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Ask QuestionPosted by Vansh Saxena 2 years, 9 months ago
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Posted by Khushi Khushi 2 years, 9 months ago
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Preeti Dabral 2 years, 9 months ago
The different methods of integration include: Integration by Substitution. Integration by Parts. Integration Using Trigonometric Identities.
Posted by Dilbag Singh 2 years, 10 months ago
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Pranjal Aggarwal 2 years, 9 months ago
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Preeti Dabral 2 years, 10 months ago
Given : Three friends Ravi, Raju and Rohit were doing buying and selling of stationery items in market.
To Find : Value of x , y and z
Solution:
The price of per dozen of Pen, Notebook and toys are Rupees x. y and z respectively
Ravi purchases 4 dozens of notebooks and sells 2 dozens of pens and 5 dozens of toys and earn 1500
=> 2x - 4y + 5z = 1500
Raju purchases 2 dozens of toy and sells 3 dozens of pens and 1 dozen of notebooks and earn 100
=> 3x + y - 2z = 100
Rohit purchases one dozen of pens and sells 3 dozens of notebooks and one dozen of toys and earn 400
-x + 3y + z = 400
2x - 4y + 5z = 1500 Eq1
3x + y - 2z = 100 Eq2
-x + 3y + z = 400 Eq3
Eq1 + 2Eq3
=> 2y + 7z = 2300
Eq2 + 3Eq3
=> 10y + z = 1300
2y + 7z = 2300 => 10y + 35z = 11500
10y + z = 1300
=> 34z = 10200
=> z = 300
10y + z = 1300
=> y = 100
on solving further x= 200 , y = 100 and z = 300
Posted by Sweety Rani 2 years, 10 months ago
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Preeti Dabral 2 years, 10 months ago
- (a) {tex}\frac{dx}{dt}{/tex} = k( 200,000 - x)
- (c) ₹ 1,55,555
- (d) ₹ 180,246
- (a) 200,000 - 150,000{tex}\left(\frac23\right)^{\mathrm t}{/tex}
- (b) {tex}\frac23{/tex}
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Dhairya Sharma 2 years, 10 months ago
Himanshu Mishra 2 years, 10 months ago
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Rachna Yadav 2 years, 9 months ago
Posted by Ajay Alahar 2 years, 10 months ago
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Preeti Dabral 2 years, 10 months ago
- (c) Let C(x) be the maintenance cost function, then C(x) = 5000000 + 160x - 0.04x2
- (b) We have, C(x) = 5000000 + 160x - 0.04x2
Now, C{x) = 160 - 0.08x
For maxima/minima, put C'(x) = 0
{tex}\Rightarrow{/tex} 160 = 0.08x
{tex}\Rightarrow{/tex} x = 2000 - (b) Clearly, from the given condition we can see that we only want critical points that are in the interval [0, 4500]
Now, we have C(0) = 5000000
C(2000) = 5160000 and C(4500) = 4910000
{tex}\therefore{/tex} Maximum value of C(x) would be ₹5160000 - (a) The complex must have 4500 apartments to minimise the maintenance cost.
- (a) The minimum maintenance cost for each apartment woud be ₹1091.11
Posted by Baba Tushir 2 years, 9 months ago
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Preeti Dabral 2 years, 9 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}
Posted by Subhodeep Mitra 2 years, 11 months ago
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Rashmi Ahuja 2 years, 11 months ago
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Posted by Devendra Kumar 2 years, 9 months ago
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Preeti Dabral 2 years, 9 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}
Posted by Shivam Mandloi 2 years, 11 months ago
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Posted by Laksh Aggarwal 2 years, 11 months ago
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Posted by Ajit Shrivastav 11 A 2 years, 9 months ago
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Preeti Dabral 2 years, 9 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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