May 2011 - Career Point

(C)25 / 6πI = x dxn ∫sec πl 327 / 6π5 / 6I = ln| sec πx+ tan πx|πln37 / 6πI = ⋅ ln3= πln3(D) ∴ |z| = 1z = cos θ + i sin θ. ∀ θ∈ (– π. π] and θ ≠ 0.1Arg(1 − z)⎛ θ ⎞⎜ icot⎟⎛ 1 ⎞⎜1= Arg ⎜⎟ = Arg + 2 ⎟⎝1− cosθ − isinθ ⎠ ⎜ 2 2 ⎟⎝ ⎠=π − θ2so maximum value is π.60. Match the statements given in Column-I with theintervals/union of intervals given in Column-II.Column-I(A) The set⎧ ⎛ 2iz ⎞⎫⎪Re⎜⎟;z is a complex number,2⎪⎨ ⎝1− z ⎠⎬ is⎪⎪⎩| z | = 1, z ≠ ± 1 ⎭(B) The domain of the function⎛ x−2⎞−1⎜8(3)f ( x)= sin⎟ is2( x−1)⎝1−3 ⎠1(C) If f ( θ)= − tan θ−1tan θ1− tan θ1tan θ , then the⎧π⎫set ⎨ f ( θ) : 0 ≤ θ < ⎬ is⎩2 ⎭3/ 2(D) If f ( x)= x (3x−10),x ≥ 0 , then f(x) isincreasing inColumn-II(p) (– ∞, –1) ∪ (1, ∞)(q) (– ∞, 0) ∪ (0, ∞)(r) [2, ∞)(s) ( – ∞, –1] ∪ [1, ∞)(t) (– ∞, 0] ∪ [2, ∞)Ans. [A → p, r, s ; B → r, t ; C → r ; D → r]1Sol.(A) Let z = cos θ + i sin θ2iz2i(cos+isinθ)so == −cosecθ21−z 1−cos2θ − isin 2θ∀ θ ≠ (2n + 1) 2π⎛ 2iz ⎞so Re⎜⎟ = − cos ec θ∈(−∞,−1]∪ [1, ∞)2⎝1− z ⎠x−28×3 8×3(B) =2x−22x1−3 9 − 3Let 3 x = t⎛So f(x) = sin –1 ⎜8 3⎝ 9 − 3x⎞⎟ = sin⎠×x−12x8t−1≤ 129 − t≤ on solvingx ∈ (– ∞, 0] ∪ [2, ∞) ∪ {1}(C) f(θ) = 2 sec 2 θso f(θ) ∈[2, ∞)(D) f(x) = 3x 5/2 – 10x 3/215 xf'(x) = ( x − 2)2So f(x) is increasing for f '(x) ≥ 0x ∈ [2, ∞)Random Facts⎛ 8t⎜⎝ 9 − t• As a gas' temperature is raised to over10,000°, its molecules collide so violentlythat they are broken apart into individualatoms.• When the tsunami reaches the coast andmoves inland, the water level can risemany meters. In extreme cases, water levelhas risen to more than 15 m (50 ft) fortsunamis of distant origin and over 30 m(100 ft) for tsunami waves generated nearthe earthquake’s epicenter.• Some minerals, notably quartz, arepiezoelectric--that is, they produceelectricity when subjected to pressure orstress. This same phenomenon is probablyalso responsible for "earthquake lights,"the luminescence sometimes reported (and,on occasion, photographed) in the skyduring earthquakes.2⎟⎠⎞XtraEdge for IIT-JEE 96 MAY 2011