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1 Question Booklet Version (Write this number on your Answer Sheet) Day and Date : Thursday, 0th May, 08 QUESTION BOOKLET (MHT-CET - 08) Subjects : Paper I : Mathematics MH-CET 08 Roll No. Question Booklet Sr. No. Answer Sheet No (Write this number on your Answer Sheet) Duration : hours Total Marks : 00 This is to certify that, the entries of MH-CET Roll No. and Answer Sheet No. have been correctly written and verified. Candidate s Signature Invigilator s Signature Instructions to Candidates. This question booklet contains 50 Objective Type Questions in the subjects of Physics (50), Chemistry (50) and Mathematics (50). The question paper and OMR (Optical Mark Reader) Answer Sheet is issued separately at the start of the eamination.. Choice and sequence for attempting questions will be as per the convenience of the candidate.. Candidate should carefully read the instructions printed on the Question Booklet and Answer Sheet and make the correct entries on the Answer Sheet. As Answer Sheets are designed to suit the OPTICAL MARK READER (OMR) SYSTEM, special care should be taken to mark the entries correctly. Special care should be taken to fill QUESTION BOOKLET VERSION, SERIAL No. and MH-CET Roll No. accurately. The correctness of entries has to be cross-checked by the invigilators. The candidate must sign on the Answer Sheet and Question Booklet. 5. Read each question carefully. 6. Select the correct answer from the four available options given for each question. 7. Mark the appropriate circle completely like this, for answering a particular question. Mark with Black ink ball point pen only. 8. For Physics and Chemistry each answer with correct response shall be awarded one () mark. For Mathematics each answer with correct response shall be awarded () marks.there is no Negative Marking. No mark shall be awarded for marking two or more answers of same question, scratching or overwriting. 9. Use of whitener or any other material to erase/hide the circle once filled is not permitted. 0. Avoid overwriting and/or striking of answers once marked.. Rough work should be done only on the blank space provided on the Question Booklet. Rough work should not be done on the Answer Sheet.. The required Log-Antilog table will be provided along with the Question Booklet.. Immediately after the prescribed eamination time is over, the Question Booklet and Answer sheet is to be returned to the Invigilator. Confirm that both the Candidate and Invigilator have signed on question booklet and answer sheet.. No candidate is allowed to leave the eamination hall till the end of eamination. 5. No marks will be deducted if a particular question is not attempted.

2 Kalrashukla Classes MATHEMATICS K d. If, 8 then the value of K is 0 K K K d d d 8 9 ( ) K tan () tan K or K K 0. The cartesian coordinates of the point on the parabola y = 6, whose parameter is, are (, ) (, ) (, ) (, ) Given y = 6 = () In usual notations a =, if t is the parameter t = /, then its cartesian coordinates are ( at, at) = ( t, 8t) or, 8 (, ). d sin cos sec + log sec + tan + c sec. tan + c sec + log sec tan + c sec + log cosec cot + c I = sin cos d sin cos sin cos d sin cos sin cos sin cos = sec tan d cosec d sec log cosec cot c d y. If log0, then y y y y y

3 MH-CET - 08 y y 0 00 y 00 00y or y 99 but 0 99 y 0 dy 99 d 0 or y dy 99 y y d 0y y, 5. If f : R {} R is a function defined by f() = then its range is R R {} R {} R {, } f : R {} R f () f() = + As < < and < < < + < and < + < f() R {} 6. If f() = + for 0 = for < 0 is continuous at = 0 and f then + is Given f() is continous = + or = Now, f d y d 7. If y (tan ), then ( ) ( ) d dy 0 Consider y = (tan ) dy (tan ) dy d y dy ( ) (tan ) ( ) d d d d ( ) d y dy ( ) ( ) d d

4 Kalrashukla Classes 8. The line 5 + y = 0 coincides with one of the lines given by 5 + y k y + = 0 then the value of k is Given 5 + y = 0 coincides with one of the lines 5 + y k y + = 0 coefficient of = 5, coefficient of y = 0, coefficient of y =, constant term = The other line can be ( ) = 0 Their combined equation is (5 + y ) ( ) = y y + = 0 comparing it with 5 + y k y + = 0 k = 9. If A then (A 5A)A = (A 5A)A = A A 5AA = A 5I = 0. The equation of line passing through (,, ) and perpendicular to the lines r (i ˆ ˆj k) ˆ (i ˆ ˆj k) ˆ and r (i ˆ ˆj k) ˆ (i ˆ j ˆ k) ˆ is y z y z A (,, ) Given lines are L r (i ˆ ˆj k) ˆ (iˆ j ˆ k) ˆ b p L r (iˆ ˆj k) ˆ (i ˆ j ˆ k) ˆ c q ˆi ˆj kˆ p q i ˆ j ˆ kˆ y z y z The desired line is y z i.e. y z

5 MH-CET Letters in the word HULULULU are rearranged. The probability of all three L being together is 0 The word is HULULULU H, L s and U s 5 8! They can be re-arranged in!! ways Now when all L s are together considere them as group. Thus we have group, H and U s to arrange they can be arrnged in 6!! ways 8 6!/! 6!!! 6 Desired probability = 8!/ (!!)! The sum of the first 0 terms of the series , is 5 9 (9 0 ) 00 (0 9 ) (0 0 ) Consider the sum S = upto 0 terms = (0 ) + (0 ) + (0 ) (0 0 ) = (0 ) = 0 0 = = 00 (0 9 ) 9. If A, B, C are the angles of ABC then cot A. cot B + cot B. cot C + cot C. cot A = 0 We know that, in ABC tan A + tan B + tan C = tan A tan B tan C tan B tan C tan A tan C tan A tan B cot B cot C + cot A cot C + cot A cot B = d. If Asin (B) C, then A + B = d sin C A and B A B ()

6 6 Kalrashukla Classes 5. sin e d cos e tan + c e + tan + c e tan + c e tan + c sin sin cos e d e d e [sec tan ] e tan c cos cos 6. A coin is tossed three times. If X denotes the absolute difference between the number of heads and the number of tails then P(X = ) = Given n =, X = number of head number of tails, {HHH, HHT, HTH, THH, TTT, TTH, THT, HTT } P(X = ) = P[(H + T) + (H, T)] = If sin cos, then tan = 6 sin cos sin cos cos sin cos cos sin sin sin cos cos sin sin cos tan 8. The area of the region bounded by = y, y =, y = and the y-ais lying in the first quadrant is square units. We have = y y 8 0 / y 8 ( / ) A y dy y y (8 )

7 MH-CET e cos 9. If f() =, for 0 is continuous at = 0, then value of f(0) is e cos f (0) lim lim The maimum value of + y subject to + 5y 6 and 5 + y 0, 0, y 0 is Given z = + y y Let L ( 5y 6) Y y y L (5 + y = 0) 6 0 Solving L and L we get their point of 9 5 intersection as, OABCO is the feasible region, where O (0, 0); 9 5 A (0, 5.); B,, C (6,0) z 0 = 0, z A = 0., z B =.5, z c = Ma z at C (6, 0) X 8 6 A B 0 C X Y L L. If a, b, c are mutually perpendicular vectors having magnitudes,, respectively, then [a b c b a c] Given a, b, c [ a b c b a c ] = ( a b c ) ( b c a c ) = [ a b c ] [ b a c ] = ( a b ) c = cˆ cˆ cˆ =

8 8 Kalrashukla Classes. If points P(, 5, ), Q(, y, ) and R(5, 8, 0) are collinear, then the value of + y is 5 Drs of PQ, 5 y,, Drs of QR, y 8, 5 y = y 8 = = y 8 = 0 + y y = + y =. If the slope of one of the lines given by a + hy + by = 0 is two times the other then 8h = 9ab 8h = 9ab 8h = 9ab 8h = 9ab Given a hy by 0 y y b h a 0 bm hm a 0 h m m and b m a m b = h b ( h ) 9b a b 8h = 9ab. The equation of the line passing through the point (, ) and bisecting the angle between co-ordinate aes is + y + = 0 + y + = 0 y + = 0 + y + 5 = 0 (A, C) Note : The interpretation of bisecting the angle between coordinate ais is not to be considered passing through origin otherwise the Q vanishes. y = ( + ) y = or y = + + y + = 0 or y + = 0 5. The negation of the statement: Getting above 95% marks is necessary condition for Hema to get the admission is good college. Hema gets above 95% marks but she does not get the admission in good college Hema does not get above 95% marks and she gets admission in good college If Hema does not get above 95% marks then she will not get the admission in good college Hema does not get above 95% marks or she gets the admission in good college Given p q i.e. q is necessary for p q getting above 95% p Hema to get the admission in good college Its negation is p ~q i.e. ~q p Hema does not get above 95% but she gets admission in good college 6. Cos CosCos Cos79= 0 Cos CosCos Cos79= 0 ( cos 90º = 0)

9 MH-CET If planes cy bz = 0, c y + az = 0 and b + ay z = 0 pass through a straight line then a + b + c = abc abc abc abc Given cy bz = 0, c y + az = 0 and b + ay z = 0 They pass through a common line of intersection. The vectors (, c, b), (c,, a) and (b, a, ) are perpendicular to the desired line The vectors are coplanar c b c a b a 0 ( a ) + c( c ab) b (a c + b) = 0 a c abc abc b = 0 a + b + c = abc 8. The point of intersection of lines represented by y + + y = 0 is (, 0) (0, ) The lines are y + + y = 0, In usual notations a =, b =, g = f =, c =, h = 0 hf bg gh af Y n =, ab h ab b = / /, =,,, 9. A die is rolled. If X denotes the number of positive divisors of the outcome then the range of the random variable X is {,, } {,,, } {,,,, 5, 6} {,, 5} { Number of the divisiors of outcome}, Set of divisiors = (), (, ), (,), (,, ), (,5), (,,,6) Range of X = {,,, } 0. A die is thrown four times. The probability of getting perfect square in at least one throw is n =, 6 8 p, q P (at least once) = P( X ) P( X 0) = C 0 =

10 0 Kalrashukla Classes. 0 sec d log log log / sec d = 0 / / / tan tan d = tan log sec = log log. In ABC, with usual notations, if a, b, c are in A.P. then a cos a Given a, b, c are in A.P. b = a + c a cos c A c cos = c C A c cos b cos cos ( cos cos ) a C c A a c a C c A = abc a c b b. If = e (sin cos ), y = e (sin + cos ) then dy d at is 0 = (sin cos ) d d = e (cos sin ) e (sin cos ) = e (sin ) y = e (sin cos ) dy d dy d = cot dy d = e (cos sin ) e (sin cos ) = e (cos ). The number of solutions of sin + sin + sin 5 = 0 in the interval, is 5 sin sin 5 sin 0 sin cos sin 6 sin = 0 OR cos = n for n = 0, n =, n =, n =,,

11 MH-CET - 08 OR = m, m m 0,, 5. If tan + tan =, then = tan tan = 5 = 6 or = = 0 6 ( + ) ( + ) = 0 = (reject) or Matri A = 5 then the value of a A + a A + a A is 7 We know that a A + a A + a A = A 5 A = = (0 ) (5 ) + 7 ( ) = 7 7. The contrapositive of the statement: If the weather is fine then my friends will come and we go for a picnic. The weather is fine but my friends will not come or we do not go for a picnic If my friends do not come or we do not go for picnic then weather will not be fine If the weather is not fine then my friends will not come or we do not go for a picnic The weather is not fine but my friends will come and we go for a picnic Given p (q r) Its contrapositive is ~ ( q r) ~ p (~ q ~ r) ~ p p weather is fine, q my friends will come, r we go for a picnic 8. If f() = is increasing function then the value of lies in R (, ) (, ) (, ) ( )() ( ) f() = f () = ( ) < (, ) = ( ) f () > 0 > 0

12 Kalrashukla Classes 9. If X = { n n : n N} and Y = {9 (n ) : n N}, then X Y = X Y {0} X {n n : n N} 0, 9, 5,, 008,... Y {9(n ): nn} {0, 9, 8, 7,...}, X Y X Y = X 0. The statement pattern p (~ p q) is a tautology equivalent to p q p (~p q) = (p ~p) q = f q f a contradiction equivalent to p q. If the line y = 5 touches to the curve y = a + b at the point (, ) then 7a + b = 0 Given y = 5 touches the curve y = a + b y dy = a dy d d = a y dy d (,) (, ) lies on y = + b 9 = 6 + b b = = a 6 = a = 7 7a + b = = 0. The sides of a rectangle are given by = a and y = b. The equation of the circle passing through the vertices of the rectangle is + y = a + y = a + b + y = a b ( a) + (y b) = a + b y = b y = b + y = (a + b ) + y = a + b = a = a

13 MH-CET The minimum value of the function f() = log is e e Given f() = log f ( ) log log, f ( ) 0 e f ( ) = + e > 0 min of f at e e and min. f e log e e e. If X ~ B (n, p) with n = 0, p = 0. the E (X ) =? n = 0, p = 0., q = 0.6 E(X) = np =, Var(X) = npq =. (E(X )) (E(X)) =. E(X ) =. + 6 = The general solution of differential equation d = cos ( + y) is dy y tan y c y tan c y cot y c y cot c d dy = cos ( + y) d dt Put + y = t dy dy t y tan y c tan y c dt dy = cos t or dt dy = cos t t sec dt dy 6. If planes r (pi ˆ ˆj k) ˆ 0 and r (i ˆ pj ˆ k) ˆ 5 0 include angle then the value of p is,, Give angle between (p,, ) and (, p, ) is p p p 5 p 5 6p = p + 5 or p 6p + 9 = 0 (p ) = 0 p =

14 Kalrashukla Classes 7. The order of the differential equation of all parabolas, whose latus rectum is a and ais parallel to the -ais, is one four three two Its equation is (y k) = a ( h), h and k are two arbitrary constants second order differential equation 8. If lines y z y k and z intersect then the value of k is 9 The point of intersection is P (r +, r, + ) (p +, p + k, p) r + = r + r = r + r = P,, 5 5 (, 0 + k, 5) 0 + k = 7 p = r = = 5 k = =.5 = 9 9. If a line makes angles 0and 60with the positive directions of X and Z aes respectively then the angle made by the line with positive Y-ais is Given = 0º, = 60º cos = cos 0º cos 60º = = cos 50. L and M are two points with position vectors a b and a b respectively. The position vector of the point N which divides the line segment LM in the ratio : eternally is b b 5b a b l a b, m a b N is point on LM ratio : is m l = a b ( a b ) = 5b

SUBJECT : PAPER I MATHEMATICS

Question Booklet Version SUBJECT : PAPER I MATHEMATICS Instruction to Candidates. This question booklet contains 50 Objective Type Questions (Single Best Response Type) in the subject of Mathematics..