May 2011 - Career Point

Students' ForumExpert’s Solution for Question asked by IIT-JEE AspirantsMATHS1. Let S be the coefficients of x 49 in given expressionf (x) and if P be product of roots of the equationf (x) = 0, then find the value of PS , given that :f (x) = (x – 1) 2 ⎛⎜⎝x2⎞ ⎛ 1 ⎞ ⎛− 2⎟ ⎜ x − ⎟ ⎜⎠ ⎝ 2 ⎠ ⎝x3⎞ ⎛ 1 ⎞− 3⎟ ⎜ x − ⎟ ,⎠ ⎝ 3 ⎠⎛ x ⎞ ⎛......... ⎜ − 25⎟ ⎜ x⎝ 25 ⎠ ⎝Sol. Here we can write f(x) as :⎧ ⎛ x ⎞⎛x ⎞ ⎛ x ⎞⎫f (x) = ⎨(x −1)⎜ − 2⎟⎜− 3⎟...⎜ − 25⎟⎬⎩ ⎝ 2 ⎠⎝3 ⎠ ⎝ 25 ⎠⎭1− 25⎧ ⎛ 1 ⎞⎛1 ⎞ ⎛ 1 ⎞⎫× ⎨( x −1)⎜ x − ⎟⎜x − ⎟...⎜ x − ⎟⎬⎩ ⎝ 2 ⎠⎝3 ⎠ ⎝ 25 ⎠⎭Now roots of f (x) = 0 are;1 2 , 2 2 , 3 2 , ..... , 25 2 1 1 1and 1, , , ....., 2 3 25Now f (x) is the polynomial of degree 50,So coefficient of x 49 will be :S = – (sum of roots)= – (1 2 + 2 2 + ... + 25 2 ⎛ 1 1 1 ⎞) – ⎜1+ + + .... + ⎟⎝ 2 3 25 ⎠⎧25×26×51 ⎫= – ⎨ + K⎬where, K =⎩ 6 ⎭ ∑n=⇒ S = –(K + 5525).Product of roots :1 2 . 2 2 . 3 2 .... 25 2 1 1 1. 1 . . .... = 1 . 2 . 3 ...252 3 25∴ P = 25 !S −( K + 5525)Hence = P 25!2511n25, where K = ∑n=2. From an external point P(α, 2) a variable line is2 2x ydrawn to meet the ellipse + = 1 at the points9 4A and D. Same line meets the x-axis and y-axis at thepoints B and C respectively. Find the range of valuesof 'α' such that PA. PD = PB.PC.11n⎞⎟⎠Sol. We have been given,yOCDP(α,2)APA.PD = PB.PCEquation of any line through point 'P' is :x − α y − 2= = rcosθsin θor x = α + r cos θ, y = 2 + r sin θPutting this point in the equation of given ellipse, we get4(r cos θ + α) 2 + 9(2 + r sin θ) 2 = 36⇒ r 2 (4 cos 2 θ + 9 sin 2 θ) + 4r (9 sin θ + 2 α cos θ)+ 4 α 2 = 0Since PA and PD are the roots of this quadratic in r,we get24αPA.PD =...(i)2 2(4cos θ + 9sin θ)Similarly, putting x = r cos θ + α, y = r sin θ + 2 inthe equation of coordinate axis i.e. xy = 0(r cos θ + α). (r sin θ + 2) = 0⇒ r 2 sin θ cos θ + r (2 cos θ + α sin θ) + 2α = 0Since PB and PC and the roots of this quadratic in 'r',2α4αwe get, PB.PC = =...(ii)sin θcosθsin 2αThus, we get24α4α={from (i) and (ii)}2 2sin 2θ4cos θ + 9sin θ4α8α2⇒ =sin 2θ4(1 + cos 2θ)+ 9(1 − cos 2θ)B1 2α⇒ =sin 2θ 13 − 5cos 2θ⇒ 13 = 5 cos 2θ + 2α sin 2θwhere; 5 cos 2θ + 2α sin 2θ ≤∴ 4α 2 + 25 ≥ 13 ⇒ α 2 ≥⇒ α ∈ (–∞, – 6] ∪ [6, ∞)25 + 4α169 − 254x2= 36XtraEdge for IIT-JEE 40 MAY 2011