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This is one of our many quizzes on Graphing sec, csc, cot functions. Click on this to access more such competitive quizzes on Trigonometric Functions.

Graphing sec, csc, cot functions Quiz 2

mathematics

The function of the form $f\left( x \right)=\tan \left( ax \right)+b$ for the graph plotted below is <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ff684cf7-585c-4da1-819b-772605bd82658782381939965893856.png"/>

 Try to see the length of the interval in which the function is repeating itself by analyzing the given vertical dashed line in the figure to find the time period and hence get the value for $a$ . To find the value of $b$ , use the point mentioned in the graph that the given function is satisfying.

From the graph shown the time period is $\frac{\pi }{3}$ . $\Rightarrow a=3$ And from the point given, $f\left( 0 \right)=1\Rightarrow b=1$ . $\therefore \text{The required function is }f\left( x \right)=\tan \left( 3x \right)+1$

 Number of points of non-differentiability of $\left| \sec x-2 \right|$ in the interval $\left( 0,2\pi  \right)$.

 Plot the graph for $\sec x-2$ . To make the graph of $\left| \sec x-2 \right|$ , rotate the part of the graph in the negative $y-axis$ in the positive side by leaving the positive side as it is. Then calculate the points of non-differentiability.

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/aa8405c4-e1e8-436d-979b-8cdf6b2959037846106904624566360.png"/> <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/aa5976d8-c92b-4dbc-8ec5-5d0d00860f9c297861150618422357.png"/> In the solution, the first graph is for $\sec x-2$ . So in $\left| \sec x-2 \right|$ , we rotate the part of the graph in the negative $y-axis$ in the positive side by leaving the positive side as it was. In the second graph obtained, there are two points in $\left( 0,2\pi \right)$, which have sharp edges and two points are approaching infinity. $\therefore $There are four points of non-differentiability in the interval $\left( 0,2\pi \right)$.

 Find the range of the function $\left| \sec x-2 \right|$ in the interval $\left[ \dfrac{\pi }{2},\dfrac{3\pi }{2} \right]$.

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/3114294c-b7a3-455f-8b6f-70603901b5da5746894458965105746.png"/> <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ad1f13b7-8aec-4abd-834f-9ea0be27c8d36976108093090769824.png"/> In the solution, the first graph is for $\sec x-2$. So in $\left| \sec x-2 \right|$, we rotate the part of the graph in the negative $y-axis$ in the positive side by leaving the positive side as it was. In the second graph obtained, there are two points in $\left( 0,2\pi \right)$ which have sharp edges and two points are approaching infinity. $\therefore $There are four points of non-differentiability in the interval $\left( 0,2\pi \right)$.

The below graph is of the form $a\operatorname{cosec}\left( bx \right)+c$ . The value of $\left( a+b-c \right)$ is <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a115bc7d-8b8c-46e6-aa1b-04feb5a758bf44161138650919745.png"/>

 Try to compute the time period, vertical shift in the graph by using the data given in the graph like vertical and horizontal lines given in the question.

The period of the graph is $2\pi $ as seen from the given graph $\Rightarrow b=1$ If the function would be $cosec\left( x \right)-1$ , $B$ would have passed through $y=-2$ . But $B$ is passing through $y=-4$ . So, the graph of the function is $2cosec\left( x \right)-2$ . $\therefore a+b-c=2+1-\left( -2 \right)=5$

 The number of solutions of the equation $\cos x=\sec x+1$ in the interval $\left[ 0,\pi  \right]$.

 The graph for $y=\cos x$ is standard . To make a graph of $y=\sec x+1$ , make the graph for $y=\sec x$ , then shift the whole graph by one unit in positive $y-direction$. Then calculate the common points between these two graphs.

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ce78f2cd-f963-42c7-b06c-3391b52c7b294594289332547591595.png"/> Make the graph for $y=\cos x$ and to make a graph of $y=\sec x+1$ , make the graph for $y=\sec x$ , then shift the whole graph by one unit in positive $y-direction$ . As seen from the graph there is one common point between $y=\cos x$ and $y=\sec x+1$ in the interval $\left[ 0,\pi \right]$ . $\therefore $There is one solution of the equation $\cos x=\sec x+1$ in the interval $\left[ 0,\pi \right]$.