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Ask QuestionPosted by Baba Tushir 1 year, 11 months ago
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Preeti Dabral 1 year, 11 months ago
{tex}\int {x\log 2xdx} {/tex}
{tex} = \int {\left( {\log 2x} \right)xdx} {/tex}
{tex}= \left( {\log 2x} \right)\int {xdx - \int {\left[ {\frac{d}{{dx}}\log 2x\int {xdx} } \right]dx} } {/tex}
[Applying product rule]
{tex}= \left( {\log 2x} \right)\frac{{{x^2}}}{2} - \int {\frac{1}{{2x}}.2.\frac{{{x^2}}}{2}dx} {/tex}
{tex} = \frac{1}{2}{x^2}\log 2x - \frac{1}{2}\int {xdx} {/tex}
{tex}= \frac{1}{2}{x^2}\log 2x - \frac{1}{2}\frac{{{x^2}}}{2} + c{/tex}
{tex}= \frac{{{x^2}}}{2}\log 2x - \frac{{{x^2}}}{4} + c{/tex}
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Preeti Dabral 1 year, 11 months ago
Given: {tex}y = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right){/tex}
To simplify the given Inverse Trigonometric function,we put, {tex}x = \tan \theta{/tex}
{tex}\Rightarrow y = {\sin ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) = {\sin ^{ - 1}}\left( {\sin 2\theta } \right) = 2\theta{/tex}
{tex}\Rightarrow y = 2{\tan ^{ - 1}}x{/tex}
{tex}\Rightarrow \frac{{dy}}{{dx}} = 2.\frac{1}{{1 + {x^2}}} = \frac{2}{{1 + {x^2}}}{/tex}
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Preeti Dabral 1 year, 11 months ago
{tex}\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} 2x + 1{/tex}
{tex}\mathop {\lim }\limits_{h \to 0} \left[ {2\left( {2 - h} \right) + 1} \right]{/tex} = 5
{tex}= \mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {3x -1} \right){/tex}
{tex} = \mathop {\lim }\limits_{h \to 0} 3\left( {2 + h} \right) - 1{/tex} = 5
In given that question f(x) is continuous at x=2,therefore
{tex} \mathop {\lim }\limits_{x \to {2^ - }} f(x) = f(2) = \mathop {\lim }\limits_{x \to {2^ + }} f(x){/tex}
5 = k
{tex}\Rightarrow k = 5{/tex}
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Preeti Dabral 2 years ago
Now, C{x) = 160 - 0.08x
For maxima/minima, put C'(x) = 0
{tex}\Rightarrow{/tex} 160 = 0.08x
{tex}\Rightarrow{/tex} x = 2000
Now, we have C(0) = 5000000
C(2000) = 5160000 and C(4500) = 4910000
{tex}\therefore{/tex} Maximum value of C(x) would be ₹5160000
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