MATHEMATICS SOLUTION
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1 MATHEMATICS SOLUTION MHT-CET 6 (MATHEMATICS). (A) (A) If the school bus does not come) (I will not go to school) ( I shall meet my friend) (I shall go out for a movie) ~ p ~ q r s ~ p q r s [De Morgan s law]. (C) p : is an even prime number q : 6 is a divisor of and r : HCF of and 6 is ~ p q r is true.. (B) C centre of the balloon, r = radius, O eye of the observer. POQ By geometry, POC QOC COH Height of the centre h CH OC sin CP cosec sin r cosec sin h r cosec sin...() CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI #
2 5. (C) A.(cos sin ) Now, A adja adja A 6. (B) We have sin x 0 cos x 6 0 cos x cos x 0 cos x or x n cos cos x [ cos x ] 7. (B) Let OAP be the tree. When broken by the wind at A, let its top P strike the ground at M so that OM = 0 m. OMA 0, AP AM in rt. d AOM, we have, 0 OA tan 0 OM 0 OA OM tan 0 0 m AM AM Also, sec0 OM 0 0 AM m Height of the tree = OP = OA + AP = OA + AM m Height 0 m..() CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI #
3 8. (B) Let y mx and y mx be the lines represented by the given equation 9. (A) 0. (B) h a Then, m m and mm b b We have m : m : m m h b h a m and m b b h a m and m b b m m and mm h The equation h a ab b b x hxy y 0 a b Will represents a pair of real and distinct lines if h h 0 0 Which is true for all h R. The distance between the parallel straight lines given by by g ac d a a b Here, a = 8, b =, c = 5 and g = Required distance units (C) We have sin (x x) sin x (x x) n ( ) x either x x m x x m z Where K is odd OR x x k x x x k 0 n 8K x For least positive non-integral solution, ax hxy by gx fy c 0 is given CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI #
4 8K q x, where q is an odd integer.. (D) Given, a, b and c are three non-coplanar vectors and p, q and r defined by the relations b p c, q c a a b and r abc abc abc b c abc c a abc a b c a p a abc And. (D) a ca a q a 0 abc Similarly, b q cr and a r b q cq b r 0 a bp b cq c a r a p b p bq cq c r a r We have, a ˆi ˆj kˆ And b i ˆ a i ˆ ˆ ja ˆ j k ˆ a k ˆ a a a ˆi j ˆ kˆ b (A) 5. (D) Let a ˆi ˆj k, ˆ b i ˆ j ˆ kˆ And x ˆi ˆj kˆ a b c CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI #
5 6. (A) 7. (A) Vectors are collinear tan h ab a b y cosec x 5 8 dy 7 cosec x. cos ecx cot x. x dx x cosec 5 x cosec x.cot x 8. (D) The direction ratios of the given line are, 7,. The given line is parallel to i ˆ 7ˆj kˆ From the cartesian equation of the given line, it is clear that it passes through the point The vector equation of the line is r 5i ˆ j ˆ 6kˆ i ˆ 7j ˆ kˆ 9. (A) If l, m, n are the direction cosines of the line, then l m n 0 And l m 0n 0 l m n 0 0 Hence, the direction ratios of the line are,, 0. (A) q Short cut. p ,,6. (A) Let the DR s of a required line be a, b and c. Since, the normal to the given planes x y z and 6x 8y z are perpendicular to the line. a b c 0 CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI # 5
6 a b c a b c a b c Or Also, line passes through (,, ). Equation is x y z. (C) Equation of any plane passing through given line is a x by cz 0... Above plane is perpendicular to the plane x y z a b c 0 Also, normal to the plane is perpendicular to the line a b c 0 a b c 8 a b c 9 5 9x y 5z 0 9x y 5z 0 a 9, b, c 5. (C) Direction ratios of given planes are a, b, c and a, b, c cos 6 cos 7. (A) In a right < 5. (A) cos cos cos CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI # 6
7 6. (C) We want to check the continuity of the function at x = For this, we have to prove that LHL = RHL = f() f h LHL lim f x limf h lim x h0 h0 RHL lim f x lim f h x h0 lim h h h0 Thus, LHL = RHL = f() = Hence, function is continuous at x =. 7. (C) Put x h as x, h 0 8. (B) Given limit h tan cos h lim h0 h h tan h h sin lim tan h0 8h h h tan sin lim tan h0 h h h h tan tan lim x0 h h We have, f x cos x..(i) On differentiating Eq. (i) w.r.t.x, we get f ' x sin x cos x cos x f ' x (x) sin x cos x x..(ii) CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI # 7
8 Put x in Eq. (ii) we get sin sin f ' (B) 0. (A) Given, x y y Applying componendo and dividendo, we get y y x x y y x x y x y x On differentiating w.r.t. we get x x x x x x x x x x x x x dy dx x y mx b m is a tangent to the circle x y b for all values of m. If it also touches second circle x a y b then the length of perpendicular from its centre (a, 0) on this line is equal to the radius b of circle ma b m m b taking positive value m ma b m b b m a b a b. (A) x sin a.cot (a + y) Now, differentiating w.r.t. y, we get dx sin a.cos ec a y dy CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI # 8
9 dy sin (a y) dx sin a. (D) Given, x t, y t On differentiating w.r.t. t, we get dx dy and t dt dt dy dy dt t dx dx t dt. (C) Given x xy y 7 On differentiating w.r.t. x, we get dy dy x x y y 0 dx dx dy x y dx x y dy dx, 5. (C) As the curve crosses Y axis i.e. x = 0 0 y e y Given, x y e dy e e dx dy 0 e dx 0, x x Equation of tangent at (0, ) is y x 0 x y 5. (C) If X and Y are number of heads obtained by A and B, then : X B( n, p / ) and Y B( n, p / ). P( x) P( y) C r chance that both get the same no. of heads P( X 0) P( Y 0) P( X ) P( Y ) P( X ) P( Y ) P( X ) P( Y ) CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI # 9
10 [ P( X 0)] [ P( X )] [ P( X )] [ P( X )] [( C0) ( C) ( C) ( C ) ] 0 5 [ 9 9 ] (A) x x x x x 7 x x x dx x x x x dx C (C) sin x cos x dx Let sin x t dt cos x dx dt dx cos x sin x cos x dx dt t cos x cos x t dt 8. (A) / t / C t C sin x C Given, log log x dx x f x g x C log x log log x dx dx II I log x LHS x log log x dx dx log x / log x x x log log x dx dx C log x log x log x x log log x C log x f x loglog x, g x log x CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI # 0
11 9. (D) Let 0. (B) x dx dx x x dx x x I x x x x x x x x I log tan C x x But x a, b Let I a log b tan C I dx dx x x dx 5 x x Put t dx dt 5 x x x x x t t / / I f dt C t C / / C x x / C x. (C) Let / / sin x I dx... i 0 I / / sin x cos x / 0 / / sin x cos x / sin x dx a a f xdx f a xdx 0 0 CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI #
12 0 x / / cos I dx... ii / / cos x sin x Adding Eqs. (i) and (ii), we get / I dx I 0. (C). (B). (A) Required area (shown in shaded region) 8 8 5x 7 ydx dx 8 5x x = 96 sq. units CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI #
13 5. (A) x y... i x y... ii Given curve is And For intersection point put the value of y from eq. (i) in Eq. (ii), we get x x x x 0 x x 0 x, When x, When x, y then from Eq. (ii), we get y y y then from Eq. (ii) we get The line meets the parabola at the points B, A,. and Required area = (Area under the line x = y ) (Area under the parabola x x dx dx x x x sq. units 8 8 Therefore, required area is 9 sq. units. 8 x y ) 6. (B) Let the numbers be a, ar, ar, where r > Then, a, ar, ar are in A.P. ar a ar ( a, b, c are in A.P. b = a+) CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI #
14 6r r r 6r 0 r ( 6) ( 6) 7. (A) Given, x x x y Ae Be Ce... i y Ae Be Ce From Eq. (i), x x x x x x Ae y Be Ce x x y y Be Ce.. ii x y y y Ce... iii Again, differentiating w.r.t., we get x y y y 6Ce From Eq. (iii), x Ce y y y y y y y y y y 6y y 6y 0 8. (D) The given distribution will be a probability distribution, if PX 0 PX PX PX k k k k k 8 CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI #
15 9. (B) p th, q th and n th terms are in G.P. aq ar a a () p q using a n = a + (n ) d, a (q )d a (r )d a (p )d a (q )d By subtracting from both sides (q p)d (r q)d a (p )d a (q )d q p r q a (p )d a (q )d q p r q a a p p a : a (q p) : (r q) q q () becomes: 50. (A) Let the probability of success and failure are p and q respectively. p q and p q q q and Required probability p C C5 C CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI # 5
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