","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" Both","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" None","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" ","position":1,"answerExplanation":{"@type":"Comment","text":"Here, checking the signs of $$f\\left( a \\right),f\\left( b \\right)$$ where both having opposite signs the only option is option $$(b)$$ that is $$f\\left( a \\right) < 0$$ and $$f\\left( b \\right) > 0$$. By using the intermediate mean value theorem the function shown in option $$(b)$$ will have a root in interval $$\\left[ a,b \\right]$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Intermediate value theorem Quiz 1","text":" Find the value of $$'b'$$ such that $$f(x)={{x}^{3}}-3x+91$$ has a root in the interval $$(0,b)$$","comment":{"@type":"Comment","text":" First, find the value of $f(0)$. Then determine the sign of $f(0)$. Now find the value of $f(b)$. By using intermediate mean value, find the interval of $'b'$ where it lies. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$(-7,13)$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$(7,-13)$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$(-13,-7)$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$(-13,7)$$","position":2,"answerExplanation":{"@type":"Comment","text":"By using the intermediate value theorem, to have a root in the interval $$(0,b)$$ , we need to have different signs for $$f(0)$$ and $$f(b)$$ $$ \\\\ f(0)={{0}^{2}}+6(0)+91=91 > 0 \\\\ f(b)={{b}^{2}}+6b-91 < 0 \\\\ \\Rightarrow (b+13)(b-7) < 0 \\\\ \\Rightarrow -13 < b < 7 $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Intermediate value theorem Quiz 1","text":" A function $$f(x)=a{{x}^{3}}-30{{x}^{2}}+28x-47$$ has a root in the interval $$(0,1)$$. Then the value of a can be equal to","comment":{"@type":"Comment","text":" Initially, find the value of $f(0)$ and $f(1)$. By using the intermediate value theorem, to have a root in the interval $(0,1)$, $f(0)$ and $f(1)$ have different signs. Based on this concept, find the range of $'a'$ lies. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 48 ","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" 49","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 47","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 50","position":2,"answerExplanation":{"@type":"Comment","text":"By using the intermediate value theorem, to have a root in the interval $$(0,1)$$, we need to have different signs for $$f(0)$$ and $$f(1)$$. $$ \\\\ f(0)=a{{(0)}^{3}}-30{{(0)}^{2}}+28(0)-47=-47 \\\\ f(1)=a{{(1)}^{3}}-30{{(1)}^{2}}+28(1)-47=a-49 \\\\ $$ As $$f(0)$$ is less than zero, the value of $$f(1)$$ should be greater than zero. $$ \\\\ \\Rightarrow f(1) > 0 \\\\ \\Rightarrow a-49 > 0 \\\\ \\Rightarrow a > 49 $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Intermediate value theorem Quiz 1","text":" The function $$f(x)=5{{x}^{103}}+2{{x}^{97}}-3{{x}^{91}}-93{{x}^{5}}$$ ","comment":{"@type":"Comment","text":" Firstly, find the value of $f(1),f(10),f(-1),f(-10)$. Now find the signs of the values of $f(1),f(10),f(-1),f(-10)$. Now by using the intermediate mean value theorem, find whether a root lies between the required interval."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" Has a root in the interval $$\\left( -10,-1 \\right)$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" Both of these","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" None of these","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" Has a root in the interval $$\\left( 1,10 \\right)$$","position":0,"answerExplanation":{"@type":"Comment","text":"$$f(1)=5{{\\left( 1 \\right)}^{103}}+2{{\\left( 1 \\right)}^{97}}-3{{\\left( 1 \\right)}^{91}}-93{{\\left( 1 \\right)}^{5}} < 0 \\\\ f(10)=5{{\\left( 10 \\right)}^{103}}+2{{\\left( 10 \\right)}^{97}}-3{{\\left( 10 \\right)}^{91}}-93{{\\left( 10 \\right)}^{5}} > 0 \\\\ $$ By using intermediate value theorem, it is clear that $$ \\\\ $$ $$f(x)=5{{x}^{103}}+2{{x}^{97}}-3{{x}^{91}}-93{{x}^{5}} $$ has a root in the interval $$ \\\\ \\left( 1,10 \\right)$$. $$ \\\\ f(-1)=5{{\\left( -1 \\right)}^{103}}+2{{\\left( -1 \\right)}^{97}}-3{{\\left( -1 \\right)}^{91}}-93{{\\left( -1 \\right)}^{5}} < 0 \\\\ f(10)=5{{\\left( -10 \\right)}^{103}}+2{{\\left( -10 \\right)}^{97}}-3{{\\left( -10 \\right)}^{91}}-93{{\\left( -10 \\right)}^{5}} < 0 \\\\ $$ By using intermediate value theorem, it is clear that, $$ \\\\ $$ $$f(x)=5{{x}^{103}}+2{{x}^{97}}-3{{x}^{91}}-93{{x}^{5}}$$ does not have a root in the interval $$\\left( -10,-1 \\right)$$.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}