Q A. A B C A C D A,C,D A,B,C B,C B,C Q A. C C B SECTION-I. , not possible

Transcription

1 PAPER CODE SCORE-I LEADER & ENTHUSIAST COURSE TARGET : JEE (Main + Advanced) 04 PART- : MATHEMATICS SECTION-I SECTION-II SECTION-IV. Ans. (A) 0 C T 0 7 SECTION-I For x > & >, [x] - [] - + is not possible []- [x]- For 0 < x < & 0 < < -x x Þ +- x +- [ ] [ x], not possible For < x < 0 & < < 0 Þ 0 < x < & 0 < < Þ +- x x, possible For < x < and < < Þ < x < and < < x TEST # 06 PATTERN : JEE (Advanced) Path to success KOTA (RAJASTHAN) Date : ANSWER KEY Q A. A B C A C D A,C,D A,B,C B,C B,C Q. A. C C B Q. A B C D S T R Q Q A SOLUTION Þ +- x x, possible x \ required area. Ans. (B) (g(x) 4) (g (x) g(x) + 4) 0 Þ g(x) 4 (i) & (g(x) 4) (g(x) ) ³ 0 Þ g(x) ³ 4 & g(x) (ii) & (g(x) 4) (g(x) ) 0 Þ g(x) 4 (iii) From (i), (ii) & (iii) Þ g(x) 4 Þ g(cos(x )) 4 \ 4x Þ equation of normal drawn from (, 0) to the curve 4x is 4x + \ a + b 5 Corporate Office : ALLEN CAREER INSTITUTE, SANKALP, CP-6, INDRA VIHAR, KOTA-4005 PHONE : , Fax : , info@allen.ac.in Website: KOTA / HS - /

2 Path to success KOTA (RAJASTHAN). Ans. (C) (x ) (x 5) d x 6 0 dx - x. Maximum value of 4. Ans. (A) TARGET : JEE (Main + Advanced) 04 uuur uuur uuur uuur CA CB. CA CB uuur uuur.. CA CB sin 90 Equation of tangent is xsec q tan q -. S sum of intercepts S cosq-cot q Þ ds d q - q - - q ( sin ) ( cosec ) 0 Þ sin q cosec q Þ sin q Þ sinq é æö sin êæ ö ç Þq ê ç è ø è ø ë 6-6 ù ú ú û Now, ds dq - q- q q< cos 4cosec cot 0 \ a 5. Ans. (C) A (a +, a +, 4a + ) B (b +, b, b ) as qî(0, p/). \ a-b- a-b+ 4a-b+ 5 (i) ( ii) ( iii) Solving (i) & (ii) Þ 4a b Solving (ii) & (iii) a b + 0 Þ a, b Þ A (,, ) & B (,, ) uuuv AB 6 KOTA / HS - /

3 Path to success KOTA (RAJASTHAN) 6. Ans. (D) LEADER & ENTHUSIAST COURSE a 0 b a 0 f (0) a b+ 0 b b 0 a f (0) a + b. 7. Ans. (A,C,D) f (x) g (x) & g (x) h (x), f (x) h (x) Þ f (x) g (x) + C Þ g (x) h (x) + C Þ f (x) h (x) + c Þ C C Þ c 0 Þ f (x) g (x) +, g (x) h (x), f (x) h (x) Interate Integrate Integrate Þ f(x) g(x) + x + k, g(x) h(x) x + k f(x) h(x) + k k Þ k Þ k Þk Þ k 0 Þ f(x) g(x) + x+, g(x) h(x) x, f(x) h(x) (A) ò ò ò (f(x) + h(x) - g(x))dx (4x + 4)dx 4 (x + )dx (C) f(x) - g(x) x + Þ x+ x + Þ x 0, (D) f(x) & h(x) are indentical function. 8. Ans. (A,B,C) f(x) k + e -e + x x x e 4 x x x x x x (e + 4)(e -e ) -(k + e -e )e f'(x) 0 x (e + 4) x x Þ e + 8e + k for local max and minimum D > 0 Þ 64-4(k - 8) > 0 Þ6 - k + 8 > 0 Þ k < 8Þ - < k < 9. Ans. (B,C) f() -f(0) -c 4-5 f'(c) Þ Þ c f() -f() & f '(c ) Þ c - KOTA / HS - /

4 Path to success KOTA (RAJASTHAN) TARGET : JEE (Main + Advanced) Ans. (B,C) f(x) ( + [ a]) x + ( + [ a]) x [ a] Þ f() < 0 Þ 4(4 + [ a] + 4[ a]) + +[ a] [ a] < 0 Þ 4[ a] + 9[ a] + < 0 Þ 4[ a] + [ a] + 8 [ a] + < 0 Þ [ a] (4[ a]+) + (4[ a] + ) > 0 æ ö Þ ([- a] + ) ç [- a] + < 0 è 4 ø Þ- < [ - a] <- 4. Ans.(C) r r r r r r r r r r r r ( ) ( ) P a + a b + c + a. a b + c (a b) + a c + sin q+ + a.c rr + sin q+ + a.c rr r rr q + sin q- + a.c r r p+ q rr rr 5 Þ + sin q+ a.c + + sin q-a.c -. > 0. Ans.(C) Þ sin q > 0 Þ æp pö sin q> ÞqÎ ç, è4 4 ø cos a - q p.q sin p q (+ sin q) -4(a.c) < 0. Ans.(B) r r p- q + sin q + sin q Let r r p + q 4 + sin q + sin q - q q Þ 0 d sin cos dq + sin q Þ q 0, p, p \ Range is é ö ê, ë ø KOTA / HS - 4/

5 Path to success KOTA (RAJASTHAN) SECTION II. Ans. (A) S ; (B) T ; (C) R ; (D) Q (A) log cos x x px Y LEADER & ENTHUSIAST COURSE no of solution 4 (0, 0) O p p X x p (B) Let mx + m 4 \ 4m + m Þ 4 4m 4 + m Þ m 4 + m 0 Þ m - ± Þ (4m + ) (C) [568] M [ ] M [0 0] [ 45] Þ [4 45] M [ 4 4] Þ a + b c - x Þ log x x- graph solutions (0, 0) Y x, 0 log x X SECTION IV. Ans. Equation of line L is (0,) A L (x, ) (, ) (x - 0) Þ- - x Þ x +. Put in equation of circle - Þ x + ( x) x ( x + ) + 0 Þ x x 4x x + x 0 Þ x 4x + 0 Þ x 4 ± 6-8 +, \ x - Þ æ- + ö \ centre is, ç è ø KOTA / HS - 5/

6 Path to success KOTA (RAJASTHAN). Ans. 7 A adj A + I ( A + )I 7. Ans. 4 Line L : x 4x 4 0 Þ (x ) (x + ) 0 x x, - TARGET : JEE (Main + Advanced) \ One line is coincides with L Þ 5x + (0 k) 0-k k Þ Þ k 0 k 0 Þò p 0 ò sin0x dx 0 0 sin0x dx 4 4. Ans. 8 Let centre be (h,k) 0 p x x - k 0 4 Þ(x- h) + (- k) (h- ) + (k-) Þ x + -hx- k -h- 4k+ 5 Þ Now (h) () + k.0 h + 4k 5 + a x hx k h 4k 5 0 Þ 4 5-a 5- a again 4 Þ 5- a 4 Þ a 9, 5. Ans. - (x - e )d xdx - dx x -e Þ d x - dx x e Þ x - d KOTA / HS - 6/

7 Path to success KOTA (RAJASTHAN) - dx x e Þx - - d Put x t dx dt Þ x d dt - dt t e Þ - - d dt - Þ -.t.e d Integrating factor - I.F ò e d - e -ln - - e t.. d ò LEADER & ENTHUSIAST COURSE t - e + C - Þ t e + C - Þ x e + C 6. Ans. f(x) x x ln(x ) e - - (x -x)ln(x-) é (x -x) ù Þ f '(x) e ê(x -)ln(x - ) + 0 x- ú ë û Clearl x is a root - + \ a KOTA / HS - 7/

8 Path to success KOTA (RAJASTHAN) PART- : PHYSICS SECTION-I SECTION-II SECTION-IV TARGET : JEE (Main + Advanced) 04 SECTION-I. Ans. (C). Ans. (A). Ans. (C) 4. Ans. (B) Sol. (i) Length (L), mass (M) and velocit (LT ) are independent. (ii) Pressure (ML T ), densit (ML ) and velocit (LT ) are dependent as ( - 0) + ( --0)-( - 0) (iii) Force (MLT ), velocit (LT ) and time (T) are independent as ( -0) -( 0-0) -( 0-0) ¹ momentum (iv) Force time ANSWER KEY Q A. C A C B A B B,C,D A,C B,C,D B,C Q. A. C A D Q. A B C D T Q P,S R Q A SOLUTION Þ force, momentum and time are dependent. 5. Ans. (A) Sol. The force has been exerted b liquid on the tube due to change in momentum at the corners i.e., when liquid is taking turn from A to B and from B to C.As corss-section area at A is half of that of B and C, so velocit of liquid flow at B and C is half to that of velocit at A. Let velocit of flow of liquid at A be v and cross section area at A be S, the velocit of flow of liquid at B and C would be v [from continuit equation] and corss section area at B and C would be S. dsv dsv dsv dsv KOTA / HS - 8/

9 LEADER & ENTHUSIAST COURSE Path to success KOTA (RAJASTHAN) Due to flow of liquid, it is exerting a force per unit time of rsv on the tube, where r is the densit of liquid, S is cross section areaand v is velocit of flow of liquid. The force exerted b liquid on tube is shown in the figure. Which clearl shows that a net force is acting on tube due to flowing liquid towards right and a clockwise torque sets in. 6. Ans. (B) r r r r r r r r r A B C A æa B Cö A r A ç è 4 ø [Q ] 7. Ans. (B,C,D) Sol. For (A) : Displacement x 5 x m For (B) : Distance travelled For (C) : Average velocit : x - x + x - x + x - x (- ) + 6+ m Dx Dt for given interval Dx < 0 so v < 0 av Dv For (D) : aav, in given interval Dv < 0 so a D t av < Ans. (A,C) Ncos0 N...(i) & N sin0 Mg...(ii) 9. Ans. (B,C,D) 0. Ans. (B, C) E Dl léë+at ù E û l (K + K )x N cos 0...(iii) Þ x Mg K Dl éë+a t ù...() û l [ + a t] + l [ + a t] -éëd l +D lùû l + l...(). Ans. (C) & Sol. In the time interval t 0 sec to t t 0 sec, till the time. The relative velocit is not zero the nature of friction will be kinetic. r u 0i ˆ- 0i ˆ 0m / sec ˆi AB ( ) ( ) KOTA / HS - 9/

10 Path to success TARGET : JEE (Main + Advanced) 04 KOTA (RAJASTHAN) N 00 N Þ F k N r aa 4m/sec i r r r ( ˆ r ), ab 0m/sec ( ˆi) ( ˆ) aab aa - ab 4m/sec i V r r r u + a t 0 r AB AB AB Þ t 5 sec F 5N 0 g 0 kg When the relative motion between block A and belt conveor will be zero, the nature of friction will be static and its magnitude will equal to magnitude of unbalanced external force acting on the block A and its direction will be I the opposite direction of unbalanced external force. Þ F Frictional force ˆ ( ) ˆ ( ) ì5n i Þkinetic in nature ï í ï5n -i ÞStatic in nature î Friction force 5N O 5N N 5sec F k t SECTION II Ans. (A) (T) ; (B) (Q) ; (C) (P, S) ; (D) (R) Sol. (P) (Q) (R) (S) (T) M(b) mb sin 0º Mb mb Mb mb mb mb mb sin 0º mb mb 4 sin 45º b r b I. Ans.. Ans.4. Ans.6 4. Ans. 6 mr mb SECTION IV Least count mm 0. mm; Zero error (0 6) mm; 0 Reading (0.) ( 0.4) 7. mm 5. Ans. nl ( ) RD l BD n+ ( n ) R ( n+ ) B Þ d d Þ 7800n (500) (n+) Þ 6n 4 n+4 Þ n 6. Ans. KOTA / HS - 0/

11 Path to success KOTA (RAJASTHAN). Ans.(D) PART- : CHEMISTRY SECTION-I SECTION-II SECTION-IV LEADER & ENTHUSIAST COURSE SOLUTION SECTION-I O Sol. CHI -C- CHI is not formed during haloform reaction. ANSWER KEY Q A. D C D C D C A,B,C A,B,D A,B,C A,B Q. A. B C C Q. A B C D P,Q,R,S,T P,Q,R,S,T P,Q,S P,Q,R,S,T Q A Ans.(C) Sol. Neopentl chloride reacts b unimolecular nucleophilic substitution.. Ans. (D) 4. Ans. (C) 5. Ans. (D) 6. Ans. (C) Sol. Cell reaction H O O + 4H + + 4e Anode 4H + + 4e H Cathode H O H + O DGº Eº cell ÞEº cell. V 7. Ans. (A,B,C) Sol. During resonance delocalisation of p or p electron takes place while the position of atoms remains same. 8. Ans. (A,B,D) 9. Ans. (A,B,C) 0. Ans. (A,B). Ans. (B) Compound-I show geometrical isomerism. (I) KOTA / HS - /

12 Path to success KOTA (RAJASTHAN). Ans. (C) TARGET : JEE (Main + Advanced) If R is CHDCH (I) can show geometrical isomerism as well as optical isomerism.. Ans. (C) HCHO makes onl one hdrazone but CH CHO makes hdrazone due to geometrical isomerism. SECTION II. Ans. (A) (P,Q,R,S,T) ; (B) (P,Q,R,S,T) ; (C) (P,Q,S) ; (D) (P,Q,R,S,T) (A) Compound (A) soluble in NaOH, NaHCO, reacts with Na. Aromatic. CO H group shows ( I, R both) (B) Compound (B) soluble in NaOH, NaHCO, reacts with Na. Aromatic. NO group shows ( I, R both). (C) Compound (C) soluble in NaOH, reacts with Na. Aromatic. OH group shows ( I, +R) (D) Compound (D), soluble in NaHCO, NaOH, reacts with Na. Aromatic. SO H group shows ( I, R). Ans. SECTION IV It is onl observe in (ix) where saturated a H are present with CC or C +.. Ans. 4. Ans. Sol. Y æx öæp ö ç YB X ç B P è øè 0,B ø A A 0,A ç æp 4 P ö 0,A ç è 0,B ø Þ 0,A 0,B P 4 P 4. Ans. 8 Sol. Hint :. r a.r Ans. 7 P Total 6atm Sol. K P P. T a -a +a ; P T.0.6 K p ( )( ) Þ P T 6atm 6. Ans. 7 units KOTA / HS - /