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  • 1 answers

Ritesh Chakraborty 6 years, 3 months ago

For one one let us consider that f(x1) =f(x2) Now 2x1=2x2(note 2 is only to show the difference it has no work and X, is alphabetically) Thus when we divide we get X1=X2. Thus it is one one function The one to will be a very big process... It can't be written here although in theory it's is onto because if we multiply and R number with 2x we get the any number belongs to R... So thus in this case Range=Codomain Therefore it is onto
  • 1 answers

Becky Shoshannah 6 years, 3 months ago

Total surface area of cylinder, T.S.A = 2πr(r+h) Circular/Lateral surface area, C.S.A/ L.S.A = 2πrh Volume, V = πr^2h
  • 4 answers

Deepika Nagesh 6 years, 3 months ago

π/3 is ur answer

Nishu Sahu 6 years, 3 months ago

Cos^-1(sin4π/3)= 5π/6

Prashant Singh 6 years, 3 months ago

yaar galti se is maim reply kar diya waha jo dusra ques di ho n uska ans h sin^-1( underroot3 +1)

Prashant Singh 6 years, 3 months ago

sin^-1(cos 15)=sin^-1(cos (90-75))=sin^-1(sin75)=75degree ans aa gya i hope shi ho nhi hoga toh bol dena i will try
  • 0 answers
  • 1 answers

Sia ? 6 years, 3 months ago

Check NCERT Solutions here : https://mycbseguide.com/ncert-solutions.html

  • 1 answers

S.....Sharma☺??? .. 6 years, 3 months ago

Aapko iska answer ise app m mil jayega
  • 4 answers

Sudhir Patel 6 years, 3 months ago

It's simply 1

Aman Hanspal 6 years, 3 months ago

1

S.....Sharma☺??? .. 6 years, 3 months ago

1

Shatayu Ganvir 6 years, 3 months ago

1
  • 2 answers

S.....Sharma☺??? .. 6 years, 3 months ago

3πr^2

Santy Baralu 6 years, 3 months ago

c. s.a of hemisphere =2(22/7)r2. total surface area =3(22/7)r2
  • 1 answers

Vishal Kumar 6 years, 3 months ago

(-infinity,0]
  • 1 answers

Sia ? 6 years, 3 months ago

A perfect number is a positive integer that is equal to the sum of its positive divisors, excluding the number itself. For instance, 6 has divisors 1, 2 and 3, and 1 + 2 + 3 = 6, so 6 is a perfect number.

  • 0 answers
  • 4 answers

S.....Sharma☺??? .. 6 years, 3 months ago

Nahi saare to nahi but kuch chapter h jinka basic 12 m kaam aayega jase ki trigonometry ,differential

Muskan .. 6 years, 3 months ago

Nooo

Santy Baralu 6 years, 3 months ago

no it is not necessary for board exam but important for competitive exam

Priya Dharshini 6 years, 3 months ago

No... it is not necessary for boards. Only for competitive exams u need class 11 and 12
  • 2 answers

S.....Sharma☺??? .. 6 years, 3 months ago

-2√2xcosec^2 x^2/√sin2x^2

Dolly ?️ 6 years, 3 months ago

= (-2cosec^2(2x)) /under root cot2x
  • 1 answers

Sahil Kumar 6 years, 3 months ago

Angle (theeta)/2 = tan inverse x/2
  • 8 answers

Kumar Vishal Behera 6 years, 3 months ago

Follow rd sharma you will find ncert and ncert exemplar in that too.

Mithlesh Prajapati 6 years, 3 months ago

All book is best for board but to solve more questions or for concept clear you should follow RD Sharma

Priya Dharshini 6 years, 3 months ago

Rd sharma,rs agarwal and ncert

Sidharth Singh 6 years, 3 months ago

Rs Aggarwal

Rabi Sankar Roy 6 years, 3 months ago

Rd Sharma

Gautam Gupta 6 years, 3 months ago

Rd sharma it have max no. Of questions for prepretion

Rudra Singh 6 years, 3 months ago

Rs Aggrwal is the best book for the board exams...

Ayush Mishra 6 years, 3 months ago

Rd sharma
  • 0 answers
  • 1 answers

Sia ? 6 years, 3 months ago

You can check last year papers here : https://mycbseguide.com/cbse-question-papers.html

  • 1 answers

Sia ? 6 years, 3 months ago

{tex}\vec a = 2\hat i + 4\hat j - 5\hat k{/tex}

{tex}\vec b = \lambda \hat i + 2\hat j + 3\hat k{/tex}

{tex}\vec a + \vec b = \left( {2 + \lambda } \right)\hat i + 6\hat j - 2\hat k{/tex}

Unit vector along

{tex}\vec a + \vec b = \frac{{\vec a + \vec b}}{{\left| {\vec a + \vec b} \right|}}{/tex}

{tex}= \frac{{\left( {2 + \lambda } \right)\hat i + 6\hat j - 2\hat k}}{{\sqrt {{{\left( {2 + \lambda } \right)}^2} + {{\left( 6 \right)}^2} + {{\left( { - 2} \right)}^2}} }}{/tex}

{tex}= \frac{{\left( {2 + \lambda } \right)\hat i + 6\hat j - 2\hat k}}{{\sqrt {{{\left( {2 + \lambda } \right)}^2} + 40} }}{/tex}

ATQ , {tex}\vec c.\left( {\vec a + \vec b} \right) = 1{/tex}

{tex}\left( {\hat i + \hat j + \hat k} \right).\left( {\frac{{\left( {2 + \lambda } \right)\hat i + 6\hat j - 2\hat k}}{{{{\left( {2 + \lambda } \right)}^2} + 40}}} \right) = 1{/tex}

{tex}\frac{{\left( {2 + \lambda } \right) + 6 - 2}}{{\sqrt {{{\left( {2 + \lambda } \right)}^2} + 40} }} = 1{/tex}

{tex}2 + \lambda + 4 = \sqrt {{{\left( {2 + \lambda } \right)}^2} + 40} {/tex}

Sq both sides,

{tex}{\lambda ^2} + 36 + 12\lambda = {\left( {2 + \lambda } \right)^2} + 40{/tex}

{tex}\lambda = 1{/tex}

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