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  • 1 answers

Rajat Barwar 6 years, 2 months ago

It is K=2. By simply equating the coefficients.
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  • 1 answers

Payal Goyal 4 years, 6 months ago

answer dijiye above question ka
  • 1 answers

Sohan .G 6 years, 2 months ago

Split denominator 1+x+x^2+x^3 as =(1+x)+x^2(x+1) =(1+x)+(x^2+1) Now the sum would be like integration of 1/(x+1)(x^2+1)dx No by using partial fraction,formula no.5 (refer pg.317 in ncert textbook) 1/(x+1)(x^2+1)dx=A/(x+1)+Bx+C/(x^2+1).........(1) A/(x+1)+Bx+C/(x^2+1)=A(x+1)+(Bx+C)(x^2+1)/(x+1)(x^2+1) Solving and cancelling the denominator an equating and finding the values of a,b,c gives us A=1/2; B= -1/2; C=1/2 substituting the values of a,b,c in (1) ,splitting the integration terms and integrating gives us the answer ANSWER: (1/2)log|x+1|-(1/4)log|x^2+1|+(1/2)tan^(-1)x+c
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Tisha Garg 6 years, 2 months ago

Question is wrong so please dont answer and sorry for wrong question
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  • 1 answers

Pari Goyal 6 years, 2 months ago

It is incomplete question for this an particular value of fx and gx should be given
  • 2 answers

Pari Goyal 6 years, 2 months ago

How would we get the fofmulas in this app ples tell

Ɽøⱨł₮ ₲Ʉ₱₮₳ 6 years, 2 months ago

U are get in this app all formula, related to continuity and differtiability
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Sia ? 6 years, 2 months ago

Same syllabus as before.

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  • 1 answers

Sia ? 6 years, 2 months ago

A matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.

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Rajat Barwar 6 years, 2 months ago

Friend,if u are asking for whole root on x+a,then the answer by aman is correct. If the root is only on x and not a,then it is. -1/2(√x +a){√(√x + a )}(2√x). I hope this may help you. I am sorry for any error.

Aman Kumar 6 years, 2 months ago

Ans :------- -1/ (2√x +a )(x + a)
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Rajat Barwar 6 years, 2 months ago

Friend I think it is 0.(these have a unique sol.) hope it may help you. If I am wrong please inform me.
  • 2 answers

Rajat Barwar 6 years, 2 months ago

I think the answer is 1/28. You can let 2^x as t and then find the value of t,it will be1/3. Now put 8^x as t^3 and solve by putting the value of t in this. Hope it may help you.

Hardik Singhaniya 6 years, 2 months ago

Solutions h but kese du???
  • 1 answers

Sia ? 6 years, 2 months ago

we have, xy =1

{tex} \Rightarrow y = \frac{1}{x}{/tex}
{tex}\therefore\ \frac{dy}{dx}=-\frac{1}{x^2}{/tex}

The slope of the normal ={tex}x^2{/tex}

If ax+by+c=0 is normal to the curve xy=1,then

{tex}x^2=-\frac{a}{b} \ [\because slope\ of\ normal\ =-\frac{coeff.\ of\ x}{coeff. of\ y}]{/tex}

{tex}\therefore \; - \frac{a}{b} > 0{/tex}
{tex}\Rightarrow\ a>0,b<0 \ or\ a<0,b>0{/tex}

  • 3 answers

Lucky Trivedi 6 years, 2 months ago

1/2√xSin√x

Nithish Kumar Sekar 6 years, 2 months ago

For cos diff -sin and for /x is /x/2 -sin /x/2/x

Jagjeet Singh 6 years, 2 months ago

-sin√x/2√x
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Dolly ?️ 6 years, 2 months ago

0

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