","comment":{"@type":"Comment","text":" Let the required function be $a\\sin \\left( bx+c \\right)+d$ . Then use the time period, values at specific points, jump in the curve to find the variables."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\text{sin}\\left( x-\\dfrac{\\pi }{4} \\right)+2$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\text{2sin}\\left( x-\\dfrac{\\pi }{4} \\right)+1$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\cos \\left( x-\\dfrac{\\pi }{4} \\right)+1$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\text{sin}\\left( x-\\dfrac{\\pi }{4} \\right)+1$$","position":2,"answerExplanation":{"@type":"Comment","text":" The required function would be of the form $$a\\sin \\left( bx+c \\right)+d$$ $$ \\\\ $$ From the graph, the distance between maxima and minima of the function is $$2$$ units $$ \\\\ \\Rightarrow a=1 $$ $$ \\\\ $$ Graph is shifted in positive y-axis by $$1$$ unit $$ \\\\ \\Rightarrow d=1 \\\\ $$ Time period of the graph is $$2\\pi$$ $$ \\\\ \\Rightarrow b=1 \\\\ \\text{So, function is }\\sin \\left( x+c \\right)+1. \\\\ $$ At $$x=0$$, value of the given function is $$1-\\dfrac{1}{\\sqrt{2}}$$ $$ \\\\ \\Rightarrow c=-\\dfrac{\\pi }{4} \\\\ $$ $$\\therefore$$ The required function is $$\\sin\\left( x-\\dfrac{\\pi }{4} \\right)+1. $$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Graphing trigonometric functions Quiz 1","text":" The time period for the following curve is $$ \\\\ $$ ","comment":{"@type":"Comment","text":" Graph is representing an even function and an asymptote is given to find the time period of the function."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{\\pi }{4}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{\\pi }{2}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$2\\pi $$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\pi $$","position":3,"answerExplanation":{"@type":"Comment","text":" As seen from the curve, that graph is repeating at intervals $$ \\\\ T=2\\times \\dfrac{\\pi }{2}=\\pi \\\\ $$ $$\\therefore$$ Time period of the graph is $$\\pi$$. ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Graphing trigonometric functions Quiz 1","text":" Find the number of solutions of the equation $$\\cos \\left( 2x \\right)=\\cos \\left( 3x \\right)$$ in the interval $$\\left[ 0,\\pi \\right]$$","comment":{"@type":"Comment","text":" To find the curve of $\\text{cos}\\left( kx \\right)$ take a time period as $\\dfrac{2\\pi }{k}$. Later draw maxima and minima points for graphs. Give them a shape similar to the graph of $\\text{cos}\\left( x \\right)$ ."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$5$$ ","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$4$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$3$$","position":1,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ $$ The graph of $$cos\\left( kx \\right)$$ is similar to $$cos\\left( x \\right)$$ but will have different time period given by $$\\dfrac{2\\pi }{k}$$ $$ \\\\ $$ Using this to make the curve for $$cos\\left( 2x \\right)$$ and $$cos\\left( 3x \\right)$$, we obtain the above graphs $$ \\\\ $$ Both graphs have three common points in the interval $$\\left[ 0,\\pi \\right]$$ $$ \\\\ $$ $$\\therefore$$ Required number of solutions is three.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Graphing trigonometric functions Quiz 1","text":" Range of the function $$f\\left( x \\right)=\\cos \\left\\{ x\\text{ } \\right\\}$$ is, where $$\\left\\{ \\text{ }.\\text{ } \\right\\}$$ represents fractional part function.","comment":{"@type":"Comment","text":" The range of fractional part functions is the domain for $f\\left( x \\right)=\\cos \\left\\{ x\\text{ } \\right\\}$. Hence, use the range of fractional part functions to find the range for $f\\left( x \\right)=\\cos \\left\\{ x\\text{ } \\right\\}$."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left[ 0,1 \\right)$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left[ 0,1 \\right]$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left[ 1,\\cos 1 \\right]$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left[ 1,\\cos 1 \\right)$$","position":2,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ \\text{For any interval }\\left[ n,n+1 \\right), \\\\ \\text{The range of }\\left\\{ x\\text{ } \\right\\}\\text{ is }\\left[ 0,1 \\right), \\\\ \\text{Where }n\\text{ is any integer}\\text{.} \\\\ \\Rightarrow \\text{For any interval }\\left[ n,n+1 \\right), \\\\ \\text{The range of cos}\\left\\{ x\\text{ } \\right\\}\\text{ is }\\left[ 1,\\cos 1 \\right). \\\\ \\text{Using this we get the above graph}\\text{.} \\\\ \\therefore \\text{The range of the function is }\\left[ 1,\\cos 1 \\right).$$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Graphing trigonometric functions Quiz 1","text":" Number of points of discontinuity $$f\\left( x \\right)=\\cos \\left\\{ x\\text{ } \\right\\}$$ has in the interval $$\\left( -1,2 \\right]$$.","comment":{"@type":"Comment","text":" The range of fractional part functions is the domain for $f\\left( x \\right)=\\cos \\left\\{ x\\text{ } \\right\\}$. Hence, use the range of fractional part functions to find the range for $f\\left( x \\right)=\\cos \\left\\{ x\\text{ } \\right\\}$. Later see where the graph is breaking in the interval $\\left( -1,2 \\right]$."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$4$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$4$$ ","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$2$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$3$$","position":1,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ $$ $$\\text{For any interval }\\left[ n,n+1 \\right), \\\\ \\text{The range of }\\left\\{ x\\text{ } \\right\\}\\text{ is }\\left[ 0,1 \\right), \\\\ \\text{Where }n\\text{ is any integer}\\text{.} \\\\ \\Rightarrow \\text{For any interval }\\left[ n,n+1 \\right), \\\\ \\text{The range of cos}\\left\\{ x\\text{ } \\right\\}\\text{ is }\\left[ 1,\\cos 1 \\right). \\\\ \\text{Using this we can obtain above graph}\\text{.} \\\\ \\text{So,the range of the function }f\\left( x \\right)=\\cos \\left\\{ x\\text{ } \\right\\}\\text{ is }\\left[ 1,\\cos 1 \\right). \\\\ \\therefore \\text{The points of discontinuity of }f\\left( x \\right)=\\cos \\left\\{ x\\text{ } \\right\\}$$ $$\\text{ in the interval }\\left( -1,2 \\right] \\\\ \\text{are at }x=0,x=1\\text{ and }x=2. \\\\ \\therefore \\text{ 3points of discontinuity are there in }\\left( -1,2 \\right]. $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}