Discriminant≥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
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
thnks for contributing everyone for the beautifull solutions and logic [1][1]
I 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
Discriminant≥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
@prophet sir can u tell if my method is rong and wat shuld i do in that to correct it ??[7]
@prophet sir
and also at p=4 or p-3=1 number of solution is 1 ,but number of roots is 2
@prophet sir
can we consider zero in this case because at there number of solutions is 2 but number of roots is 4[7]
But 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.
BUT 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]
ok 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]
@manmay - that '?' was a test post. I was unable to add post just b4 that
sir answer is 21.....where am i rong ??
I also tried removing the modulus but all in vain........
waht to do [7][7][2][2]
P≥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
AFTER 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
might be pritish but i got biquadratic in x not a quadratic[7][7]
(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.