ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

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1 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) ANSWERSHEET (TOPIC DIFFERENTIAL CALCULUS) COLLECTION # Question Type A.Single Correct Type Q. (A) Sol least value is which occurs when [, 8] ] + Q. (B) Sol h ( ) h f 3+ h f 3 e e f ' 3 lim lim lim h 0 h h 0 h h 0 h ( ) 0 ( 3 h) f 3 h f 3 + 6h h f '( 3 ) lim lim lim h 0 h h 0 h h 0 h ( 6h h ) h ( h 6) 6 lim lim 3 h 0 h 0 h + 6h h + h + 6h h + + Hence f ' 3 f ' 3 B ] Page 009 Q. 3 (C) Sol Now again g ( k) k f ( k) k f ( 009 k) 009 Again g ( 009 k) () + (3) gives k 0 g k g 0 + g + g g 009? f ( k) + f ( 009 k) f k ( f ( k) ) + f ( k) f ( 009 k) [ ] ( f 009 k + f 009 k ) ().() f ( k) ( f ( k) ) + ( f ( k) ) f ( k) + ( f ( k) ) ( f ( k) ) + ( f ( k) ) g k + g 009 k g ( 0) + g ( 009) g( ) + g( 008) g + g( 007) : : :... g 00 + g 005..(3) 009 k 0 g k 005 [C] ]

2 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) Q. (C) Sol ( ) g '( ) f ( + f ( f ( ) )) + f '( f ( )).f '( ) g '( 0) f '( f ( 0 )). + f '( 0 ).f '( 0) f '( 0) + ( 3) 6 Ans. ] Q. 5 (C) Sol f ( ) g f + f f ; f 0 0; f ' Page f ' f ', y ; y ( 7, think! ) 3 y b 3 now 009 y b b 7. ( ) Ans. ] Q. 6 (D) Sol f ( ) 009 ( + ) Put + t d dt t dt I. dt 006 t t t put y dt dy t t 006 d y t I y dy C C t 005

3 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) Q. 7 (B) Sol 008 r r 008 n n n n 008 Tr n S 009 n 008 S d ] Page 3 Question Type B.Comprehension or Paragraph Q. 8 () Sol Q. A Q. B Q. 3 D [Sol. () tan y tan + C (3) 0; y π C tan y tan π + note : even π π π < tan + < ; < (A) π + is as shown. y tan tan + (D) is correct () The graph of f π π < tan < ; < < ( A) Hence range is (, ) (B) Q. 9 () Sol Q. A Q. D Q. 3 A [Sol. Since minimum value is zero hence touches the -ais and mouth opening upwards i.e., a 0 f f + 3 > given

4 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) f ( ) f ( ) f ( + ) f ( ) Hence f is symmetric about the line f ( ) a ( + ) Now given f ( ) f ( ) + f () and f ( ) in ( 0, ) From () and () f now f ( ) a ( + ) f ( ) a a ( + ) f now proceed ] Q. 0 () Sol Q. A Q. D Q. 3 C [Sol. f ( 0) ( + ) f e e cos f ' t dt tf ' t dt 0 0 f ( ) ( e + e ) cos ( f ( ) f ( 0) ) t.f ( t) f ( t) dt 0 0 f ( ) ( e + e ) cos f ( ) + + f ( ) f ( t) dt 0 f e + e cos f t dt () 0 differentiating equation () f ' f cos e e + + e + e sin dy + + Ans. (i) d f ' 0 + f Ans (ii) Hence y e ( cos sin ) e ( cos sin ) (ii) (iii) I.F. of DE () is e Page

5 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) Page 5 y.e e cos sin d cos + sin d y.e e cos sin d sin cos + C Let I e ( cos sin ) d e ( A cos + Bsin ) Solving A 3/ 5 and B / 5 and C / 5 3 y e cos sin ( sin cos ) e + e Ans. (iii) ] Question Type C.Assertion Reason Type Q. (B) Sol f log log / + + > 6 log/ + + log log log + + f is contant Hence f is many one as well into. Also range is a singleton f is constant but a constant function can be anything not the correct eplanation] π π Q. (B) Sol Domain is {, } and range is, elements range must have two elements] and domain having two Q. 3 (A) Sol Using { } + { } if I { } { } f, { } { } f + ( { }) { } { } min. ] f /. 3 3 e e { } Q. (D) Sol +. e e e Hence range is [0, ) S is false]

6 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) Page 6 Q. 5 (C) Sol. Q. 6 (B) Sol Line touches the curve at ( 0, b ) and dy d fails to eist. tangent line can be drawn. ] dy d 0 also eists but even if Q. 7 (D) Sol tan h 0 h lim ] π/ sin cot cot π lim. ; put h ( π ) cot π Q. 8 (B) Sol Range of f is singleton then angle has to be a singleton. If S and S are reverse then the answer will be B. ] Q. 9 (A) Sol y ln not differentiable at y cos is not differentiable at and domain of f is { 0 }. Hence if domain of f is π 3π, 0, π ] differentiable y cos sgn cos 0 ln Q. 0 (A) Sol f '( ) ; note that f f ( ) f is increasing ( 0, e) and f is decreasing e, ] Q. (B) Sol + > f ' 3 0 a > 0 ab < 0 b

7 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) Page k 3 + / 3 3 g( ) + + k 3 3 g '( ) ] 3 3 Q. (A) Sol. Q. 3 (C) Sol. Q. (A) Sol Let f ( ) 0 f ( r ) f ( r ) Hence there must eist some c ( r, r ) where but f '( ) for for has two roots say r and r, f ' > > 3 5 6, f ' 0 f ' c 0 where r, r [ a, b] hence f '( ) > 0 for all Rolles theorem fails f ( ) 0 can not have two or more roots.] Q. 5 (D) Sol f ( ) f ( ) f ( ) Area < 0 area > 0 Case - I : for 0 < <

8 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) Page y 7 3 If < < 0 f 3 + now f ' 0 f ' 0 0 for > f f ( ) ( ) note lim f ( ) f ( ) f ' f ' + f is continuous. Also f is derivable at ] Q. 6 (D) Sol Let b > 0, then f ( ) b > 0 and f ( 5) a + 3b 6 ( a + b) b 6 b 6 ( + b) < 0 Hence by IVT, some c (, 5) s.t. f ( c) 0 If b 0 then a f ( )( 5) 0 5 < < and Hence f ( ) 0 if If b 0, f ( ) b 0 f a + b 3 3 ( a + b) + ( 3 ) b 3 ( 3) ( 3) b which lies in (, 5 )

9 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) ( 3)( b) > 0 ( as b < 0) Hence f ( ) as f ( ) have opposite signs some c (, ) (, 5) for which f c 0 Statement - is valid for all b R statement - is false. D ] Statement - is obviously true Q. 7 (D) Sol f ( ) Page 9 f ( ) f ( ) Area for < 0 area of > 0 Case-I : for 0 < < y For > f f ( ) ( ) note lim f ( ) f ( ) f is continuous R. + Also f ' f ' f is derivable at y f and +ve -ais is Area bounded by the Area 6 3 d Area bounded by the f ( ) and -ais 3 3 Ans.] Question Type D.More than one may corect type Q. 8 () Sol A, B, D [Hint. A; A ; B ; C aperiodic; D π ]

10 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) Q. 9 () Sol Page B, C, D [Sol. The graph of f ( ) f and f ( 0) Verify alternatives y f + Q. 30 () Sol Q. B, C, D f g [Sol. Solving and n n We have 0 and 3 n n + A d 3 n n + 3 n + hence, 3 n + 3 n + 3 n + 6 n 5 6 n + Hence n is a divisor of 5, 0, 30 B, C, D] Q. 3 () Sol Q. A, B, D dy [Sol. y f ( ) d + I.F. e ye e f d + C now if 0 then 0, y 0, C ye +.() ye e e d + C ye + C

11 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) y ; y ( ) Ans. e e ( A) is correct e ( + ) e y ' ; e e e e y '( ) Ans. e e e ( B) is correct if > ye e d + ye e C y e + Ce as y is continuous + lim lim ( e + Ce ) e 3e e + Ce C for > 3 + hence y( 3) e + e e ( e + ) y e e y ' e 3 y ' 3 e Ans. D is correct ] Question Type E.Match the Columns Q. 3 () Sol Q. (A) P, S, (B) Q, R; (C) Q, R (D) P.S. Page π tan f ( ) f ( ) (, ) π g tan f ( ) f ( ) (, ) 3 π tan f ( ) f ( ) (, ) d( ) f ( ) ± 5 d( ) + f ( ) 3 6, 0 6,0 P,S [Sol. Let (A) (B) refer to graph of y f ( ) Q, R

12 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) Page (C) k ( 3, ) k (, 3) Q, R f '( ) (D) + f ( ) g ' < 0 f ' > 0 6, 0 P,S ] Q. 33 () Sol (A) Q; (B) S; (C) P; (D) R ln [Sol. f ( ) a ; f '( ) a () f '( ) ( ) If a, f '( ) 0 8 Hence / a C P is the point of inflection and now f '( ) 0 gives 6 8a 0 + or 6 8a a ± 6a 6 a + a a a a > or a > 3 and f ''( ) 8 a + a 6 f '' a > 8 a + a a + a Hence for a a + a > and, f ( B ) ly for a > and has a local minima S we have local miima A Q finally for 0 a < 6 8a + f '( ) 8 6a 6 < 0 f ' 0 a a Hence > f is monotonic ( D) ( R ) 6 8a + 8 Q. 3 () Sol (A) S; (B) Q; (C) P; (D) R [Sol.

13 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) Page (A) st verte n C way and n can not be taken. Remaining vertices are 3,, 5... ( n ) OOOO four tobetaken ( n 3) vertices... ( n 7) not tobeta ken number of gaps ( n 6 ) n 6 C.n Hence required number of ways 36 5 n Ans. S which is satisfied by 3 3 (B) + a + b + c + + k + k + + k out of which can be selected in n 6 C ways. b and a c Now a can be taken in 0 ways and as a c hence c can be only in one way Also b. Q Hence total 0 Ans. Alternatively: i a + bi + c 0 + 0i c a + b i 0 + 0i a c and b ] 6 (C) z ( + i) z( i )...( ) 6 6 z + i z + i z z z 0 or z if z 0 z 0 if z then zz z z hence equation () becomes ( + ) ( + ) z i i z 7 + i ( + i)( i) z i + i π π mπ + mπ + z cos + isin 7 7 Where m 0,,,...,6 are the other solutions 6 Total solutions 8 Ans. ( P ) (D) f ( ) + g( )

14 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) Page f ( ) g( ) + Put f + g g( ) f ( ) 0 f ( ) < < f < f ( ) f (as f ( ) is a non negative integer) again put f ( 000) + g( 000) 000 f g 000 log 000 g( 000) log ( 000) f ( 000) 0 log 000 f ( 000) < log 000 f ( 000) < ( log000) ( log 000) f ( 000) log 000 f ( 000) 9 as f is an integer Hence f ( ) + f ( 000) Ans. ( R )]