SOLUTIONS SECTION I: CHEMISTRY

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1 P. CH CH CH CH H CH CH CH CH SLUTINS SECTIN I: CHEMISTRY S N NaH yellow I. CH CH CH CH C CH CH CH C Na CHI H H. H Ea 4..9 = loga.r769 Ea. = loga.r667 Ea = cal mol - CH CH Cl H C CN H C CH 5. Cl h KCN CH5H H 6. BeCl, N, N, N,, SCl, ICl, I, XeF BeCl sp linear N sp linear N sp linear N sp linear sp bent SCl sp bent I sp d linear ICl sp d linear XeF sp d linear So among the following only four (4) has linear shape and no d-orbital is involved in hybridization. 7. As covalent character increases then solubility decreases 8. C6H5NCl gives scarlet red coloured dye with - naphthol. 4. H C NH (A) Ac H C NH C CH (B) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 6594

2 Ai TS - (XII)-SET-A-APT--(CMP-IITJEE/9)- 5. H C NH HN Cl 5 C + H C N Cl HP H C (C) (B) PART A SECTIN II: MATHEMATICS. If a boy is selected then number of ways = 4 C 6C If a boy is not selected then number of ways = 6 C 4 Captain can be selected in 4 C ways Required number of ways = 4 C 6C 4C + 6 C 4 4C = 8. For ma. or min. f() = 8 + ( + ) = for one maima and minima D > < 4. (a + (b + c)) n = a n + n C a n (b + c) + n C a n (b + c) + + n C n (b + c) n. Further epanding each term of R.H.S., First term on epansion gives one term. Second term on epansion gives two terms. Third term on epansion gives three terms and so on. (n ) (n ) Total no. of terms = (n + ) =. 4. The required numbers are,,,,,,..,. Let us calculate how many numbers are these. There are one-digit such numbers. There are two-digit such numbers and so on. There are 8 eight-digit such numbers. All the digit numbers beginning with and written by means of and are smaller than. 8. Thus, there are 8 such nine-digit numbers. Hence the required number of numbers is = = = 766. t t (t ) (t ) ( ) (since > ) 5. t = log c t 6. The area bounded by the lines y =, is shown in the fig. Area A() =. ( ) A() = ( ) = 6 4 = = / Maimum area of the rectangle occurs when = /. Maimum area =. 9 sq. units. y = + y y = FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 6594

3 7. Let 8 7 I f, where f = fractional part and I = integral part then g I Also let 8 7 g 8 Here I f g C C 7 I f g = even integer But f g So, I even Integer f g I odd Integer Ai TS - (XII)-SET-A-APT--(CMP-IITJEE/9)- 8. 9! 56!! 9. F/ / lnt ln / u du u u t t u So statement- is not true. If F() = lnt then t / lnt lnt F F/ t t u t lnt ln / u du t u = du F u u u lnu ln t ln t t t t t t ln lnt t.. sin, f() sin, lim f() lim f() f() f is continuous at = f( h) f() Rf () lim h h h hsin h lim h h lim h sin does not eist h h h Similarly Lf() does not eists f() is continuous but not differentiable at =. [] ; f() ( )[] ; FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 6594 Website: Mail :aiits@fiitjee.com

4 Ai TS - (XII)-SET-A-APT--(CMP-IITJEE/9)-4 ; ; f() ( ) ; 6 ; Lf() = and Rf() = Lf() Rf() Lf() = and Rf() = Lf() Rf() Both f() and f() does not eist. f( h) f(). Lf() = lim h h sinhh Lf () lim h h =. If f() touches -ais at only one irrational point, then f() = ( ) g(), where is irrational. coefficients of f() can t be rational for f() with rational coefficients, then point of touching is rational. 4. The point of touching has to be rational the two roots of f() = are rational third root is also rational. 5. f() = ( ) ( ) f() = ( ) ( ) ( ( + )) f() has roots, and a root between, and,. PART B A cos sin has a local maimum at and (B) tan sin cos is strictly increasing in, (C) 5 6 for all d f = d.d f f D I d d d f f f 9 I. Given differential equation is ( + y) dy d = a Put + y = t + dy d t d = a a t t d, we get d t d = a + t (i) in the interval, FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 6594

5 Ai TS - (XII)-SET-A-APT--(CMP-IITJEE/9)-5 t Integrating, d a t a d a t t a tan - t c a + y a tan - y = + c a y = a tan - y c is the required solution. a (A) (s) (B) sec y tan dy = -sec tan y d sec y sec dy d tan y = tan log tan y + log tan = logc log tan tany = log c Hence tan tan = c is the required solution (B) (r) (C) dy d = e e 4y e -4y dy = e d + c 4y e e C 4 Putting =, we have = C 4 4y 7 C e e 7 Hence 4 7 = e -4y + 4e (C) (q) (D) 5 + C 5 5 y (D) (p) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 6594 Website: Mail :aiits@fiitjee.com

6 Ai TS - (XII)-SET-A-APT--(CMP-IITJEE/9)-6 PART - A SECTIN III: PHYSICS. Magnetic field at is I I k i I i 4R 4R 4R. The magnetic dipole moment of the current carrying coil is given by m NIAnˆ =.5 (.8 ).4 î = 6 Am (î) The torque acting on the coil is mb mb (i ˆ ˆj).5 6 k ˆ = 5.66 (N - m) ˆk.. Let and be the etensions of the two rods F / A F Y / A 4. Consider a cylindrical element of radius r and length. According to Gauss s Law E ds q in E(r ) E = r ( r ) r R 5. Inside the cavity, B = utside the cylinder, B I r In the shaded region I a B r r(b a ) r at r = a, B = at r = b I, B b a b 6. d T sin BiRd Td = BiRd T = BiR = BiL (for small) T cos (d/) T T cos (d/) T T sin (d/) 7. Statement does not confirm first statement. 8. Total energy is always negative for such systems. 9. Angular momentum is conserved. So velocity is variable.. Apply F qv B. ve charge accumulates on face ABCD. Charge will drift due to magnetic force only. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 6594

7 Ai TS - (XII)-SET-A-APT--(CMP-IITJEE/9)-7. Energy must be less than V 4. [] = ML - T - nly (B) option has dimension of time Alternatively d m k ka 4 4 d k A m 4 4 m A d 4 T k 4 4 A m du 4 T k A 4 u Substitute = Au 5. As potential energy is constant for > X, the force on the particle is zero hence acceleration is zero. PART - B. Uniform E constant acceleration so straight line or parabola. Uniform B initial v along B -straight line. Uniform B initial v B -circle Uniform B initial v < B -uniform right circular cylindrical heli. Uniform B uniform E initial velocity along B or E -straight line. Uniform B uniform E initial velocity to B or E -non uniform line. Uniform B uniform E q v B =-q E straight line.. At steady state capacitor behave like open circuit. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 6594 Website: Mail :aiits@fiitjee.com

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