}^{2}}-1\] |
| \[=1+2\sin (co{{t}^{-1}}x)cos(co{{t}^{-1}}x)-1\] |
| \[f(x)=sin(2co{{t}^{-1}}x)=sin(2ta{{n}^{-1}}x)=sin2\theta \] |
| As \[\left| x \right|<|{{\tan }^{-1}}x<-\frac{\pi }{4}or{{\tan }^{-1}}x>\frac{\pi }{4}\] |
| \[\therefore 2<-\frac{\pi }{2}\,\text{or}\,2\theta >\frac{\pi }{2}\] |
| Hence \[{{\sin }^{-1}}(f(x))=si{{n}^{-1}}(sin(2ta{{n}^{-1}}x))=\pi -2ta{{n}^{-1}}x\] |
| or \[-\pi -2{{\tan }^{-1}}x\] |
| \[\therefore \frac{dy}{dx}=\frac{1}{2}\cdot \frac{-2}{1+{{x}^{2}}}\Rightarrow y={{\cot }^{-1}}x+c\] |
| or \[y(\sqrt{3})=\frac{\pi }{6}\] we get c = 0 |
| \[\therefore y={{\cot }^{-1}}x\,\,\text{if}\,\,x>1\] |
| But if \[x<-1\] |
| \[y={{\cot }^{-1}}x+d\] |
| Where constant 'd' cannot be found to less data given. |
Solution :
| (Bonus) |
| \[f(x)={{\left\{ 2\sin \left( \frac{{{\tan }^{-1}}x+{{\cot }^{-1}}x}{2} \right)\cos \left( \frac{{{\tan }^{-1}}x-{{\cot }^{-1}}x}{2} \right) \right\}}^{2}}-1\]\[={{\left\{ 2\frac{1}{\sqrt{2}}\cos \left( \frac{\pi }{4}-{{\cos }^{-1}}x \right) \right\}}^{2}}-1\] |
| \[={{\left( \sqrt{2}\left( \frac{1}{\sqrt{2}}\cos (co{{t}^{-1}}x)+\frac{1}{\sqrt{2}}\sin (co{{t}^{-1}}x) \right) \right)}^{2}}-1\] |
| \[=1+2\sin (co{{t}^{-1}}x)cos(co{{t}^{-1}}x)-1\] |
| \[f(x)=sin(2co{{t}^{-1}}x)=sin(2ta{{n}^{-1}}x)=sin2\theta \] |
| As \[\left| x \right|<|{{\tan }^{-1}}x<-\frac{\pi }{4}or{{\tan }^{-1}}x>\frac{\pi }{4}\] |
| \[\therefore 2<-\frac{\pi }{2}\,\text{or}\,2\theta >\frac{\pi }{2}\] |
| Hence \[{{\sin }^{-1}}(f(x))=si{{n}^{-1}}(sin(2ta{{n}^{-1}}x))=\pi -2ta{{n}^{-1}}x\] |
| or \[-\pi -2{{\tan }^{-1}}x\] |
| \[\therefore \frac{dy}{dx}=\frac{1}{2}\cdot \frac{-2}{1+{{x}^{2}}}\Rightarrow y={{\cot }^{-1}}x+c\] |
| or \[y(\sqrt{3})=\frac{\pi }{6}\] we get c = 0 |
| \[\therefore y={{\cot }^{-1}}x\,\,\text{if}\,\,x>1\] |
| But if \[x<-1\] |
| \[y={{\cot }^{-1}}x+d\] |
| Where constant 'd' cannot be found to less data given. |
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