331st condition D≥0
and
for m+5>0 ..(f(0)>0)
-b/2a>0
________
and for m+5<0
all m will do..
Parabola is upward opening... so if f(0)<0 and also as x→+∞ f(x)→∞ so one positive root is compulsory...
1)Find the smallest natural number which leaves remainder 4,6,10,1 when divided by 5,7,11 and 13 respectively.
2) xy+yz+zx=1
then prove that (1+x^2)(1+y^2)(1+z^2)=(x+y)^2(y+z)^2(z+x)^2
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18 Answers
33
6213m+1 comes from the fact that if a number divided by 13 leaves a remainder of 1, then it has to be of the form
13k+1 where k is an integer.
11sir ne 2nd post edit kiya hai
i didnt saw that form earlier
please explain his 4m where 13m+1 came
wont that be 14??
341i meant that if the remainder is 12, the problem is uncomplicated as x+1 will be divisible by 5,7,11 and 13. Otherwise we have to proceed as Nishant sir has said in Post #2
11sir question 1 wohi tha
sir but y r u asking this question[7][7][7]
341are you sure you have typed the 1st one correctly? Specifically, the remainder when the number is divided by 13 - pls confirm its 1
625k-1
7l-1
11m-1
13m+1
The number has these forms together..
does it help?
Think thoda sa.. you might end up getting the oslution without any help :)
Adding More hint
the number is of the form
385t-1 and 13m+1
so for both to happen,
385t-13m=2
find the smallest t and m for which it is possible.!
hence you will be done
1x^2+2(m-1)x+m+5=0 has at least one +ve root.Determine the range for m.
11For the second one..
I have a feelingthat putting
x=tan a
y=tan b
z=tan c
will make a hell lot of difference
try that :)
sir daal ke dekh liya
mere se aur phas hi raha hai
not getting it[17]
62For the second one..
I have a feelingthat putting
x=tan a
y=tan b
z=tan c
will make a hell lot of difference
try that :)
1manipal in d second wala....IS d RHS ((x+y)(y+z)(z+x))2???
then y did u use so many powers daa??