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This is one of our many quizzes on Graphing inverse trigonometric functions. Click on this to access more such competitive quizzes on Calculus.

Graphing inverse trigonometric functions Quiz 2

mathematics

 What is the correct equation for the graph shown in the figure? <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b8a8f58c-bf95-48ae-bdd5-874e1ca5d1d65472045191411419256.png"/>

 Use the standard form $y = a\sin bx\cos cx$ and determine the values of the variables.

 It is clear from the graph that the maximum value of $y$ is $2$ and minimum value $- 2$. $ \\ a = \dfrac{{2 - \left( { - 2} \right)}}{2} = 2 \\ $ There is one cycle of the curve in the interval $\dfrac{{ - \pi }}{3} \leqslant x \leqslant \pi$ $ \\ $  The period is the difference between the endpoints of the interval for one cycle. $ \\ $ The period is $\pi  - \left( {\dfrac{{ - \pi }}{3}} \right)$ or $\left( {\dfrac{{4\pi }}{3}} \right) $ $ \\  \Rightarrow \dfrac{{2\pi }}{b} = \dfrac{{4\pi }}{3} \\  b = \dfrac{3}{2} \\ $ Similarly for value $c$, $ \\ \dfrac{{2\pi }}{c} = \dfrac{{4\pi }}{3} \\  c = \dfrac{3}{2} \\   \Rightarrow y = 2\sin \dfrac{3}{2}xcos\dfrac{3}{2}x \\  $

 Identify the correct graph of the function $y = 3\sin 2\left( {x + \dfrac{\pi }{6}} \right)$.

 Apply $x = \dfrac{{ - \pi }}{6},\dfrac{\pi }{6}$ in the given function to match the graph from the given options.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/186912bf-5678-40e4-b964-133135832d8d6646378935468498621.png"/> $ \\ {\text{Given, }}y = 3{\sin ^2}\left( {x + \dfrac{\pi }{6}} \right) \\ \text{Apply } {\text{ x = }}\dfrac{\pi }{6} \\  y = 3\sin 2\left( {\dfrac{\pi }{6} + \dfrac{\pi }{6}} \right) \\   \Rightarrow 3\sin \left( {\dfrac{{2\pi }}{3}} \right) \\   \Rightarrow 3\sin \left( {\pi  - \dfrac{\pi }{3}} \right) \\   \Rightarrow 3 \times 0.866 \\   \Rightarrow 2.59 \\ \text{Apply }{\text{ x = }}\dfrac{{ - \pi }}{6} \\  y = 3\sin 2\left( {\dfrac{{ - \pi }}{6} + \dfrac{\pi }{6}} \right) = 3\sin 2 \times 0 \\   \Rightarrow y = 0 \\  $ 

 Which is the correct graph for the function $y = \cot \left( {x - \dfrac{\pi }{4}} \right)$?

 Apply given interval to find out coordinate then plot the graph.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/cb09b760-e4e1-4683-a8ec-33542c441aca7215175537080152719.png"/> $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/35983d95-0b70-4950-a2a0-cdeb4850ac697753707890754189000.png"/> $ \\  y = \cot \left( {x - \dfrac{\pi }{4}} \right) \\  \text{Apply }{\text{ }}x = \dfrac{\pi }{2} \\  y = \cot \left( {\dfrac{\pi }{2} - \dfrac{\pi }{4}} \right) \\   \Rightarrow \tan \dfrac{\pi }{4} = 1 \\ \text{Apply }{\text{ }}x =  - \dfrac{\pi }{2} \\  y = \cot \left( {\dfrac{{ - \pi }}{2} - \dfrac{\pi }{4}} \right) \\   \Rightarrow y =  - 1 \\  $

 What is the value of $x$ in radians such that $\operatorname{secx}  =  - 1.300$ in the interval $0 \leqslant x \leqslant 2\pi $

 Apply $\sec \theta  = \dfrac{1}{{\cos \theta }}$ then further solve.

 $  \operatorname{secx}  =  - 1.300 \\   \Rightarrow x = {\sec ^{ - 1}}\left( { - 1.300} \right) = \dfrac{1}{{{{\cos }^{ - 1}}\left( {1.300} \right)}} = \dfrac{1}{{0.9997}} = 1.00025rad \\  $

 In the interval $\left[ { - 2\pi ,0} \right]$, for which  values of $x$, does $y = \tan x$ decrease

 Use intervals to construct graphs and make tables to check the pattern of the function in the given intervals.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7d125f81-3bad-42f9-9b3e-a06d0c5123493558820962662438642.png"/><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/2a9d3e75-edb5-4141-b213-e45f4c7be66c4497284461758210380.png"/>$ y = \tan x \\  x \in \left[ { - 2\pi ,0} \right] \\ \text{Apply }{\text{ }}x = 0 \\   \Rightarrow y = \tan {\text{ 0}} = 0 \\  {\text{Apply }}x =  - \dfrac{\pi }{4} \\   \Rightarrow y = \tan \left( { - \dfrac{\pi }{4}} \right) =  - 1 \\  {\text{Apply }}x =  - \pi  \\   \Rightarrow y = \tan \left( { - \pi } \right) = 0 \\  {\text{Apply }}x = \dfrac{{ - 3\pi }}{4} \\   \Rightarrow y = \tan \left( {\dfrac{{ - 3\pi }}{4}} \right) = 1 \\ \text{Apply }{\text{ }}x = 2\pi  \\   \Rightarrow y = \tan \left( {2\pi } \right) = 0 \\  $