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  • 1 answers

Pranay Jaiswal 4 years, 7 months ago

What do u mean???
  • 5 answers

Student Of The Year 4 years, 7 months ago

NCERT

Manish Karwasra 4 years, 7 months ago

Rd

Zeeshan Younis 4 years, 7 months ago

Cingage

Bholu Tonger 4 years, 7 months ago

R s Aggarwal

Shriya ? 4 years, 7 months ago

Arihant all in one is best
  • 5 answers

Twinkle . 4 years, 7 months ago

1/(√1-x²)

Adarsh Ninja 4 years, 7 months ago

1/√1-x²

Jyotika Upadhyay 4 years, 7 months ago

1/✓1-x²

Jyotika Upadhyay 4 years, 7 months ago

1- ✓1-x²

Ashish Tanwar 4 years, 8 months ago

1/1-x²
  • 2 answers

Twinkle . 4 years, 7 months ago

(x-1/x²+1)dx = (x/x²+1)dx - (1/x²+1)dx Let (x/x²+1)dx be I1 and (1/x²+1)dx be I2 I2=tan-¹x I1=(x/x²+1)dx Let x²=u 2xdx=du On substitution, I1=(1/2)(1/u+1)du =log|u+1|/2 =log|x²+1|/2 Ans. I1-I2 = log|x²+1|/2 - tan-¹x

Adarsh Ninja 4 years, 7 months ago

1/2 log|x²+1| - tan—¹x
  • 1 answers

Nitish Singh 4 years, 8 months ago

Go through the solution section
  • 2 answers

Twinkle . 4 years, 7 months ago

a^(1/t).t^(a-1).log a.(t+1)^(1-a)

Gurharman Kaur 4 years, 8 months ago

dy/dx=(-a^t+1/t.loga)/a.(t+1/t)^a-1.1/t^2
  • 5 answers

Yash Nagwal 4 years, 7 months ago

5/4

Twinkle . 4 years, 7 months ago

P(B/A) = 0.4 P(B∩A)/ P(B) = 0.4 P( B ∩ A) / 0.5 = 0.4 P( B ∩ A) = 0.4×0.5 P( A ∩ B) = 0.20

Ashish Tanwar 4 years, 8 months ago

??

Rajaneha Rajasekar 4 years, 8 months ago

P(B/A)= 0.4= p(A intersection B) /p(A) . Then p( A intersection B) = p(B/A) × P(A) .that is p(A intersection B) = 0.4×0.8= 0.32

Parth Pathak 4 years, 8 months ago

P(B/A) = 0.4 P(B intersection A)/ P(B) = 0.4 P( B intersection A) / 0.5 = 0.4 P( B intersection A) = 0.4*0.5 P( A intersection B) = 0.20
X^y
  • 3 answers

Gurharman Kaur 4 years, 8 months ago

=dy/dxlogx+y/x

Ashutosh Kesarwani 4 years, 8 months ago

G

Gourav Chhapola 4 years, 8 months ago

Taking log Y×logX
  • 2 answers

Gungun Gupta?? 4 years, 7 months ago

Sach me kya ye remove ho gya h

Student Of The Year 4 years, 8 months ago

Chapter nhi h topic h relation and function chapter ka. And this topic is removed in revised mathematics sallybus.
  • 2 answers

Gourav Chhapola 4 years, 8 months ago

3/√(1-x² ) - 1/√(1-x² ) =2/√(1-x² )

Mansi Maheshwari 4 years, 8 months ago

X=1/root2
  • 4 answers

Twinkle . 4 years, 7 months ago

2/x³

Gurharman Kaur 4 years, 8 months ago

-2/x^3

Gourav Chhapola 4 years, 8 months ago

x+1/x

Mansi Maheshwari 4 years, 8 months ago

2/x3
  • 0 answers
  • 2 answers

Gurharman Kaur 4 years, 8 months ago

Equation show nii ho rhi properly

Guvv Rona 4 years, 8 months ago

Can I delete it
  • 1 answers

Gurharman Kaur 4 years, 8 months ago

Question ch -2 whole de product ch hai
  • 1 answers

Alex Pal 4 years, 8 months ago

The sum of three no is 6 if we multiply third no by three
  • 0 answers
  • 0 answers
  • 2 answers

Shraddha ✨✰✰ 4 years, 8 months ago

For every real number (or) valued function f(x), the values of x which satisfies the equation f¹(x)=0are the point of it's local and global maxima or minima. This occus due to the fact that, at the point of maxima or minima, the curve of the function has a ,zero slope. We have function f(x)=(1​/x)^x We will be using the equation, y=(1/x)^x Taking in both sides we get ln y=−xlnx Differentiating both sides with respect to x. y.dy​/dx = −lnx − 1 dy​/dx=−y (ln x + 1) Equating dy​/dx  to 0, we get −y(ln x + 1)=0 Since y is an exponential function it can never be equal to zero, hence ln x + 1 = 0 ln x = −1 x=e^−1 So, for the maximum value we put x=e^−1 in f(x) to get the value of f(x) at the point. f(e^−1) = e^1/e Hence the maximum value of the function is e^1/e. ∴ So, the answer is option →B. {e^1/e}

बेबि बीच 4 years, 8 months ago

1
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  • 2 answers

Elina ❤️ 4 years, 8 months ago

Only Practice

Sushant Adhana 4 years, 8 months ago

Firstly read all ncert chapter and then solve last year paper and sample paper
  • 3 answers

Sushant Adhana 4 years, 8 months ago

Calculas

Aruthra A 4 years, 8 months ago

Calculas

Chetna Yadav 4 years, 8 months ago

Calculas unit

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