Solution - Career Point

56.[D]OH Zndust ⎯⎯→H3C–Cl⎯⎯⎯→CH 3AlCl3alk. KMnO 4⎯ ⎯⎯ → COOH64.[C] f(x) = sin x – cos x – kx + bf '(x) = cos x + sin x – kf '(x) = 2 sin(π/4 + x) – kif f(x) is decreases for all x ∴ f '(x) is –vei. e. k > max. of 2 sin(π/4 + x)i. e. k > 257.[D]58.[A] 4 HCl + O 2 ⎯⎯→ 2 H 2 O + 2 Cl 2Chlorine is in the form of cloud.59.[A] Coordination no. = 6Oxidation no. = 3no. of d electron = 6no. of Unpairedd electron = 060.[C] Resonance structure should have same numberof electron pairs.61.[A]MATHEMATICS30º 60ºAOLet the height of the tower is h1000 = h cot 30º – h cot 60º1000⇒ h =cot 30º −cot60ºh = 500 3 m62.[D] since variance is independent of change inorigin. Hence variance of observations 101,102, ...... 200 is same as variance of 151, 152,.....250.∴ V A = V BV A⇒ = 1VB63.[C] centroid of triangle is given by⎛⎞⎜c 2 + 3 a + b − 3, ⎟⎝ 3 3 ⎠if centroid lie on y axis ⇒ abscissae = 0c 2 + 3 = 0 ⇒ no real c existif centroid lie on x axis ⇒ ordinate = 0⇒ a + b – 3 = 0⇒ a + b = 3Ph65.[A] x lim → 0= x lim → 0⎡⎢⎢⎣⎛⎜⎜⎝xe 2− cos x ⎤⎥2x ⎥⎦−11−cos x⎟ ⎟ ⎞+22x x⎠xe 221 + x lim 2sin x / 2 1 3→ 0= 1 +2=x2 266.[A] Second determinant has been obtainedfrom the first by the operationC 1 → C 1 + 2C 2 – 3C 3 . so its value remainsunchanged67.[C] given 3sin x – 4sin 3 x – k = 0⇒ 3sin x – 4sin 3 x = k⇒ sin3x = k.........(i)angle A and B satisfy the equation (i)∴ sin 3A = sin3B = k ⇒ sin3A = sin3BBut A > B ⇒ A BNow, sin3A = sin(π – 3B)3A = π – 3Bπ 2πA + B = ⇒ C = 3 368.[B]69.[C]given statement(p q) ∨ ~ r → (p ∧ r)(F ↔ F) ∨ F → (F ∧ T)T ∨ F → FT → F = FpTTFFqTFTFp ∨ qTTTF~ pFFTTq∧~ pFFTF(p ∨ q) ↔ q∧~ pHence neither tautology nor contradiction70.[D] Given sequence can be written as5 20 10 20 , , , , ..............2 13 9 2320 20 20 20or , , , ..........8 13 18 23FFTTXtraEdge for IIT-JEE 106APRIL 2010