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  • 1 answers

Gayatri Guntreddi 4 years, 3 months ago

Write tanx as sinx/cosx and cotx as cosx/sinx I= ✓((sin^2x/sinxcosx) + ✓(cos^2x/sinxcosx))dx I= (sinx+cosx dx)/✓(sinxcosx) sinx-cosx=t (sinx+cosx)dx=dt (sinx-cosx)^2=t^2 1–2sinxcosx=t^2 sinxcosx=(1-t^2)/2 Therefore, I=dt/✓((1-t^2)/2) I=(✓2)sin^-1(t) + C I=(✓2)sin^-1(sinx-cosx) + C
  • 1 answers

Sushant Dwivedi 4 years, 3 months ago

Let x^3=t Differentiate above 3x^2=dt/dx x^2 dx=dt/3 -------------(1) From equation(1) Int. x^2 sin x^3 dx =Int. Sint dt/3 =1/3 Int. Sint dt =1/3 (- cos t) +c =-1/3 (cos t) +c. [ANS]
  • 4 answers

Thakkar Dhruv 4 years, 3 months ago

Unitless Quantity

Sandipan Samanta 4 years, 3 months ago

Unit of slope = unit of y-axis/ unit of x-axis

Gaurav Srivastav Gs 4 years, 3 months ago

In Mathematics, slope =tanx And trigonometry value value has no unit so slope is may be unitless.

Sachi Singh 4 years, 3 months ago

Meter/second..
  • 1 answers

Hiral Solanki 4 years, 2 months ago

0.9
  • 1 answers

Watch It 4 years, 3 months ago

Log y^x=logx^y Xlogy=ylogx Differntiating with respect to x X×(1/y)×(dy/DX) + logy.1=y.(1/x) + logx×(dy/dx) Dy/DX[x/y - log x]= y/x - log y dy\dx[(x-ylogx)/y]=(y-xlogy)/x Agae khud krlena...
  • 0 answers
  • 3 answers

Rajesh Kumar 4 years, 3 months ago

[-1/4 , ∞]

Tannu Rao 4 years, 3 months ago

Open interval on infinity

Tannu Rao 4 years, 3 months ago

f'(x)=2x+1=0 x=-1/2 f(-1/2)=1/4 - 1/2 = -1/4 Therefore range of f(x) is [-1/4 , infinity]
  • 2 answers

Sharayu . 4 years, 2 months ago

As per the latest cbse pattern of 2021 4.2 ex or the properties od determinants has been reducted

Vedant Kumar Prajapati 4 years, 3 months ago

Why shouldn't you go for pdf of this exercise?
  • 1 answers

Smile Josan 4 years, 3 months ago

Can you plz give some tips!.. I am in 12th right now!
  • 2 answers

Shailesh Singh 4 years, 3 months ago

44.71

Gaurav Srivastav Gs 4 years, 3 months ago

47.876
  • 2 answers

Tannu Rao 4 years, 3 months ago

U can draw graphs of cosx and sinx in interval [0,π/2]. From the graphs, we can infer that both the functions are one one. Or f'(x)=cosx and g'(x)=-sinx. Derrivatives of both the functions are only +ve and -ve respectively in the given interval. Hence both are one one. (f+g)(x)=sinx +cosx (f+g)'(x)=cosx-sinx which is both +ve as well as -ve in the given interval. Hence it is many one.

Priyanka Birat 4 years, 3 months ago

Can anyone solve this question in good explanation .
  • 3 answers

Nikhil Kumar 4 years, 3 months ago

Hor Kuchh

Nikhil Kumar 4 years, 3 months ago

When gof is one one then f is always one one but not necessarily that g is one one. When gof is onto then g always one one but not necessarily f is one one.

Aryan Choudhary ?? 4 years, 3 months ago

When gof is one one then f is always one one but not necessarily that g is one one. When gof is onto then g always one one but not necessarily f is one one.
  • 1 answers

Tannu Rao 4 years, 2 months ago

0.87
  • 2 answers

Anjali ? 4 years, 3 months ago

3 (tan^-1)x+(cot^-1)x=π 2(tan^-1)x+(tan^-1)x+(cot^-1)x=π 2(tan^-1)x+π/2=π 2(tan^-1)x=π/2 (tan^-1)x=π/4 X=1

Sumit Malviya 4 years, 3 months ago

We can write it as . 2tan-1x+tan-1x + cot-1x=π Now,we know tan-1x+cot-1x=π÷2 So 2tan-1x+π÷2= π 2tan-1x=π÷2 tan-1x=π Putting (tan-1 0 =π tan-1x=tan-1 0 On comparing (x=0)
  • 1 answers

Mohd Atif 4 years, 2 months ago

(x)^cosx{cosx/x--sinx.logx}+cos2x=y'
  • 3 answers

Tannu Rao 4 years, 3 months ago

4π-10 is also in range of cos-1(x)

Tannu Rao 4 years, 3 months ago

Cos(10) can be written as cos(4π-10). Therefore cos-1(cos10)=cos-1(cos(4π-10)) = 4π-10

Ravi Raj 4 years, 3 months ago

10
  • 1 answers

Tannu Rao 4 years, 3 months ago

(1/(7Log7x)).(1/x).7
  • 1 answers

Tannu Rao 4 years, 3 months ago

(e^√2x).(√2)

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