the answer of 1st one is probably 4....
1. Let f(x)=\frac{x+1}{2x-1} and g(x)=\left|x \right| + 1. Then the number of elements in the set \left\{{x:f(x)\geq g(x)} \right\}\sim \left[\left\left(1/2,3/5 \right)\cup\left(3/5,7/10 \right)\cup\left(7/10,4/5 \right)\cup\left(4/5,1 \right) \right] is ?
Integer type!
2. If 'a' is the natural number for which \sum_{k=1}^{n}{f(a+k)}=2n(32-n) where the function f satisfies the relation f(x+y)=f(x)+f(y) for all natural numbers x,y and further f(1)=4, then \sqrt{a}= ? (Again, integer type)
For the second question,
I proved f(k)=kf(1)=4k for any natural number k.
Hence,
\sum_{k=1}^{n}{f(a+k)}=\sum_{k=1}^{n}{f(a)+f(k)}= nf(a) + \sum_{k=1}^{n}{f(k)}
\Rightarrow \sum_{k=1}^{n}{f(a+k)} = nf(a) + 4( 1 + 2 + 3 + ...+ n)= 4an + 2n(n+1)
\Rightarrow \sum_{k=1}^{n}{f(a+k)} = 2n(2a+1+n)
\Rightarrow a+n=\frac{31}{2}
Now isnt there a contradiction. Sum of two natural numbers giving a fraction? Or am I doing something incorrect?
Nishant sir are you listening?
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6 Answers
621st question is simpler than it initially looks..
since all options are along the +ve x axis, we put |x|=x and then solve the inequality
\\\frac{x+1}{2x-1}\ge x+1 \\\Rightarrow\frac{1}{2x-1}\ge1\text{ as x+1 is positive} \\\Rightarrow\frac{2-2x}{2x-1}\ge0
using wavy curve, the solution is (1/2,1]
but i understand nothing from the second notation of the question!
Unfortunately havent seen it ever!
62\\\sum_{k=1}^{n}{f(a+k)}=2n(32-n) \\\Rightarrow\sum_{k=1}^{n-1}{f(a+k)}=2(n-1)(32-n-1) \\\Rightarrow{f(a+n)}=2n(32-n)-2(n-1)(31-n)=64n-2n^2+2n^2-64n+62
so f(a+n)=62 = constant
so the answer should be independent of n
Asish: Hint: The expression is linear... so what u have done is correct... basically the question is not well framed (or there is a typing mistake) can u tell what?
30@krishna - How are you arriving at the conclusion?
For the first question my doubt was abt the notation given. I didnt get what that set represents.
And sir these questions were there in this term's maths paper!
actually the notation doesnt hav much to say
it means the elements of set which satisfy
f(x)>=g(x) but do not belong to the set
(12 ,35) U (35,710) U (710,45) U(45,1)
since set satisfyin the inequality is (1/2,1] hence 4 points are left out