MOCK TEST 0 SUBJECT: MATHEMATICS ANSWERS: COMMON ENTRANCE TEST 0 ANSWERS. () π π Tke cos - (- ) then sin [ cos - (- )]sin [ ]/. () Since sin - + sin - y + sin - z π, -; y -, z - 50 + y 50 + z 50 - + + 9 9 9 y z + +. () pπ qπ pπ π qπ pπ π qπ tn cot tn tn nπ + pπ nπ +π -qπ (p + q)π (n+)π p+q (n+). () 5. () π i π π i cos isin e 6 + 6 6 log i π log + i 6 π Principle mplitude tn - 6 0 tn - π (- ) π z z z..to (..to )( y y y..to ) (CiS CiS π π CiS π.to )(CiS CiS π π CiS.to ) π π π CiS + + +... to π π π cis + + +...... to cisπ π π cis π CiS i CiS 6. () The distnce s moved by prticle in time t is given by s cos t + b sin t. ds vel sin t + b cos t ; dt d s Its ccelertion is ; cos t bsin t ( cos t + bsin t) -s dt 7. () Let the given line divided into prts of lengths nd y nd let + y k, given. Are of rectngle is A y (k ) k da k 0 k nd y k k k k or y k. Therefore rtio in which given line be divided into two prts so tht the rectngle contined by them is mimum is : 8. ( ) y sin + At (0, 0) + cos.0 + cos 0 The eqution of norml is (y-0) -(-0) or + y 0
9. () y ------- () nd y ------ () + y 0 y solving () nd () we get nd y therefore m - m m m tnθ The ngle is tn - + m m + ( )() 0. () Number of cut vertices is p nd number of cut edges is q.. () The given eqution is ( y + )+k(5 6y ) 0 Solve equtions y + 0 & 5 6y 0. () m + m m m. () h b b h h- Let (, y) be the mid point of BC y + y + Then (, -),, + +, y - (, -) is the required point. () Put 0 we get 0 d 0 -(0 6) ( 0) 6-6 0 (-, ) G (, -) 5. () The given eqution is (sina sinb)(sinb sinc)(sinc sina) 0 Therefore either A B or B C or C A b or bc or c. 6. () Apply Cyley Hmilton therem. 7. () A + B AA + BB A(BA) + B(AB) (AB)A + (BA)B BA + AB A + B 8. () It is dvised to cross check with simple integer sy n 8 0 we get the vlue of ( 9 ). (mod0) therefore the lst digit is + 5 which is (d). dr (/ ) 9 9. () S + + r ( r) (/ ) / 9 0. () Substitute n 7 is divisible by 7 only.. () If A B + C +, then + + ) A B C b) A B C c) A B C d) A B C. () If in the epnsion of vlue of n is d) 5 n, n N, sum of the coefficient of 5 nd 0 is 0, then. Ans: () y ) + log( sin ) + log(sin ) +... to sides log(sin squring on both ( sin ) + log(sin ) +... to log(sin ) y y log(sin ) + log(sin ) + log + differentiting w. r. t. we get y cot +. () + Sol: Given y sec - + sin + ( y ) cot cos - + sin [using sec - cos - /] + +
π 0. 5. () d y Sol: y e + be e + be e + be d y - + y (e + be ) (e + be ) + (e + be ) 0. 6. () ( ) + Sol: y + +. 7. () π + cos 8 π cos 8 π + cos 8 π π π cos 8 sin sin 8 8 π π cos cos. 8 8. () In tringle ABC, b nd A B, then ngle A is 90 o A B C 9. () If sin A + sin B + sin C, then A B C 90 o sin + sin + sin 0. () If Secθ m nd Tnθ n, then ( ) ( ) m + n + m m + n. () ddr l(dr) + m 5 cot 5 cos sin cot + cos + sin Dr c + bd d bc 9 l m c + d c + d 9 solution is l + m log(dr.) log( cos + sin ) + + C. () e e e e sinh cosh + + + e e e sinh + cosh e. () Integrtion by prts tke u sec - nd dv sec sec - - cosh -. () ( cos )( cos ) cos sin / sin / cos / cos sin tn / log sec 5. ()For every, b R, -b R (b) (c) re lso b.o. * b not b.o. becuse b cn be -. is binry opertion b + 6. () b i j k -5i -5j -5k b 5 + 5 + 5 5 Vector of mgnitude 5 perpendiculr to both nd b is ± 5n ^ b ±5 ± b 5 (-i j k) 7. ()Vector in the direction of hving mgnitude equl to the mgnitude of b b
(i + j + k) 9 + 6 + 9 7 (i + j + k) 8. () OA i + j + k; OB -i + j + k, OC i + j - k AB -i k; BC i + j - k; AC i + j - 7k AB 8 ; BC 6; AC 5 AC AB + BC Hence it is right ngled tringle 9. () In {, 5, 7}, 5 7 {, 5, 7} In {5, 7, } nd {7, } identity is not there Hence {, 5} is subgroup. 0. ()If one root of 5 + 9 5 0 is + I, then the other root will be i (Q compleroots with rel coefficients occur in conjugte pirs) Let α be the third root Sum of the roots ( + i) + ( i) + α 5 α The other roots of the given eqution re i nd. () Let α be the other root of + k 0 + α k; α - α - k α - 8 k 8 5 Alterte: by putting, k 5 in the choices, we observe tht 5 + 6 0 hs roots nd. is lso root of k + 6 + 0, where k 5. + log 6 5 0 log 0 0 log 0. () 0 0 0 0. () The negtion of ~(y) ~y The negtion of the sttement If it rins then you get wet is it rins nd you don t get wet sec sec.() sec cos ec ( + tn ) ( ) tn + tn tn ( + ) dt t cot + tn cot t 5. () Put, b nd c 0 Then π 0 ( + )( + 9)( + 0) ( + )( + 0)(0 + ) 5 i.e., 0 ( + )( + 9) 6. () 6 60 π Are 7. () 6 6 9 + 7 Squring both sides: + 7 Order, Degree 8. () y ( + ) (y + ) y y + + y + log ( + ) log (y + ) + y + ( + ) k (y + ) log k
9.() 50. () lim n n n ( + ) m{, } n The function f() + is discontinuous t the origin is discontinuous there 5. () f () is defined if nd > 0 nd + 0 i. e. 0 nd > nd - i. e. - < nd 0 i. e. [-, ) ecluding 0 5. () Number of proper subsets 6 6 5. () Only (, - ) stisfies the eqution + y 8 + 6y + 0 0 5. () Eqution of the circle pssing through the intersection is S + ks 0. + y 6 + k( + y 6 + 8) 0 -----------------------() Since the required circle psses through (, ) + 6 + k ( + 6 + 8) 0 k. From () eqution is + y + 0. 55. () The centres re A (, - ) nd B (-, ) Since the circles cut orthogonlly, AB r + r i.e ( + ) + (- ) r + r 8 r r 56. () The locus is circle concentric with the given circle. But only lst option represents circle concentric with given circle 57. () ( ) + 8 - + 8 6 0 58. () 8 +6 0 ( ) 0, y Point of contct is (, ) The tngent to prbol y 8 is y m + m. If this mkes n ngle 5 o with y + 5 tn 5 o m + m m ± + m m or m m - + m Eqution of tngent is y - i. e. + y + 0 or y + i. e., y + 8 0 59. () y + 8y 0 + (y y + ) - ( ) (y ) - which is hyperbol with verticl is Trnsverse is b 60. () b Given.b ------- () nd --------- () From () b nd hence from ()
b b 6 b Distnce between foci b 6 9 6