[amp_mcq option1=”1″ option2=”e” option3=”\\[{{\\text{e}}^{\\sqrt 2 }}\\]” option4=”\\[\\infty \\]” correct=”option1″]<\/p>\n
The maximum value of $\\frac{{{{\\text{e}}^{\\sin {\\text{x}}}}}}{{{{\\text{e}}^{\\cos {\\text{x}}}}}}$ is $\\text{e}^{\\sqrt{2}}$.<\/p>\n
To see this, let $y = \\frac{{{{\\text{e}}^{\\sin {\\text{x}}}}}}{{{{\\text{e}}^{\\cos {\\text{x}}}}}}$. Then, $y^2 = \\text{e}^{\\sin x + \\cos x} = \\text{e}^{\\sqrt{2}}$, where the last equality follows from the trigonometric identity $\\sin x + \\cos x = \\sqrt{2} \\sin \\left( \\frac{\\pi}{4} + x \\right)$. Taking the square root of both sides, we get $y = \\text{e}^{\\sqrt{2} \\sin \\left( \\frac{\\pi}{4} + x \\right)}$.<\/p>\n
Since $\\sin \\left( \\frac{\\pi}{4} + x \\right)$ is a periodic function with period $2 \\pi$, it takes on all values in the interval $[-1, 1]$. Therefore, $y$ takes on all values in the interval $[1, \\text{e}^{\\sqrt{2}}]$. Since $y$ is continuous, it must have a maximum value. This maximum value occurs when $\\sin \\left( \\frac{\\pi}{4} + x \\right) = 1$, which is when $x = \\frac{\\pi}{4}$. Therefore, the maximum value of $y$ is $\\text{e}^{\\sqrt{2}}$.<\/p>\n
The other options are incorrect because they are not the maximum value of $\\frac{{{{\\text{e}}^{\\sin {\\text{x}}}}}}{{{{\\text{e}}^{\\cos {\\text{x}}}}}}$. For example, $1$ is not the maximum value because $y = 1$ when $x = 0$, but $y$ takes on values greater than $1$ when $x$ is not equal to $0$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”1″ option2=”e” option3=”\\[{{\\text{e}}^{\\sqrt 2 }}\\]” option4=”\\[\\infty \\]” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20242","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n