FT4 dbt 1

Q1 If f(x)={x}g(x){x}g(x) is a periodic fn with pd 1/4 where g(x) is diff fn. then prove g(x)=0 at x=k/4 k ε I

Q2 f: R→R,f(x)=x2+bx+1x2+2x+b ,if the function f(x) and 1/f(x) have same bounded set as their range then find value of b

5 Answers

62
Lokesh Verma ·

This function is equal to zero except when x=n or g(x) = 0

If g(x) is never zero, then THe period is 1...

So we want g(x) to be zero at x=I+1/4, x=I+2/4 and x=I+3/4 for this function to be periodic..

But the question says that g(x)= for alll x=I+k/4

I thnk the problem is wrong because we will want x=I+k/4 .. but x=I+0 is not necessary..

4
UTTARA ·

@Nishant Sir : I dint get this one

If g(x) is never zero, then THe period is 1...

62
Lokesh Verma ·

if g(x) is never zero then g(x)/g(x) = 1 (always)

Therefor the given function is {x}/{x}

Which is 1 except when x is integer...

Only when x is integer the function is not defined..

So the graph repeats after an interval of 1 unit.

4
UTTARA ·

Yes sir got it now

Thanks

24
eureka123 ·

so ques is wrong ??

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