$$ \\\\ $$ Orange graph is for $$1+cot\\left( \\theta \\right)$$ and green graph is for $$cot\\left( \\dfrac{\\theta }{2} \\right)$$ $$ \\\\ $$ Two graphs are touching at points $$A$$ and $$B$$ between $$\\left( -2\\pi ,\\pi \\right)$$. $$ \\\\ $$ Given equation has two solutions for $$\\theta$$.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Graphing sec, csc, cot functions Quiz 1","text":" Number of solutions of equation $${{e}^{x}}=\\cot \\left( x \\right)$$ in the interval $$\\left[ 0,\\dfrac{\\pi }{2} \\right]$$ is","comment":{"@type":"Comment","text":" Make the graph of ${{e}^{x}}\\text{ and }\\cot \\left( x \\right)$ in the interval $\\left[ 0,\\dfrac{\\pi }{2} \\right]$ . Then count the number of points where these two curves are intersecting in the given interval."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$3$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$2$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$4$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1$$","position":2,"answerExplanation":{"@type":"Comment","text":" $$ \\\\ $$ As we can see from the graph, the blue curve is the graph for $${{e}^{x}}$$ and the red curve is the graph for $$\\cot x$$. $$ \\\\ $$ It has only one intersection point in $$\\left[ 0,\\dfrac{\\pi }{2} \\right]$$. Therefore, $${{e}^{x}}=\\cot \\left( x \\right)$$ has one solution.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Graphing sec, csc, cot functions Quiz 1","text":" Find the value of $$a+b$$ if the following graph as the form $$\\sec ax+b$$. $$ \\\\ $$
","comment":{"@type":"Comment","text":" See the time period of the graph from the figure to find the value of $a$. Then use shift of the graph along $y-axis$ to find the value of $b$ ."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{3}{2}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{5}{2}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$5$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{7}{2}$$","position":2,"answerExplanation":{"@type":"Comment","text":"From the graph, it can be seen that time period of the function is $$4\\pi $$ $$ \\\\ \\dfrac{2\\pi }{a}=4\\pi \\Rightarrow a=\\dfrac{1}{2} \\\\ $$ Also, graph has been shifted in $$+y$$ direction by $$3$$ units. $$ \\\\ \\Rightarrow b=3 \\\\ \\therefore a+b=\\dfrac{7}{2}$$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Graphing sec, csc, cot functions Quiz 1","text":" The number of solutions of the equation given by $$\\sin \\theta ={{\\sec }^{2}}4\\theta $$ in $$\\left[ 0,\\pi \\right]$$ is/are:","comment":{"@type":"Comment","text":" Find the range and time period for the function ${{\\sec }^{2}}\\left( 4\\theta \\right)$ to plot its graph. Find the common point between ${{\\sec }^{2}}\\left( 4\\theta \\right)$ and $\\sin \\left( \\theta \\right)$ to find the number of solutions."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$3$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$2$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$0$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1$$ ","position":2,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ $$ Orange graph is for $$\\sin \\theta$$ and green graph is for $${{\\sec }^{2}}\\left( 4\\theta \\right)$$. $$ \\\\ $$ Due to $$\\sec 4\\theta$$, the time period becomes $$\\dfrac{\\pi }{2}$$. After squaring this function graph on the negative side of the y-axis will come on the positive side and the time period further reduces by half. $$ \\\\ $$ So, it becomes $$\\dfrac{\\pi }{4}$$. Graphs just become steeper. Range becomes $$[ 1, \\infty )$$ $$ \\\\ $$ There is one solution to the given equation.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Graphing sec, csc, cot functions Quiz 1","text":" Number of solutions of the $$\\text{max}\\left\\{ \\operatorname{cosec}x,\\sec x \\right\\}=\\sqrt{2}$$ in $$\\left[ 0,\\pi \\right]$$ is/are","comment":{"@type":"Comment","text":" To plot $\\text{max}\\left\\{ \\operatorname{cosec}x,\\sec x \\right\\}$ , use $\\text{max}\\left\\{ \\operatorname{cosec}x,\\sec x \\right\\}=\\dfrac{\\left( \\operatorname{cosec}x+\\sec x+\\left| \\operatorname{cosec}x-\\sec x \\right| \\right)}{2}$ $ \\\\ $ Then find the common points between the curves $y=\\text{max}\\left\\{ \\operatorname{cosec}x,\\sec x \\right\\}\\text{ and }y=\\sqrt{2}$ in $\\left[ 0,\\pi \\right]$"},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$4$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$3$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$1$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$2$$ ","position":2,"answerExplanation":{"@type":"Comment","text":" $$\\mathbf{First}\\text{ }\\mathbf{graph}$$ $$ \\\\ $$
$$ \\\\ \\mathbf{Second}\\text{ }\\mathbf{graph}$$ $$ \\\\ $$
$$ \\\\ $$ In the first graph shown the , green curve is $$y=\\operatorname{cosec}x$$, blue curve is $$y=\\sec x$$ and red curve is $$y=\\sqrt{2}$$. In the second graph the curve is max $$\\left\\{ \\operatorname{cosec}x,\\sec x \\right\\}$$ $$ \\text{max}\\left\\{ \\operatorname{cosec}x,\\sec x \\right\\}=\\dfrac{\\left( \\operatorname{cosec}x+\\sec x+\\left| \\operatorname{cosec}x-\\sec x \\right| \\right)}{2}. \\\\ $$ There are two common points between the curves $$y=\\text{max}\\left\\{ \\operatorname{cosec}x,\\sec x \\right\\}\\text{ and }y=\\sqrt{2}$$ $$ \\\\ $$ So, two solutions are there for the given equation. ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}