CUET UG Mathematics Applied Inverse Trigonometrical Function Previous Year Questions (PYQs)
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$\tan^{-1}\left[2\sin\left\{2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}\right]$
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Solution
$\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}$
$2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$
$\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$
$2\sin\left\{2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}=\sqrt{3}$
$\tan^{-1}\left[2\sin\left\{2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}\right]=\frac{\pi}{3}$
Match List I with List II
Choose the correct answer from the options given below:
| List I | List II |
| A. The range of $\sin^{-1}x$ is | I. $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-{0}$ |
| B. The range of $\tan^{-1}x$ is | II. $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ |
| C. The range of $\csc^{-1}x$ is | III. $[0,\pi] - \{\frac{\pi}{2}\}$ |
| D. The range of $\sec^{-1}x$ is | IV. $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ |
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Solution
Range of $\sin^{-1}x$ is
$[-\frac{\pi}{2},\frac{\pi}{2}]$
So A → IV
Range of $\tan^{-1}x$ is
$(-\frac{\pi}{2},\frac{\pi}{2})$
So B → II
Range of $\csc^{-1}x$ is
$[-\frac{\pi}{2},\frac{\pi}{2}] - {0}$
So C → I
Range of $\sec^{-1}x$ is
$[0,\pi] - {\frac{\pi}{2}}$
So D → III
If $y = e^{\log(\sin^{-1}(x))} + e^{\log(\cos^{-1}(x))},\; 0 < x < 1$ then
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Solution
Using identity
$e^{\log a}=a$
So
$y=\sin^{-1}(x)+\cos^{-1}(x)$
But
$\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}$
Hence
$y=\frac{\pi}{2}$
Therefore
$\frac{dy}{dx}=0$
CUET UG Mathematics Applied
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CUET UG Mathematics Applied
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and More.
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