JEE Advanced 2016 (6)

Algebra Level 5

Say a real number r r is repetitive if there exist two distinct complex numbers z 1 , z 2 z_{1},z_{2} with z 1 = z 2 = 1 |z_{1}| = |z_{2}| = 1 and z 1 , z 2 z_{1},z_{2} is not equal to i , i {-i,i} such that z 1 ( z 1 3 + z 1 2 + r z 1 + 1 ) = z 2 ( z 2 3 + z 2 2 + r z 2 + 1 ) z_{1}(z_{1}^{3} + z_{1}^{2} + rz_{1} + 1) = z_{2}(z_{2}^{3} + z_{2}^{2} + rz_{2} +1) There exist real numbers a , b a,b such that a real number r r is repetitive if and only if a < r b a < r ≤ b . If the value of a + b |a| + |b| can be expressed in the form p q \frac{p}{q} for relatively prime positive integers p , q p,q . Find 100 p + q 100p +q .

Try my set JEE ADVANCED 2016

Source: OMO Spring 2016

2504 2604 None of these 2000 0 2576

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

n. Let the argument of z, with |z| = 1, be θ. Note that f(z) = z

4 + z

3 + rz2 + z + 1 is in the

direction of argument 2θ with a signed magnitude of r + 2 cos θ + 2 cos(2θ) = 4 cos2

θ + 2 cos θ −2 +r =

4(cos θ + 1 4 ) 2 − 9 4 + r. For r > 9 4

, then note that this signed magnitude is always positive. Then, f(z1) = f(z2) can only

happen for z1 6= z2 if their arguments θ1, θ2 satisfy 2θ1 = 2θ2 =⇒ θ1 = θ2 + π. But then their signed

magnitudes can only be equal if cos θ1 = cos θ2, but that implies {z1, z2} = {i, −i}. Hence f(z1) 6= f(z2)

for all distinct z1, z2. For r < −4, the signed magnitude is always negative since | cos θ + 1 4 | ≤ 5 4 over all θ.

Once again, this implies that f(z1) 6= f(z2). However, if −4 ≤ r ≤ 9 4 ,

then there exists a unique value for cos θ with − 1 4 ≤ cos θ ≤ 1 such that the signed magnitude is equal to 0

. For all cos θ 6= 1, there are two values of θ giving the desired value

of cos θ, showing there exist distinct z1, z2 giving f(z1) = f(z2) = 0.

However, cos θ = 1 only when θ = 0, meaning for r = −4 we see that the signed magnitude is always negative except for θ = 0, when the signed magnitude is exactly 0. Hence r = −4 is also repetitive.

Hence a = −4 and b = 9 4 so 100p + q = 2504.

Pls explain what is f (z).the function?

rajdeep brahma - 3 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...