Click to Chat
1800 2000 838
CART 0
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
JEE Advanced
JEE Main
BITSAT
View complete IIT JEE section
AIPMT
AIIMS
View Complete Medical section
Medical Exam Calendar
NTSE
KVPY
Olympiads
CBSE
ICSE
UAE
Saudi Arabia
Qatar
Oman
Bahrain
Kuwait
Indonesia
Malaysia
Singapore
Uganda
View complete NRI section
Physics
Chemistry
Maths
Revision notes
View complete study material
Be the first one to answer
for these type of questions best approach is SERIES EXPANSION the answer will be 3 as u expand the series of cosx and e x and multiply the atarting coefficient u get x 3 and then so on but we consider...
Thanks :)
the answer is e^((-2)/pi) its simple because on putting x=a its 1^infinity form we have a method to solove dis type of limnits it can be wriiten as e^(tan(pi x/2a))(2-a/x -1) with limit as x=a now put...
Dear student, Attached Image is missing.. Thanks.
no the answer is option c
its a que from differential que we have to solve it
Hello student, Your question is not clear and seems to be incomplete. Please recheck your question and post it again so that I can provide a meaningful answer.
Hello student, First of all, let us tabulate the data in a systematic form: Class Interval f cf 161-167 79 79 167-173 92 171 173-179 60 231 179-185 22 253 185-191 5 258 191-197 2 260 Q 1 = 161 + 6/79...
Hello student, You are right that for finite series, we can simply use the sum rule for differentiation. But, in case of infinite series, one has to be a bit cautious about the general conditions...
dont understand whats the limit of integration?? not written correctly
Prime minister drinks the hotter coffee. The PM doesn’t gives time for heat transfer to happen, hence temperature is still the same, while for President has decreased.
[tanx= x + 1/3x^3 + 2/(15)x^5 + (17)/(315)x^7 + (62)/(2835)x^9 +...]
tanx= x + 1/3x^3+ 2/(15)x^5 + (17)/(315)x^7 + (62)/(2835)x^9+...] I hope u needed this only.
substitue y' =uy''=u'we get u'=2x/(x*x+1)uln(u)=ln(x^2+1)+C=> u=c(x^2+1)dy/dx=c(x^2+1)y=c((x^3)/3+x)+c1
dy/dx= cosx + sinx In first quadrant both are positive so increasing In second quadrant sinx is positive but cos x is negative till 135 magnitude of sinx > cosx so it will be positive Increasing...
Hi, Tangent to first parabola is y=mx+1/m If this is also the tangent to second parabola than put this equation in second parabola. Form a quadratic in x and put its d=0 as tangent and curve has only...
d/dx[ax+b] = d/dx[ax]+d/dx[b] =ad/dx[x]+0 a(1) =a
Yes
1529