1. Xtra Edge February 2012 - Career Point

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sin A a sin B = b = sin C c 1 = 2R Where R is the radius of circumcircle of triangle ABC. 2. Cosine formulae : 2 2 2 2 2 2 b + c − a a + c − b cos A = , cos B = , 2bc 2ac 2 2 2 a + b − c cos C = 2ab It should be remembered that, in a triangle ABC If ∠A = 60º, then b 2 + c 2 – a 2 = bc If ∠B = 60º, then a 2 + c 2 – b 2 = ac If ∠C = 60º, then a 2 + b 2 – c 2 = ab 3. Projection formulae : a = b cos C + c cos B, b = c cos A + a cos C c = a cos B + b cos A Trigonometrical Ratios of the Half Angles of a Triangle: a + b + c If s = in triangle ABC, where a, b and c are 2 the lengths of sides of ∆ABC, then A s ( s − a) B s ( s − b) (a) cos = , cos = , 2 bc 2 ac C cos = 2 A (b) sin = 2 C sin = 2 A (c) tan = 2 s ( s − c) ab ( s − b)( s − c) bc ( s − a)( s − b) ab ( s − b)( s − c) , s( s − a) B , sin = 2 ( s − a)( s − c) , ac B ( s − a)( s − c) C ( s − a)( s − b) tan = , tan 2 s( s − b) 2 s( s − c) Napier's Analogy : B − C b − c A C − A c − a B tan = cot , tan = cot 2 b + c 2 2 c + a 2 A − B a − b C tan = cot 2 a + b 2 Area of Triangle : 1 1 1 ∆ = bc sin A= ca sin B = ab sin C 2 2 2 ∆ = a sin B sin C b sinC sin A c sin Asin B = = 2 sin( B + C) 2 sin( C + A) 2 sin( A + B) 1 2 1 2 1 2 2 2 ∆ sin A = s( s − a)( s − b)( s − c) = bc bc 2 ∆ 2 ∆ Similarly sin B = & sin C = ca ab Some Important Results : A B 1. tan tan = 2 2 s − c A B ∴ cot cot = s 2 2 A B c C c 2. tan + tan = cot = (s – c) 2 2 s 2 ∆ A B a − b 3. tan – tan = (s – c) 2 2 ∆ A B 4. cot + cot = 2 2 A B tan + tan 2 2 A B tan tan 2 2 XtraEdge for IIT-JEE 46 FEBRUARY 2012 = s s − c c s − c cot C 2 5. Also note the following identities : Σ(p – q) = (p – q) + (q – r) + (r – p) = 0 Σp(q – r) = p(q – r) + q(r – p) + r(p – q) = 0 Σ(p + a)(q – r) = Σp(q – r) + aΣ(q – r) = 0 Solution of Triangles : 1. Introduction : In a triangle, there are six elements viz. three sides and three angles. In plane geometry we have done that if three of the elements are given, at least one of which must be a side, then the other three elements can be uniquely determined. The procedure of determining unknown elements from the known elements is called solving a triangle. 2. Solution of a right angled triangle : Case I. When two sides are given : Let the triangle be right angled at C. Then we can determine the remaining elements as given in the following table. Given Required (i) a, b (ii) a, c a tanA = , B = 90º – A, c = b a sin A a sinA = , b = c cos A, B = 90º – A c Case II. When a side and an acute angle are given – In this case, we can determine Given Required (i) a, A B = 90º – A, b = a cot A, c = a sin A (ii) c, A B = 90º – A, a = c sin A, b = c cos A
a PHYSICS Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. 1. In the circuit shown, the cell is ideal, with e.m.f. = 15 V. Each resistance is of 3Ω. The potential difference across the capacitors in steady state – R=3Ω C=3µF R R + – R R (A) 0 (B) 9 V 15V (C) 12 V (D) 15 V 2. In a young double slit apparatus the screen is rotated by 60º about an axis parallel to the slits. The slits separation is 3mm, slits to screen distance (i.e AB) is 4 m & wavelength of light is 450 nm. The separation between the 3 rd dark fringe on the either side of B. 60º A B screen Based on New Pattern IIT-JEE 2012 XtraEdge Test Series # 10 Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 10 to 13 are Reason and Assertion type question with one correct answer. +3 marks will be awarded for correct answer and –1 mark for wrong answer. • Question 14 to 19 are passage based questions. +4 marks will be awarded for correct answer and –1 mark for wrong answer. • Question 20 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly matched answer (i.e. +1 marks for each correct row) and No Negative marks for wrong answer. (A) 6 mm (B) 8 mm (C) 4 3 mm (D) 2 3 mm 3. A black body emits radiation at the rate P when its temperature is T. At this temperature the wavelength at which the radiation has maximum intensity is λ0. If at another temperature T' the power radiated is P' & λ 0 wavelength at maximum intensity is 2 then – (A) P'T' = 32 PT (B) P'T' = 16 PT (C) P'T' = 8 PT (D) P'T' = 4 PT 4. If two identical string are stretched such that there is fractional increase in their length, the fractional increase in length of first string is f and second string is 2f. Then the ratio of their fundamental frequency is. (Assume both obey the Hook Law i.e. tension ∝ elongation in string) - (A) 1 2 1+ 2f 1+ f (B) 2 1+ 2f 1+ f XtraEdge for IIT-JEE 47 FEBRUARY 2012 (C) 1 2 1+ f 1+ 2f (D) 1+ f 1+ 2f 5. A infinite line charge of charge density λ lies along the x axis and let the surface of zero potential passes through (0, 5, 12) m. The potential at point (2, 3, –4) is – z V = 0 (0,5,12) y
Page 3 and 4: Volume - 7 Issue - 8 February, 2012
Page 5 and 6: 25% increase in number of girls fro
Page 7 and 8: Dr Amitabha Ghosh was the only Asia
Page 9 and 10: Sol. (i) The capacitor A with diele
Page 11 and 12: Sol. (i) Let m be the mass of elect
Page 13 and 14: Step 5. 1 gram equivalent acid neut
Page 15 and 16: = 2π ∫ π 2 / 3 2 t sec dt 2 t 1
Page 17 and 18: (A) (C) A 2B0πqR 3 B 0 PR 3 2 ×
Page 19 and 20: PHYSICS 1. AB and CD are two ideal
Page 21 and 22: 4. In a Young's experiment, the upp
Page 23 and 24: Matter Waves : Planck's quantum the
Page 25 and 26: The intercept of V0 versus ν graph
Page 27 and 28: PHYSICS FUNDAMENTAL FOR IIT-JEE The
Page 29 and 30: etween total mass and number of mol
Page 31 and 32: KEY CONCEPT Organic Chemistry Funda
Page 33 and 34: KEY CONCEPT Inorganic Chemistry Fun
Page 35 and 36: 1. It is possible to supercool wate
Page 37 and 38: z 1 K2 = = y( 0. 48M − z) 1. 26×
Page 39 and 40: MATHEMATICAL CHALLENGES 1. as φ (a
Page 41 and 42: 9. Point P (x, 1/2) under the given
Page 43 and 44: 10. 11 ∴ n(E) = 10 + 9 + 8 + ...
Page 45 and 46: dx (ii) ∫ 2 ax + bx + c This can
Page 47: MATHS Functions with their Periods
Page 51 and 52: R P ⊗ B O ω0 14. Magnitude of cu
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Page 67 and 68: (D) P A T B (S) Pressure is increas
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Page 73 and 74: (a) temperature and (b) Pressure ?
Page 75 and 76: should be bought to meet the requir
Page 77 and 78: This work done is stored inside the
Page 79 and 80: Use : It can be used to accelerate
Page 81 and 82: 10. (i) (ii) Benzene H CH3 - C = O
Page 83 and 84: NH3 NH3 NH3 Co +3 NH3 NH3 NH3 No un
Page 85 and 86: 7. f(A) = A 2 - 5A + 7 I2 ⎡ 3 1
Page 87 and 88: y 2 = - (2 - x) 2 + 9 This is the e
Page 89 and 90: (0, 4) ( 2 3, 2) (4, 0) ⎡ 2 4 2
Page 91 and 92: 15. X = ? Y = 20 Ω l1 = 40 cm l -
Page 93 and 94: (i) Emission : - If radiations emit
Page 95 and 96: OH (ii) + Br2 Phenol H2O 13. I st M
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OR Let A → total of 8 in first th
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= x 2 ⎡ ⎢ ⎢x ⎢ ⎣ 0 1 2 1
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28. x x y ∴ x → side of square
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XtraEdge for IIT-JEE 105 FEBRUARY 2
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