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Match for each functions in List - I to its derivative given in List-II
\(\begin{array}{ll}
\hline \text { List I } & \text { List II } \\
\hline \text {(A) } \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) & \text { (I) } \cos x-\sin x \\
\hline \text {(B) } \tan ^{-1}\left(\frac{1-x}{1+x}\right) & \text { (II) } \frac{-1}{1+x^2} \\
\hline \text {(C) } e^{\log (\sin x+\cos x)} & \text { (III) } \frac{2}{1+x^2} \\
\hline \text {(D) } \sqrt{1-\sin 2 x}\left(0 < x < \frac{\pi}{4}\right) & \text { (IV) } \cos x+\sin x \\
\hline & \text { (V) }-\sin x-\cos x \\
\hline
\end{array}\)
The correct match is
Options:
\(\begin{array}{ll}
\hline \text { List I } & \text { List II } \\
\hline \text {(A) } \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) & \text { (I) } \cos x-\sin x \\
\hline \text {(B) } \tan ^{-1}\left(\frac{1-x}{1+x}\right) & \text { (II) } \frac{-1}{1+x^2} \\
\hline \text {(C) } e^{\log (\sin x+\cos x)} & \text { (III) } \frac{2}{1+x^2} \\
\hline \text {(D) } \sqrt{1-\sin 2 x}\left(0 < x < \frac{\pi}{4}\right) & \text { (IV) } \cos x+\sin x \\
\hline & \text { (V) }-\sin x-\cos x \\
\hline
\end{array}\)
The correct match is
Solution:
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Verified Answer
The correct answer is:
\(\begin{array}{cccc}\text { A } & \text { B } & \text { C } & \text { D } \\ \text { III } & \text { II } & \text { I } & \text { IV }\end{array}\)
(A) Let \(y=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
Again, let \(x=\tan \theta=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\)
\(\begin{array}{rlrl}
& =\sin ^{-1}(\sin 20)=20 \\
& y =2 \tan ^{-1} x \\
& \Rightarrow \frac{d y}{d x} =\frac{2}{1+x^2}(\text { differentiating w.r. t. } x)
\end{array}\)
(B) Let \(\tan ^{-1}\left(\frac{1-x}{1+x}\right)=y \Rightarrow \tan ^{-1}(l)-\tan ^{-1}(x)=y\)
\(0-\frac{1}{1+x^2}=y \text { (differentiating w.r.t } x \text {) }\)
\(\Rightarrow \quad y=\frac{-1}{1+x^2}\)
\(\mathrm{B} \rightarrow \mathrm{II}\)
(C)
\(\begin{aligned}
e^{\log (\sin x+\cos x)} & =\sin x+\cos x=y \\
y & =\sin x+\cos x
\end{aligned}\)
Differentiating w.r.t \(x\)
\(\frac{d y}{d x}=\cos x-\sin x\)
\(\mathrm{C} \rightarrow \mathrm{I}\)
(D) Let \(y=\sqrt{1-\sin 2 x}\)
\(\begin{aligned}
& =\sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x} \\
& =\sqrt{(\sin x-\cos x)^2} \Rightarrow y=\sin x-\cos x
\end{aligned}\)
Differentiating w.r. t. \(x\), we get
\(\frac{d y}{d x}=\cos x+\sin x\)
\(\mathrm{D} \rightarrow \mathrm{IV}\)
Hence, option (d) is correct.
Again, let \(x=\tan \theta=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\)
\(\begin{array}{rlrl}
& =\sin ^{-1}(\sin 20)=20 \\
& y =2 \tan ^{-1} x \\
& \Rightarrow \frac{d y}{d x} =\frac{2}{1+x^2}(\text { differentiating w.r. t. } x)
\end{array}\)
(B) Let \(\tan ^{-1}\left(\frac{1-x}{1+x}\right)=y \Rightarrow \tan ^{-1}(l)-\tan ^{-1}(x)=y\)
\(0-\frac{1}{1+x^2}=y \text { (differentiating w.r.t } x \text {) }\)
\(\Rightarrow \quad y=\frac{-1}{1+x^2}\)
\(\mathrm{B} \rightarrow \mathrm{II}\)
(C)
\(\begin{aligned}
e^{\log (\sin x+\cos x)} & =\sin x+\cos x=y \\
y & =\sin x+\cos x
\end{aligned}\)
Differentiating w.r.t \(x\)
\(\frac{d y}{d x}=\cos x-\sin x\)
\(\mathrm{C} \rightarrow \mathrm{I}\)
(D) Let \(y=\sqrt{1-\sin 2 x}\)
\(\begin{aligned}
& =\sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x} \\
& =\sqrt{(\sin x-\cos x)^2} \Rightarrow y=\sin x-\cos x
\end{aligned}\)
Differentiating w.r. t. \(x\), we get
\(\frac{d y}{d x}=\cos x+\sin x\)
\(\mathrm{D} \rightarrow \mathrm{IV}\)
Hence, option (d) is correct.
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