341Q1:\left[\frac{n}{2^{r-1}} + \frac{1}{2} \right] = \left[\frac{n}{2^{r-2}}\right ]- \left[\frac{n}{2^{r-1}} \right]
Q1. [] represents the GINT. Then the value of the infinite series
[20082] + [20094] + [20118] + [201516] + [202332] + ......
is?
I need a short method for this..
All i could do was calculate tr = [2007+2r-12r]
Q2. If {x} represents the least integer, not less than x, then total number of solutions of the equation (x-1)2 + {x} = 4 is equal to?
Q3. The domain of the function f(x) = √cos-1(cosx) - [x] where [x] = GINT is
(a) (-∞,2pi+3] (b) (-∞,pi-3] (c) (-∞,pi+3] (d) (-∞,2pi-3]
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7 Answers
1the answer to question no 2 is only 1 solution at x=-√5+1.
take {x}=4-(x-1)2.draw the graph and get only one point of intersection.
the answer to question no 3 is option d)
just take cos-1cosx>[x] draw the graphs and see the two curves intersect at x=2π-3.hence (-∞,2π-3]
341
106Thanks sir, and TITLI
Q4. *deleted*
Q5. lim(n→∞) xnn!
Q6. lim(x→0) ([f(x)] + x2)1/{f(x)} where f(x) = tanx/x and [] represents GINT and {} represents fractional part
62Q5)
even though this requires real analysis for proof.. but a very simple thing to say is that
if lim n-> infinity tntn-1 < 1
Then limit goes to zero..
you can otherwise take n>x and prove it by simpler arguements.
66Q6) In the close vicinity of x=0, tan xx is just greater than 1. So [f(x)] = 1 and
{f(x)} = f(x)-1= tan xx -1
Accordingly the given limit is
\lim_{x\to 0} (1+x^2)^\frac{x}{\tan x-x}
=\mathrm{exp}\left(\lim_{x\to 0}\ \frac{x^3}{\tan x-x}\right)=e^3