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Posted by Vinay Gautam 4 years, 4 months ago
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Gaurav Seth 4 years, 4 months ago
The Syllabus in the subject of Mathematics has undergone changes from time to time in accordance with growth of the subject and emerging needs of the society. Senior Secondary stage is a launching stage from where the students go either for higher academic education in Mathematics or for professional courses like Engineering, Physical and Biological science, Commerce or Computer Applications. The present revised syllabus has been designed in accordance with National Curriculum Framework 2005 and as per guidelines given in Focus Group on Teaching of Mathematics 2005 which is to meet the emerging needs of all categories of students. Motivating the topics from real life situations and other subject areas, greater emphasis has been laid on application of various concepts.
Click on the given link :
<a href="http://cbseacademic.nic.in/web_material/CurriculumMain21/SrSecondary/Mathematics_Sr.Sec_2020-21.pdf">http://cbseacademic.nic.in/web_material/CurriculumMain21/SrSecondary/Mathematics_Sr.Sec_2020-21.pdf</a>
Click on the given links for revised syllabus and deleted topics of all the subjects:
<font color="#FF6600"><font style="box-sizing: border-box;">Revised Curriculum for the Academic Year 2020-21</font></font>
<div class="panel-group" id="accordion" style="margin-bottom:5px; text-align:start; -webkit-text-stroke-width:0px"> <div class="panel panel-default" style="border:1px solid #dddddd; margin-bottom:0px; border-radius:4px"> <div class="panel-heading" style="border-bottom:0px #dddddd; padding:10px 15px; border-top-left-radius:3px; border-top-right-radius:3px; border-top-color:#dddddd; border-right-color:#dddddd; border-left-color:#dddddd"><a data-toggle="collapse" href="http://cbseacademic.nic.in/Revisedcurriculum_2021.html#collapse2" style="box-sizing:border-box; color:inherit; text-decoration:none; display:block; font-weight:bold">Revised Secondary Curriculum (IX-X)</a>
</div> </div> <div class="panel panel-default" style="border:1px solid #dddddd; margin-bottom:0px; border-radius:4px; margin-top:5px"> <div class="panel-heading" style="border-bottom:0px #dddddd; padding:10px 15px; border-top-left-radius:3px; border-top-right-radius:3px; border-top-color:#dddddd; border-right-color:#dddddd; border-left-color:#dddddd"><a data-toggle="collapse" href="http://cbseacademic.nic.in/Revisedcurriculum_2021.html#collapse1" style="box-sizing:border-box; color:inherit; text-decoration:none; display:block; font-weight:bold">Revised Senior Secondary Curriculum (XI-XII)</a>
</div> </div> </div>Posted by Suneha Rana 4 years, 4 months ago
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Shivam Verma 4 years, 4 months ago
Ã.P.W.B.Đ ⚡ 4 years, 4 months ago
Gaurav Seth 4 years, 4 months ago
Solution :
It is given that,
= 3 + 4n_____(1)
Now,
put n = 1, 2, 3 turn wise. So that we can find ,
and
.
★For finding ,
put n = 1 in (1)
= 3 + 4 (1)
= 3 + 4
= 7
★For finding ,
put n = 2 in (1).
= 3 + 4 (2)
= 3 + 8
= 11
★For finding ,
put n = 3 in (1).
= 3 + 4 (3)
= 3 + 12
= 15
So, the AP will be :
7, 11, 15 .........
Here, = 7
= 11
= 15
Now,
common difference (d) = -
common difference (d) = 11 - 7
common difference (d) = 4
Posted by Shahanaj Parbin 4 years, 4 months ago
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Posted by Khushboo Verma 4 years, 4 months ago
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Ayush Vishwakarma?? 4 years, 4 months ago
Posted by ?????? ???? . 4 years, 4 months ago
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Posted by Supriya Tiwari 4 years, 4 months ago
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Posted by Anamika Singh 4 years, 4 months ago
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Saurabh Singh 4 years, 4 months ago
Posted by Prince Singh 4 years, 4 months ago
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Sanya Bhatia 4 years, 3 months ago
Posted by Ayush Vishwakarma?? 4 years, 4 months ago
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Posted by Ankit Kumar 4 years, 4 months ago
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Gaurav Seth 4 years, 4 months ago
∫√<font face="Verdana"><font style="margin: 0px; padding: 0px; box-sizing: inherit; color: rgba(0, 0, 0, 0.54); font-size: 16px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; background-color: rgb(255, 255, 255); text-decoration-style: initial; text-decoration-color: initial;"><font size="3"><font style="margin: 0px; padding: 0px; box-sizing: inherit;">(tan x) dx
Let tan x = t2
⇒ sec2 x dx = 2t dt
⇒ dx = [2t / (1 + t4)]dt
⇒ Integral ∫ 2t2 / (1 + t4) dt
⇒ ∫[(t2 + 1) + (t2 - 1)] / (1 + t4) dt
⇒ ∫(t2 + 1) / (1 + t4) dt + ∫(t2 - 1) / (1 + t4) dt
⇒ ∫(1 + 1/t2 ) / (t2 + 1/t2 ) dt + ∫(1 - 1/t2 ) / (t2 + 1/t2 ) dt
⇒ ∫(1 + 1/t2 )dt / [(t - 1/t)2 + 2] + ∫(1 - 1/t2)dt / [(t + 1/t)2 -2]
Let t - 1/t = u for the first integral ⇒ (1 + 1/t2 )dt = du
and t + 1/t = v for the 2nd integral ⇒ (1 - 1/t2 )dt = dv
Integral
= ∫du/(u2 + 2) + ∫dv/(v2 - 2)
= (1/√2) tan-1 (u/√2) + (1/2√2) log(v -√2)/(v + √2)l + c
= (1/√2) tan-1 [(t2 - 1)/t√2] + (1/2√2) log (t2 + 1 - t√2) / t2 + 1 + t√2) + c
= (1/√2) tan-1 [(tanx - 1)/(√2tan x)] + (1/2√2) log [tanx + 1 - √(2tan x)] / [tan x + 1 + √(2tan x)] + c</font></font></font></font>
Posted by Ankit Kumar 4 years, 4 months ago
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Posted by Sahil Chauhan 4 years, 4 months ago
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Posted by Aditya Darne 4 years, 4 months ago
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Yogita Ingle 4 years, 4 months ago
2/3 = 2 x 5 / 3 x 5
= 10 / 15
4/5 = 4 x 3 / 5 x 3
12 / 15
we can find 5 rational numbers betwen 2/3 and 4/5 by multiplying them with 5 + 1 = 6
10 / 15 X 6 /6
= 60 / 90
12/ 15 X 6 /6
= 72 / 90
the rational numbers = 61 / 90 62 /90 63 /90 64/90 65/90 66/90 67 /90 .....
Mishti ???? 4 years, 4 months ago
Posted by Aditya Darne 4 years, 4 months ago
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Posted by Muskan Chabrs 4 years, 4 months ago
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Ayush Vishwakarma?? 4 years, 4 months ago
Posted by Ayush Vishwakarma?? 4 years, 4 months ago
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Khushi....? ??? 4 years, 4 months ago
Mishti ???? 4 years, 4 months ago
Mishti ???? 4 years, 4 months ago
Khushi....? ??? 4 years, 4 months ago
Posted by Kunal Prasad 4 years, 4 months ago
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Posted by Ayush Vishwakarma?? 4 years, 4 months ago
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Subhankar Sharma 4 years, 4 months ago
Subhankar Sharma 4 years, 4 months ago
Posted by Yuvraj Yadav 4 years, 4 months ago
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Posted by Ram Singh 4 years, 4 months ago
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Posted by Sonali Patel 4 years, 4 months ago
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Khushi....? ??? 4 years, 4 months ago
Khushi....? ??? 4 years, 4 months ago
Ayush Vishwakarma?? 4 years, 4 months ago
Posted by Kiran Verma 4 years, 4 months ago
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Posted by Gayathri Gayu 4 years, 4 months ago
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Posted by Shalu Yadav 4 years, 4 months ago
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Gaurav Seth 4 years, 4 months ago
f(x) = sin 3x - cos 3x , 0 < x < π,
or f'(x) = 3 cos 3x + 3 sin 3x
= 3(cos 3x + sin 3x)
Put f'(x) = 0
or cos 3x + sin 3x = 0
Posted by Abhishek Tanwar 4 years, 4 months ago
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Posted by Vaibhav Kulkarni 4 years, 4 months ago
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Gaurav Seth 4 years, 4 months ago
2cos2x-cos4x
=2 cos2x -2cos22x +1
=-2(cos2x-1/2)2 +3/2
minimum value occur when cos2x-1/2 will be maximum
=-2(-1-1/2)2 +3/2
=-3
and minimum value is 3/2
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Yogita Ingle 4 years, 4 months ago
If α, β be the zeros of the quadratic polynomial ,then
(x−α)(x−β) is the quadratic polynomial.
Thus, (x−α)(x−β) is the polynomial.
= x2−αx−βx+αβ
=x2−x(α+β)+αβ(i)
(α+β)=−10 and αβ=-39
Now putting the value of (α+β),αβ in equation (i) we get,
x2−x(−10)+(-39)
=x2 + 10x - 39
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