106really great sir..
Q1. Remainder when the number 111....1 (123 ones) is divided by 271 is
Q2. The period of the function such that f(x-1) + f(x+1) = √2f(x) is
Q3. Let S=\sum_{k=1}^{80}{\frac{1}{\sqrt{k}}}. Then [S] equals where [] is GINT
Q4. Let x be a positive real number. Then the infinite sum \sum_{n=1}^{\infty }{\frac{(n-1)!}{(x+1)(x+2)...(x+n)}} equals
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18 Answers
1You almost did it Karna.
Using : xTn+1 = nTn - (n+1)Tn+1
xT2 = T1 - 2T2
xT3 = 2T2 - 3T3
..... and so on
adding n such equations : x(T2 + T3 + .....+ Tn+1 ) = T1 - (n+1)Tn+1
x(T1 + T2 + T3 + .....+ Tn+1 ) = (1+x)T1 - (n+1)Tn+1
xSn+1 = 1 - (n+1)!(x+1)(x+2)....(x+n+1) (since T1 = 1x+1 )
xSn+1 = 1 - (1x+1) (2x+2) .... (n+1x+n+1)
all these bracketed fractions are smaller than 1 and if we multiply infinite such terms their limiting value is 0
xS∞ = 1
S∞ = 1/x
1this sum can be done by method of difference
T_{n}=\frac{(n-1)!}{(x+1)(x+2)..........(x+n)}
T_{n+1}=\frac{n(n-1)!}{(x+1)(x+2)..........(x+n)(x+n+1)}
T_{n+1}=\frac{n T_{n}}{(x+n+1)}
hence
xT_{n+1}= n T_{n}-(n+1)T_{n+1}
hence answer is
\frac{1}{x}\lim_{n\rightarrow infinity}(\frac{n!}{(x+1)(x+2)(x+3).........(x+n)})
1\sum{\frac{1}{\sqrt{k}+\sqrt{k+1}}} < \sum{\frac{1}{2\sqrt{k}}} <\sum{\frac{1}{\sqrt{k-1}+\sqrt{k}}}
{\frac{1(\sqrt{k})-\sqrt{k+1}}{\sqrt{k}+\sqrt{k+1}(\sqrt{k})-\sqrt{k+1})}}
-{\frac{(\sqrt{k})-\sqrt{k+1}}{1}}
\sum{{\frac{(\sqrt{k+1})-\sqrt{k}}{1}}}=9-1=8
hence answer is 16
1Question. previously u had told the way to find generative function.
But I am unable to find the generative function of ax-1 + ax+1 = √2ax.
urs answer was Acos(Ï€x/4).
Please explain the way out sir.
341Well known. The hint is:
\frac{1}{\sqrt k + \sqrt{k+1}} < \frac{1}{2\sqrt k} < \frac{1}{\sqrt{k-1} + \sqrt k}
Now can you nest the sum between two integers?
1708for ans (1)
http://targetiit.com/iit-jee-forum/posts/binomial-11560.html
341Check out http://www.goiit.com/posts/list/differential-calculus-hard-question-951552.htm
21@ hari sir can u plz explain a bit more on how u found out the function f(x) [1]
i dun hav any idea abt generating functions
341thats precisely the point. The answer is already known to those who set the question paper, because they have lifted it from some source themselves. and they have not solved the question themselves, so how would they know what is the difficulty level.
Here, if you know generating functions, the function is f(x) = A \cos \frac{x\pi}{4} and its now obvious that f(x) = f(x+8)
21i dont think such type of q can come in the exam......btw it was jus hit and trial with which those substis came.....its difficult for someone to strike des substis in the exam.....unlees u hav done it befor.....
1nice one eragon :) . how do you think all these substitutions??
will someone have enough time to do this question in an exam?
21Q2
firstly substitute x+1 in place of x in original eq............i
then subst x-1 in the original eq........................ii
add both thes eq i.e eq i and ii.....to get eq iii
now substitute x+2 in place of x in eq iii to get eq iv
now substitute x-2 in place of x in eq iv to get eq v
now subtarce both eq iv and v to get eq vi
now sustitue x-4 in place of x in eq vi
to get
f(x)=f(x-8)
so 8 is the period
its all abt substituting nothing else [3]