CBSE > Class 12 > Mathematics | CBSE Board | Homework Help

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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Can anyone give me the new syllabus of maths

Posted by Vinay Gautam 5 years, 2 months ago

CBSE > Class 12 > Mathematics

1 answers

Gaurav Seth 5 years, 2 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>

<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>

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The nth term of an A. P is given by an =3+4n the comman difference is

Posted by Suneha Rana 5 years, 2 months ago

CBSE > Class 12 > Mathematics

4 answers

Shivam Verma 5 years, 2 months ago

We have given : an= 3+4n now replace "n" by (n-1) we get : a (n-1)= 3 + 4 (n-1) Now : an - a (n-1) i.e 3 + 4n - (3 + 4n - 4) ; 3 + 4n -3 - 4n + 4 = 4 (comman difference)

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Ã.P.W.B.Đ ⚡ 5 years, 2 months ago

'n'th term of an AP will always be a linear expression in 'n' and its coefficient will always be the common difference (d) of that AP.

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Ã.P.W.B.Đ ⚡ 5 years, 2 months ago

The above method is too long.

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Gaurav Seth 5 years, 2 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

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Cubic root y- 1 is a rational number or not???

Posted by Shahanaj Parbin 5 years, 2 months ago

CBSE > Class 12 > Mathematics

1 answers

Saurabh Singh 5 years, 2 months ago

y >=1 rational y<1 Irrational

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y-sin(log x)

Posted by Khushboo Verma 5 years, 2 months ago

CBSE > Class 12 > Mathematics

2 answers

Ayush Vishwakarma?? 5 years, 2 months ago

D(sin(log x))/dx... = cos(logx)/x .... Becoz d(sinx)dx = cosx and logx = 1/x.

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Aman Jaiswal 5 years, 2 months ago

Cos logx . 1/x

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Differentiate (tan-¹x)² ⁉️

Posted by ?????? ???? . 5 years, 2 months ago

CBSE > Class 12 > Mathematics

1 answers

Sudhanshu Mahle 5 years, 2 months ago

2tan-¹x (1/1+x²)

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ANSWER

Is the word problems have been deleted from matrix ch?

Posted by Supriya Tiwari 5 years, 2 months ago

CBSE > Class 12 > Mathematics

2 answers

Ayush Vishwakarma?? 5 years, 2 months ago

No dear

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Aditya Gupta 5 years, 2 months ago

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ANSWER

Prove that 1/tan (1+x)/(1-x)=π/4+1/tanx

Posted by Anamika Singh 5 years, 2 months ago

CBSE > Class 12 > Mathematics

1 answers

Saurabh Singh 5 years, 2 months ago

LHS = 1/tan(1+x)/(1-x) ( Put x =tany) = tan^-1 (1+x)/(1-x) = tan^-1(1+tany)/(1-tany) = tan^-1 (tanπ/4 + tany)/(1- tanπ/4.tany) = tan^-1[tan(π/4+y)] =π/4 + y = π/4 + tan^-1x (tan^-1 =y) =π/4 + 1/tanx

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The passing number in maths out of 80

Posted by Prince Singh 5 years, 2 months ago

CBSE > Class 12 > Mathematics

5 answers

Sanya Bhatia 5 years, 2 months ago

Out of 80 are 26, out of 70 are 23, out of 60 are 19 , out of 40 are 13 and out of 30 are 9.

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Aditya Gupta 5 years, 2 months ago

26.4

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Mishti ???? 5 years, 2 months ago

Hmmm 26/80 Aap pass ho gaye bhaiya ji

1Thank You

Prince Singh 5 years, 2 months ago

Ok igot 21 marks

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Ayush Vishwakarma?? 5 years, 2 months ago

26.

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Binary operation is present in the syllabus.

Posted by Ayush Vishwakarma?? 5 years, 2 months ago

CBSE > Class 12 > Mathematics

5 answers

Ayush Vishwakarma?? 5 years, 2 months ago

Thnq dear mishti

1Thank You

Mishti ???? 5 years, 2 months ago

Nooooo dear 2 years phle he hata diya tha ??

1Thank You

Ayush Vishwakarma?? 5 years, 2 months ago

Dekh kr bataye

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Khushi....? ??‍? 5 years, 2 months ago

Dont know mere time nii tha

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Sai Kalyan 5 years, 2 months ago

Jl?????????????????????

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Integrate √tan^-xdx

Posted by Ankit Kumar 5 years, 2 months ago

CBSE > Class 12 > Mathematics

1 answers

Gaurav Seth 5 years, 2 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>