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`tà{xÅtà|vtÄ V{tÄÄxÇzxá12SetThis section is designed to give IIT JEE aspirants a thorough grinding & exposure to varietyof possible twists and turns of problems in mathematics that would be very helpful in facingIIT JEE. Each and every problem is well thought of in order to strengthen the concepts andwe hope that this section would prove a rich resource for practicing challenging problems andenhancing the preparation level of IIT JEE aspirants.By : Shailendra MaheshwariSolutions published in this issueJoint Director Academics, Career Point, Kota1. Prove that, if n is a positive integer,a∫ − x0ne x dx =⎪⎧⎛⎞⎪⎫n ! ⎨ − − 2 na⎜a a1 e 1+a + + ... + ⎟⎬⎪⎩ ⎝ 2! n!⎠⎪⎭Also, deduce the value of∫ ∞e2. Let A ≡ (6, 5), B ≡ (2, –3) and C ≡ (–2, 1) be thevertices of a triangle. Find the point P in the interiorof the triangle such that ∆PBC is an equilateraltriangle.3. If S n = n C 1 + 2. n C 2 + 3. n C 3 + ....... + n n C n then findn∑n=1S . Also prove thatn0−xn C 1 . ( n C 2 ) 2 . ( n C 3 ) 3 .... ( n C n ) n ≤xndx⎛n2 ⎞⎜ ⎟n 1⎝ + ⎠4. Let z 1 , z 2 , z 3 be three distinct complex numberssatisfying|z 1 – 1| = |z 2 – 1| = |z 3 – 1|; Let A, B and C be thepoints represented in the argand plane correspondingto z 1 , z 2 and z 3 respectively. Prove that z 1 + z 2 + z 3 =3 if and only if ∆ABC is an equilateral triangle.5. Let A ≡ (r, 0) be a point on the circle x 2 + y 2 = r 2 andD be a given point inside the circle. If BC be anyarbitrary chord of the circle thorugh point D. Provethat the locus of the centroid of triangle ABC is acircle whose radius is less than r/3.n+1C2.6. A number is chosen at random from the set{1, 2, 3, …......, 2006}. What is the probability thatit has no prime factor in common with10 ! ?7. Two vertices of a triangle are a – î + 3ĵand 2 î + 5ĵand its orthocenter is at ( î + 2ĵ). Find the positionvector of third vertex.8. Show that an equilateral triangle is a triangle ofmaximum area for a given perimeter and a triangle ofminimum perimeter for a given area.9. The bottom of a tank with a capacity of 300 litres iscovered with a mixture of salt and some insolublesubstance. Assuming that the rate at which the saltdissolves is proportional to the difference betweenthe concentration at the given time and theconcentration of a saturated solution (1 kg of salt per3 litres of water) and that the given quantity of purewater disolves 1/3 kg of salt in 1 minute. Find thequantity of salt in solution at the expiration of onehour.10. An isosceles triangle with its base parallel to themajor axis of the ellipsex 2 +9y 2 = 1 is3circumscribed with all the three sides touchingthe ellipse. The least possible area of thetriangle is.XtraEdge for IIT-JEE 38 APRIL 2010
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
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CAREER POINTIn associationwithLaunc
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