Solution - Career Point

MATHEMATICAL CHALLENGESSOLUTION FOR MARCH ISSUE (SET # 11)1. Let z = x + iy so given1 ≤ x or 1 ≤ r cos θ , if z = ∑e iθ Now,1−x − iy (1 − x − iy)(1 + x − iy)=2 21+x + iy (1 + x) + y222(1 − x ) − yReal part≤ 0 as x ≥ 1 given and2 2(1 + x) + yimaginary part−(1− x)y − (1 + x)y −2y=≤ 02 22 2(1 + x) + y (1 + x) + yas y ≥ 0 given1−zso ∩ 0 is true.1+z2. As ∠ POQ = 90ºQso CP = (OC) . tan (90 – θ)= (OC) cot θ & CQ = (OC) tan θso CP. CQ = (OC) 2 = r 2when r is the radius of circle.3. f(0) = cf(1) = a + b+ c& f(–1) = a – b + cO90–θθsolving these, a = 21 [f(1) + f(–1) – 2f(0)]Cb = 21 [f(1) – f(–1)] & c = f(0)x (x + 1)so f(x) = f(1) + (1 – x 2 x(x−1)) f(0) + f(–1)222 | f(x) | ≤ |x| |x + 1| + 2 |1 – x 2 | + |x| |x – 1|; as|f(1)|, |f(0)|, |f(–1)| ≤ 1.2 |f(x)| ≤ |x| (x + 1) + 2(1 – x 2 ) + |x| (1 – x) asx ∈ [–1, 1]so 2 |f(x)| ≤ 2(|x| + 1 – x 2 5) ≤ 2 . 4so |f(x)| ≤ 45P4.=Now, as g(x) = x 2 ⎛ 1 ⎞f ⎜ ⎟⎝ x ⎠= 21 (1 + x) f(1) + (x 2 – 1) f(0) + 21 (1 – x) f(–1)so 2 | g(x) | ≤ |x + 1| + 2 |1 – x 2 | + |1 – x|2 | g(x) | ≤ x + 1 + 2(1 – x 2 ) + 1 – xas x ∈ [–1, 1]2 |g(x)| ≤ –2x 2 + 4 ≤ 4|g(x)| ≤ 21⎡⎤⎢2y1− y´⎥a ⎢ 2 2 ⎥⎣ 2 a − y ⎦=a +a12− y2.2−2y.y´a2− y2– y1 y´1⎡= y´a⎥ ⎥ ⎤⎢yy1−−⎢ 2 22 2 2 2⎣aa − y (a + a − y ) a − y y⎦2y´⎡y (a +⎢⎣y(a +a222ay(a +a2− y ) − y a − a(a +2a − y )22− ya − y22)a2a2− y− y22)a2− y= y´[y 2 a + y 2 a 2 − y2 – y 2 a – a 2 a 2 − y2 – a(a 2 – y 2 )]y (a +2y´ = –y´M =2a − y )a22ay− y−β− β22a − y = –y´(a 2 – y 2 ) (a +22222⎤⎥⎦a − y )y =−βx...(1)2 2a − βy = β ...(2)locus of intersection of these two lines isy =−yx2 2a − ya 2 – y 2 = x 2x 2 + y 2 = a 2XtraEdge for IIT-JEE 39 APRIL 2010