1thnks for contributing everyone for the beautifull solutions and logic [1][1]
Find the sum of all integral values of P for which the equation \left|x+\frac{1}{x} -3\right| = P-3
has exactly two distinct roots.
ans-----------21
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21 Answers
11I had tried like dis.
Let y=p-3
\Rightarrow \frac{dy}{dx}=0=1-\frac{1}{x^{2}}
\Rightarrow x=1 or -1
As x\rightarrow 0^{+},y\rightarrow \infty and x\rightarrow 0^{-},y\rightarrow -\infty
Also roots of x+\frac{1}{x}-3=0
\Rightarrow x^{2}-3x+1=0
\Rightarrow x=\frac{3\pm \sqrt{9-4}}{2}=\frac{3\pm \sqrt{5}}{2}
For two distinct solutions either p-3=0\Rightarrow p=3
or, 1<p-3 <5 \Rightarrow 4<p<8
Hence,p\in \left(3 \right)\bigcup{}\left(4,8 \right)
p=(3,5,6,7)
So,sum=21
341Discriminant≥0 is not a sufficient condition. You've got to ensure that the roots do not lie in [-2,2] because x + 1/x does not lie in that interval.
Worse yet, we need one root of the quadratic in the interval and one out. That's quite torturous
1@prophet sir can u tell if my method is rong and wat shuld i do in that to correct it ??[7]
1@prophet sir
and also at p=4 or p-3=1 number of solution is 1 ,but number of roots is 2
1@prophet sir
can we consider zero in this case because at there number of solutions is 2 but number of roots is 4[7]
341But to plot that graph by hand you need the ranges I arrived at in my post. I guess JEE doesnt permit use of graphing calculators.
1BUT SIR I GOT A DOUBT HERE ....
CAn we thnk of doing it by using graphs as lubu did.
If so can u please help me with that ???[7][7]
1ok i tried like this
squaring both the sides we get
\left(x+\frac{1}{x} -3\right)^{2} = (P-3)^{2} \Rightarrow \left(x+\frac{1}{x} \right)^{2}+9-6\left(x+\frac{1}{x} \right)= P^{2}-6P+9
now \Rightarrow \left(x+\frac{1}{x} \right)^{2}-6\left(x+\frac{1}{x} \right)-(P^{2}-6P)= 0
now this a quadratic in \left(x+\frac{1}{x} \right)
for distict roots D>0
\Rightarrow 36 + 4(P^{2}-6P) >0
\Rightarrow P^{2}-6P+9 >0
(P-3)^{2}>0
thus P>3
But i'm not able to thnk wat to do after that.........I'm i rong somewhere???[7][7]
341@manmay - that '?' was a test post. I was unable to add post just b4 that
1sir answer is 21.....where am i rong ??
I also tried removing the modulus but all in vain........
waht to do [7][7][2][2]
341P≥3 is obvious as LHS≥0.
When x is a root so is 1/x
Now when x>0, x+1/x -3 ≥-1
and when x<0 x+1/x -3 ≤ -5
That means when x>0, LHS takes values ranging from 0 to ∞
and when x<0, LHS takes values>5.
Hence we have two distinct real roots when 0≤P-3<5 as in all other cases we will have two +ve real roots and at least one -ve real roots. Of course when P-3=1 (corresponding to x=1), we have only one real root.
So when P is an integer P-3 has the solution set {0,2,3,4}
Hence the sum of values of P = 21
1AFTER THIS GRAPH TAKE MODE OF THIS AND YOU WILL GET
P - 3 BELONGS TO INTERVAL [1,5] FOR TWO ROOTS
BUT IN QUESTION ROOTS ARE DISTINCT SO
P - 3 ≠1,5
SO P=2+3,3+3,3+4
1might be pritish but i got biquadratic in x not a quadratic[7][7]
39(P-3)2 > 0 implies P → Reals. It's a square term. If your calcs are right, there are inf integral values of P...considering all reals.
If it's not, it means we have to make a quadratic in x I guess.