1K.. I also need answers for first and second questions..
Board questions..(so pls i need detailed answers.. we cant use graphs in boards for explanations)
1. If
f(x) = e1/x, x≠0. and 1 if x=0.
Find whether f is contiuous at x=0.
2. f(x) = (x-1) tan Πx/2, if x≠1 and k if x=1.
Find k
3. Do we have Differentiation of a function using first principles.(CBSE)
3. In Appl. of derievatives, there is a question which involves speed given in terms of km/hr, can we leave the final answer in m/s..(in maths)
-
UP 0 DOWN 0 0 14
14 Answers
1differentiation using first princi is in 11th
cant we use graph !! ??
i used [1]
ans shud be in m/s only
1Translate in english pls.. Yes i know is there in X!, but there were also CBSE questions in RD-XII , So is it there in XII syllabus?? But there were no questions in school..in class X!!
1translated in english [4]
but i dont rememeber any boards Q based on first principle
cheak in last few cabse papers
but i think its not in syllabus
1yeah i will giv first tried the easy ones [4][4]
i hav to study for my exams so i will need some time
1Q2,\lim_{x\rightarrow 1} {\frac{x-1}{cot\pi x/2 }}
now apply L'Hospital
\lim_{x\rightarrow 1} {\frac{1}{-cosec^2\pi x/2}} = -1
so k = -1
1Vivek! We donot have L'Hospitals rule in school, so cud u do in the normal way.....I need solution for 1st question please...
1357for II
(x-1)/cot(Ï€x/2)=(x-1)/tan(Ï€/2-Ï€x/2)=(x-1)/tan[(Ï€/2)(1-x)]
=[(Ï€/2)(1-x)]/tan[(Ï€/2)(1-x)]*(-2/Ï€)=-2/Ï€
so k=-2/Ï€
1for 2nd @manish sir limx→a tan(x-a)/(x-a) =1
but here it is not so you are not correct
111st ke liye teacher ko show kar diyo ki
limh-->0+e1/h
limh-->0-e-1/h
and at x=0 ; y =1
but it is clear that the limit does not exists at x=0 as LHS is not equal to RHS[1]
1for 2nd) @ manish sir here lim x→1 [∩/2(1-x)]/ tan∩/2(1-x) ≠1
1hmmm... @manipal how to apply limits here , by luking at it , it is clear.. But should we not find the value at LHS and RHS???
1357for I
lim(h→0+)e1/(0+h)=lim(h→0)e1/h=e∞=∞
lim(h→0-)e1/(0-h)=lim(h→0)e-1/h=e-∞=0