See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

0.25 119   CH3MgBr is limiting reagent. C H3 C O Cl 3 2CH MgBr   C H3 C OH CH3 CH3 Moles of alcohol = 0.25 2 Weight of alcohol = 0.25 2 × 74 = 9.25 gm AITS-PT-II (Paper-2)-PCM (Sol)-JEE(Advanced)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 13 Mathematics PART – III SECTION – A 47. Equation of 2 2 2 2 a b a b S x y 2 2 4                   x2 + y2 – ax – by = 0 And circle with centre (r, r) and radius r Touches AB and AC its equation is (x – r)2 + (y – r)2 = r2 x2 + y2 – 2xr – 2yr + r2 = 0 Both circle touches internally or externally Using 2 2 2 2 a b a b r r r 2 2 4 4                         b (r, r) A B(a, 0) a y x a b , 2 2       ABC = 1 2 r r 4 48. Equation of tangent at (i, yj) y = (2i + b)x – i2 + c Also at (i + 1, yi + 1), y = (2(i + 1) + b)x – (i + 1)2 + c Point of intersection is 2i 1 x 2    2x 1 i 2   Put this i in any tangent, we get y = x2 + bx + c – 1 4 49. Equation of tangent is 2 2 y 4x 16a b    this normal passes through (–2, 0)  2 2 0 8 16a b   ; 264 = 16a2 + b2 Giving AM  GM 2 2 2 2 16a b 16a b 2   2 2 64 16a b 2  4ab  32 ab  8 50. Equation of tangent at A is 4x – 3y – 8 = 0 Let y = m(x – 2) a line thro (2, 0) and m = –7 or 1 7 We have x 2 y 5 2 1 7 50 50      Centre of circle are (1, 7), (3, –7), (9, 1) and (–5, –1) 51. Normal at P(x1y1)   1 1 1 y y x k m     Where y = 0, x = x1 + my1 We have |x1 + my1| = 2|x1| AITS-PT-II (Paper-2)-PCM (Sol)-JEE(Advanced)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 14  x1 + my1 = 2x1 Case-I: x1 + my1 = 2x1  1 1 x m y   dy x dx y    1 1 at x y dy m dx                 x dx – y dy = 0; 2 2 x y C a 2 2    hyperbola Case-II: x1 + my1 = –2x1; 1 1 3x m y   dy y 3x 0 dx    2 2 y 3x C 2 2   (Ellipse) 52. Given equation can be written as 4 2 4x 4 0 x x x 4      x 2  53. Equation can be written as cos 2x(cos 4x + cos 2x) = 0  2 cos x cos 2x cos 3x = 0 54. Equation of normal to hyperbola xy = c2 at c ct, t       which passes through (h, k) is ct4 – ht3 + kt – c = 0 Roots of this equation is t1, t2, t3 and t4 Let equation of circle be x2 + y2 – 2gx – 2fy + p = 0 If c ct, t       lies on this c2t4 – 2gct3 + pt2 – 2fct + c2 = 0 Its roots are t1 t2 t3 and ' 4t We get ' 4 4 t t  also ' 4 4 c c k 2f t t    So locus of centre is    2 4c h 2g k 2f    Centre lies on hyperbola 2 h k x y c 2 2             h k c , 2 2       and c 2 