See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

C C C NH2 C C C C NH2 C C C C NH2 C C C C NH2 C C N C H C C C N C C H C N C C C C C N C H C AITS-PT-II (Paper-1)-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 11 Mathematics PART – III SECTION – A 37. Let F be the fixed focus and M be the moving focus and T be the varying points of mutual tangency. The tangent line at T makes equal angle with FT and with a vertical line. This and congruence of the two parabola imply that MT is vertical and FT = MT and must line on directrix of y = x2 – x + 1 So, 3 y 4  38.     2 2 2 1 2 1 DE AK r r r r      = 1 2 2 r r Similarly 2 3 EF 2 r r  APD =  and PD = x then 1r AD BE CF tan PD PE PF x     A B C K D E F P  3 2 1 2 1 2 2 3 r r x 2 r r x 2 2r r r r      Hence, 3 2 2 1 1 2 2 3 r r r r 2 r r 2 r r     2 1 3 r r r   2r 2 8 4    39. Since ln x AM and ND are concurrent a b c b c a 0 c a b  a(bc – a2) – b(b2 – ac) + c(ab – c2) = 0 abc – a3 – b3 + abc + abc – c3 3abc – a3 – b3 – c3 = 0 a3 + b3 + c3 = 3abc = (a + b + c)(a + b + c2) (a + b2 + c) = 0 40. Solve y = x with circle Circle must passes through the foci of the ellipse 41. Using length of direct common tangent 2 1 2 2 1 3 PQ 4r r PQ PR QR QR 4r r       3 1 2 1 1 1 r r r   P R Q C A B AITS-PT-II (Paper-1)-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 12 42. Shown in the figure since C(1, 5) is the orthocentre of triangle AEB, similarly for the other side the coordinates of C(3, 3) D A (1, 3) M (2, 4) B (4, 6) C(1, 5) E(0, 6) F 43. Given circle is (x – 1)2 + (y – 3)2 = 1 Let of tangent to circle is y – 3 = m(x – 1) + 2 1 1 m  (3, 4) lies on axis 2 1 2m 1 m    (1 – 2m)2 = 1 + m2 4m2 – 2m + 1 = 1 + m2 (3, 4) 3m2 – 4m = 0 m = 0, 4 3  y 4 4 m