May 2011 - Career Point

⎡323 21× ⎤=⎡32 1⎤1 C2C21 C1C12⎢ × 1+× ⎥ + ⎢ × 1+× + ×555 ⎥2 ⎣55 2⎦2 ⎣ C2C23 C23 ⎦1 ⎡31⎤1 ⎡ 3 1 2⎤2 11 23= ⎢ + ⎥ + ⎢ + + ⎥ = + =2 ⎣55⎦2 ⎣1030 5 ⎦ 5 30 3062. Given that the drawn ball from U 2 is white, theprobability that head appeared on the coin is1711(A) (B) 23231512(C) (D) 2323Ans. [D]Sol. Required probability =⎡ ⎛ W1⎞ ⎛ R1⎞ ⎤P(H) ⎢P⎜⎟P(W 2)+ P⎜⎟P(W 2)⎥⎣ ⎝ H ⎠ ⎝ H ⎠ ⎦⎡ ⎛ W1⎞ ⎛ R1⎞ ⎤ ⎡ ⎛ both W ⎞P(H) ⎢P⎜⎟P(W 2)+ P⎜⎟P(W 2)⎥ + P(T) ⎢P⎜⎟P(W 2)⎣ ⎝ H ⎠ ⎝ H ⎠ ⎦ ⎣ ⎝ T ⎠⎛ both R ⎞ ⎛ R+ P⎜⎟P(W 2)+ P⎜⎝ T ⎠ ⎝=1 ⎡32 1⎤12⎢ × + ×5 5 2⎥⎣ ⎦=233012131& W1⎞ ⎤⎟P(W 2)⎥T ⎠ ⎦SECTION – IVNumerical Response TypeThis section contains 7 questions. The answer to each ofthe questions is a single-digit integer, ranging from 0 to9. The bubble corresponding to the correct answer is tobe darkened in the ORS.63. Let a 1 , a 2 , a 3 , .., a 100 be an arithmetic progressionwith a 1 = 3 and S p = ∑=pi 1a i, 1 ≤ p ≤ 100. For anyinteger n with 1 ≤ n ≤ 20, let m = 5n. Ifnot depend on n, then a 2 isAns. [9]Sol. a 1 = 35nS[2am S1 + (5n−1)d]n= 5= 2S n S nn [2a1+ ( n −1)d]25[(6 − d)+ 5nd]=(6 − d)+ ndS5nQ is independent of n so d = 6SnSo a 2 = a 1 + d = 3 + 6 = 9SSmndoes64. Consider the parabola y 2 = 8x. Let ∆ 1 be the areaof the triangle formed by the end points of its⎛ 1 ⎞latus rectum and the point P ⎜ , 2⎟ on the⎝ 2 ⎠parabola, and ∆ 2 be the area of the triangle formedby drawing tangents at P and at the end points of∆1the latus rectum. Then is∆2Ans. [2]Sol. It is a property that area of triangle formed byjoining three points lying on parabola is twice thearea of triangle formed by tangents at these pointsAlternate : y 2 = 8x⎛ 1 ⎞P ⎜ , 2⎟ ⎝ 2 ⎠(2, 4) A•P•3/2•∆ 1 = 21 |Base × Height|= 21 × 23 × 8 = 6B(2, –4)Also⎛ 1 ⎞Equation of tangent at P ⎜ , 2⎟ ⎝ 2 ⎠•• P••⎛ 1 ⎞y (2) = 4. ⎜ x + ⎟⎝ 2 ⎠y = 2x + 1 ...(1)Tangent at A : y = x + 2Tangent at B : – y = + x + 2 ⇒ y = – x – 2Point of intersectionL(–2, 0), M (1, 3), N (–1, –1)− 2 0 11∆ 2 = 1 3 12−1−11= | 21 [–2(4) + (–1 + 3)]|XtraEdge for IIT-JEE 78 MAY 2011