80292
If \(u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) and \(v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right) \text {, then } \frac{d u}{d v} \text { at } x=0 \text { is }\)
1 \(\frac{1}{4}\)
2 \(\frac{-1}{8}\)
3 1
4 \(\frac{1}{8}\)
Explanation:
(A) : Given, \(\mathrm{u}=\tan ^{-1}\left(\frac{\sqrt{1+\mathrm{x}^{2}}-1}{\mathrm{x}}\right)\) Let, \(x=\tan \theta \Rightarrow \theta=\tan ^{-1} x\) \(u=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right)\) \(u=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right)\) \(\mathrm{u}=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)=\tan ^{-1}\left\{\tan \left(\frac{\theta}{2}\right)\right\}=\frac{\theta}{2}\) On putting the value \(\theta\) in above equation. \(\mathrm{u}=\frac{1}{2} \tan ^{-1} \mathrm{x}\) \(\therefore \quad \frac{\mathrm{du}}{\mathrm{dx}}=\frac{1}{2\left(1+\mathrm{x}^{2}\right)}\) Now, \(\mathrm{v}=\tan ^{-1}\left(\frac{2 \mathrm{x} \sqrt{1-\mathrm{x}^{2}}}{1-2 \mathrm{x}^{2}}\right)\) Let, \(x=\sin \theta \Rightarrow \theta=\sin ^{-1} x\) \(\mathrm{v}=\tan ^{-1}\left(\frac{2 \sin \theta \sqrt{1-\sin ^{2} \theta}}{1-2 \sin ^{2} \theta}\right)=\tan ^{-1}\left(\frac{2 \sin \theta \cdot \cos \theta}{\cos 2 \theta}\right)\) \(\mathrm{v}=\tan ^{-1}\left(\frac{\sin 2 \theta}{\cos 2 \theta}\right)=\tan ^{-1} \tan (2 \theta)=2 \theta\) On putting the value \(\theta\) in above equation, \(v=2 \sin ^{-1} x\) \(\therefore \quad \frac{d v}{d x}=\frac{2}{\sqrt{1-x^{2}}}=\frac{d u / d x}{d v / d x}\) \(\frac{d u}{d v}=\frac{\frac{1}{2\left(1+x^{2}\right)}}{\frac{2}{\sqrt{1+x^{2}}}} \Rightarrow \frac{d u}{d v}=\frac{\sqrt{1-x^{2}}}{1+x^{2}} \times \frac{1}{4}\) \(\therefore \quad\left|\frac{\mathrm{du}}{\mathrm{dv}}\right|_{\mathrm{x}=0}=\sqrt{\frac{1-0}{1+0}} \times \frac{1}{4}=\frac{1}{4}\)
80292
If \(u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) and \(v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right) \text {, then } \frac{d u}{d v} \text { at } x=0 \text { is }\)
1 \(\frac{1}{4}\)
2 \(\frac{-1}{8}\)
3 1
4 \(\frac{1}{8}\)
Explanation:
(A) : Given, \(\mathrm{u}=\tan ^{-1}\left(\frac{\sqrt{1+\mathrm{x}^{2}}-1}{\mathrm{x}}\right)\) Let, \(x=\tan \theta \Rightarrow \theta=\tan ^{-1} x\) \(u=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right)\) \(u=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right)\) \(\mathrm{u}=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)=\tan ^{-1}\left\{\tan \left(\frac{\theta}{2}\right)\right\}=\frac{\theta}{2}\) On putting the value \(\theta\) in above equation. \(\mathrm{u}=\frac{1}{2} \tan ^{-1} \mathrm{x}\) \(\therefore \quad \frac{\mathrm{du}}{\mathrm{dx}}=\frac{1}{2\left(1+\mathrm{x}^{2}\right)}\) Now, \(\mathrm{v}=\tan ^{-1}\left(\frac{2 \mathrm{x} \sqrt{1-\mathrm{x}^{2}}}{1-2 \mathrm{x}^{2}}\right)\) Let, \(x=\sin \theta \Rightarrow \theta=\sin ^{-1} x\) \(\mathrm{v}=\tan ^{-1}\left(\frac{2 \sin \theta \sqrt{1-\sin ^{2} \theta}}{1-2 \sin ^{2} \theta}\right)=\tan ^{-1}\left(\frac{2 \sin \theta \cdot \cos \theta}{\cos 2 \theta}\right)\) \(\mathrm{v}=\tan ^{-1}\left(\frac{\sin 2 \theta}{\cos 2 \theta}\right)=\tan ^{-1} \tan (2 \theta)=2 \theta\) On putting the value \(\theta\) in above equation, \(v=2 \sin ^{-1} x\) \(\therefore \quad \frac{d v}{d x}=\frac{2}{\sqrt{1-x^{2}}}=\frac{d u / d x}{d v / d x}\) \(\frac{d u}{d v}=\frac{\frac{1}{2\left(1+x^{2}\right)}}{\frac{2}{\sqrt{1+x^{2}}}} \Rightarrow \frac{d u}{d v}=\frac{\sqrt{1-x^{2}}}{1+x^{2}} \times \frac{1}{4}\) \(\therefore \quad\left|\frac{\mathrm{du}}{\mathrm{dv}}\right|_{\mathrm{x}=0}=\sqrt{\frac{1-0}{1+0}} \times \frac{1}{4}=\frac{1}{4}\)
80292
If \(u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) and \(v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right) \text {, then } \frac{d u}{d v} \text { at } x=0 \text { is }\)
1 \(\frac{1}{4}\)
2 \(\frac{-1}{8}\)
3 1
4 \(\frac{1}{8}\)
Explanation:
(A) : Given, \(\mathrm{u}=\tan ^{-1}\left(\frac{\sqrt{1+\mathrm{x}^{2}}-1}{\mathrm{x}}\right)\) Let, \(x=\tan \theta \Rightarrow \theta=\tan ^{-1} x\) \(u=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right)\) \(u=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right)\) \(\mathrm{u}=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)=\tan ^{-1}\left\{\tan \left(\frac{\theta}{2}\right)\right\}=\frac{\theta}{2}\) On putting the value \(\theta\) in above equation. \(\mathrm{u}=\frac{1}{2} \tan ^{-1} \mathrm{x}\) \(\therefore \quad \frac{\mathrm{du}}{\mathrm{dx}}=\frac{1}{2\left(1+\mathrm{x}^{2}\right)}\) Now, \(\mathrm{v}=\tan ^{-1}\left(\frac{2 \mathrm{x} \sqrt{1-\mathrm{x}^{2}}}{1-2 \mathrm{x}^{2}}\right)\) Let, \(x=\sin \theta \Rightarrow \theta=\sin ^{-1} x\) \(\mathrm{v}=\tan ^{-1}\left(\frac{2 \sin \theta \sqrt{1-\sin ^{2} \theta}}{1-2 \sin ^{2} \theta}\right)=\tan ^{-1}\left(\frac{2 \sin \theta \cdot \cos \theta}{\cos 2 \theta}\right)\) \(\mathrm{v}=\tan ^{-1}\left(\frac{\sin 2 \theta}{\cos 2 \theta}\right)=\tan ^{-1} \tan (2 \theta)=2 \theta\) On putting the value \(\theta\) in above equation, \(v=2 \sin ^{-1} x\) \(\therefore \quad \frac{d v}{d x}=\frac{2}{\sqrt{1-x^{2}}}=\frac{d u / d x}{d v / d x}\) \(\frac{d u}{d v}=\frac{\frac{1}{2\left(1+x^{2}\right)}}{\frac{2}{\sqrt{1+x^{2}}}} \Rightarrow \frac{d u}{d v}=\frac{\sqrt{1-x^{2}}}{1+x^{2}} \times \frac{1}{4}\) \(\therefore \quad\left|\frac{\mathrm{du}}{\mathrm{dv}}\right|_{\mathrm{x}=0}=\sqrt{\frac{1-0}{1+0}} \times \frac{1}{4}=\frac{1}{4}\)
80292
If \(u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) and \(v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right) \text {, then } \frac{d u}{d v} \text { at } x=0 \text { is }\)
1 \(\frac{1}{4}\)
2 \(\frac{-1}{8}\)
3 1
4 \(\frac{1}{8}\)
Explanation:
(A) : Given, \(\mathrm{u}=\tan ^{-1}\left(\frac{\sqrt{1+\mathrm{x}^{2}}-1}{\mathrm{x}}\right)\) Let, \(x=\tan \theta \Rightarrow \theta=\tan ^{-1} x\) \(u=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right)\) \(u=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right)\) \(\mathrm{u}=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)=\tan ^{-1}\left\{\tan \left(\frac{\theta}{2}\right)\right\}=\frac{\theta}{2}\) On putting the value \(\theta\) in above equation. \(\mathrm{u}=\frac{1}{2} \tan ^{-1} \mathrm{x}\) \(\therefore \quad \frac{\mathrm{du}}{\mathrm{dx}}=\frac{1}{2\left(1+\mathrm{x}^{2}\right)}\) Now, \(\mathrm{v}=\tan ^{-1}\left(\frac{2 \mathrm{x} \sqrt{1-\mathrm{x}^{2}}}{1-2 \mathrm{x}^{2}}\right)\) Let, \(x=\sin \theta \Rightarrow \theta=\sin ^{-1} x\) \(\mathrm{v}=\tan ^{-1}\left(\frac{2 \sin \theta \sqrt{1-\sin ^{2} \theta}}{1-2 \sin ^{2} \theta}\right)=\tan ^{-1}\left(\frac{2 \sin \theta \cdot \cos \theta}{\cos 2 \theta}\right)\) \(\mathrm{v}=\tan ^{-1}\left(\frac{\sin 2 \theta}{\cos 2 \theta}\right)=\tan ^{-1} \tan (2 \theta)=2 \theta\) On putting the value \(\theta\) in above equation, \(v=2 \sin ^{-1} x\) \(\therefore \quad \frac{d v}{d x}=\frac{2}{\sqrt{1-x^{2}}}=\frac{d u / d x}{d v / d x}\) \(\frac{d u}{d v}=\frac{\frac{1}{2\left(1+x^{2}\right)}}{\frac{2}{\sqrt{1+x^{2}}}} \Rightarrow \frac{d u}{d v}=\frac{\sqrt{1-x^{2}}}{1+x^{2}} \times \frac{1}{4}\) \(\therefore \quad\left|\frac{\mathrm{du}}{\mathrm{dv}}\right|_{\mathrm{x}=0}=\sqrt{\frac{1-0}{1+0}} \times \frac{1}{4}=\frac{1}{4}\)
