mht-cet – 5 (mathematics) mathematics test-05 mht-cet maths... · 2018. 4. 26. · mht-cet – 5...
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MHT-CET – 5 (MATHEMATICS)
CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI # 1
MATHEMATICS 1. (C)
Let P a cos , bsin be any point on the ellipse 2 2
2 2
x y 1a b
Equation of tangent is
x cos ysin 1
a b
It meets the coordinate axes at A and B
a bA ,0 .B 0,cos sin
Mid – point of AB is
a b
2cos 2sin
a bcos sin2 2
Squaring and adding
2 2
2 22 2
a bcos sin 14 4
Locus of , is
2 2
2 2
a b 14x 4y
2 2 2 2 2 2a y b x 4x y 2. (A)
p q p q ~ p (p q) ~ p T T T F T T F T F T F T T T T F F F T T
It is clear that p q ~ p is a tautology. 3. (C)
q p q → p p → (q → p) p q p (p q) T T T T T T T F F T T T F T T T T T F F T T F T
∴Statement p →(q → p) is equivalent to p p q . 4. (B)
Let three numbers be a ,a,arr
Then according to the question
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a a ar 3375r
3a 3375 a 15 (i)
a a ar 65r
21 r ra 65
r
2r r 1 65 13
r 15 3
23r 3r 3 13r 23r 10r 3 0 23r 9r r 3 0 3r 1 r 3 0
1r or33
5. (C)
Given, 2 1 3 x 91 3 1 y 43 2 1 z 10
It can be written as AX = B Now, A 2 3 2 1 1 3 3 2 9 7
1
5 7 8 5 7 81adj A 4 7 5 A 4 7 57
7 7 7 7 7 7
From Eq. (i) 5 7 8 9
1x 4 7 5 47
7 7 7 10
1
AX BX A B
x 7
1y 147
z 21
x 1, y 2 and z 3 6. (A) The given equation can be rewritten as
2 11 cos cos4
2 3cos cos 04
24cos 4cos 3 0
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4 16 48 4 8 1 3cos ,8 8 2 2
Since, 3cos2
is not possible, so we take
1cos cos 2n2 3 3
…….. (i)
For the given interval, put n = 0 and n 1 in Eq. (i), we get
5,
3 3
7. (B)
Given, 1 1sin x cos x 12 2
cos 14
2n 14
2n4
32n4
8. (D) Here, a 6,2h 1,b 4c
Now, 1 2 1 22h 1 a 6m m ,m mb 4c b 4c
One line of given pair of lines is 3x 4y 0
Slope of line 13 m4
[say]
23 1m4 4c
21 3m4c 4
And 3 1 3 64 4c 4 4c
6 41 3c3
3c 9 c 3 9. (C) (1) c a b c a b …….. (i) a b c a b c ………. (ii) Since a, b, c are in A. P.
a cb a c b 2b a c b
2
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Now a c 2a a c a ca b a
2 2 2
……… (iii)
And a c a c 2c a cb c c
2 2 2
………. (iv)
From (iii) & (iv) a b b c b c , c a , a b are in A. P.
(2) If a b c, ,
abc abc abcare in A. P.
Then a, b, c are in A. P.
1 1 1, ,bc ac ab
are in A. P.
(3) a b c, ,
abc abc abcare in A. P.
Then 1 1 1, ,bc ac ab
are in A. P.
So, 1 1 1,ab, bc ac
are not in A. P.
(4) 1 1 1, ,b c c a a b
are in A. P.
If 1 1 1 1c a b c a b c a
b a c bc a a c a b c a
b a c bb c a b
b a c b 2b c a i.e. a, b, c are in A. P., which is true.
1 1 1, ,b c c a a b
are in A. P.
10. (A) Given that, , be the angle between the lines represented by equaton 2 23x 7xy 4y 0
1 1
492 3 4 14tan tan3 4 7
…….. (ii)
And 2 be the angle between the lines represented by equation 2 26x 5xy y 0
2 2
252 6 14tan tan6 1 7
……. (ii)
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From Eqs. (i) and (ii), we get 1 2tan tan 1 2 11. (C) Given that, a i j k, a.b 1 And a b j k As, we know a a b a.b a a.a b
i j k j k i j k 3b a.a 1 1 1 3
2i j k i j k 3b 2i j k i j k 3b
3b 3i b i 12. (B) R 3,3 , 6, 2 , 9,1
Range = 3, 2,1 1, 2,3 13. (A)
To determine the angle between a and b, we can use the formula a.bcosa b
It is given that a 3, b 2 and a.b 6 Let be the required angle, then
a.b 6 1cosa b 3 2 2
1 11cos cos cos4 42
14. (C) 2 2 2sin sin sin 2 2 21 cos 1 cos 1 cos
2 2 23 cos cos cos 3 1 2 15. (A or B) Given, 1 1 1a ,b , c p,1, 2 And 2 2 2a , b , c 2, p, 1
1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2
a a b b c ccosa b c a b c
2 2 2 2 2 2
p 2 1 p 2 11cos3 2 p 1 2 2 p 1
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2 2
1 3p 22 p 5 p 5
2
1 3p 22 p 5
2p 6p 1 0
26 6 4 1 6 32p2 1 2
6 4 22
3 2 2 16. (A) The given equation of line is
5yx 2 2y 5 x 2 z 12, z 1 32 3 2 02
So, its vector equation is
5 3r 2i j k 2i 0k2 2
Hence, x = 0 17. (C) R 3, 2 , 5, 2 , 5, 4 Range = {2, 4} 18. (B) Since, the line passing through the points (4, -1, 2) And 3, 2,3 . So, the DR’s the line is 4 3, 1 2, 2 3 i.e. 7, 3, 1 Since, the line is perpendicular to the plane therefore, DR’s of this line is proportional to the normal of the plane. Required equation of plane is 7 x 10 3 y 5 1 z 4 0 7x 3y z 89 0 19. (B)
1x y 2 z 121 2 a
If any line is parallel to the plane, then 1 2 1 2 1 2a a b b c c 0 Here DR’s of a plane 1 1 1a , b , c 2, 1,1 And DR’s of a line 2 2 2a , b , c 1, 2, a
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2 1 2 1 a 1 0 a 4 20. (C)
Time taken to cover x km x h25
Time taken to cover y km y h
40
Since, he has only one hour.
x y 125 40
Total cost of petrol 2x 5y Since, he has only Rs. 100 to spend 2x 5y 100 Since, and y are distances, so x and y cannot be negative. x 0, y 0 Hence, inequalities represent the given data are
x y2x 5y 100, 1, x 0, y 025 40
21. (C) When inequality 2x y 8 At 0,0 ,0 0 8 (not true) So shaded region is away from the origin. When inequality x 2y 10 . At 0,0 ,0 0 10 (not true) So shaded region is away for the origin. 22. (D)
We know 1 1sin x cos x2
We have 1 1 bbsin x bcos x2
And 1 1a sin x bcos x c Adding the above 2 equations.
1 1 bbsin x a sin x c2
1 ba b sin x c2
1b c
2sin xa b
Similarly 1a c
2cos xa b
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1 1 ab c a b
a sin x bcos xa b
23. (D) Since, f x is continuous at x = 0 LHL RHL f 0 ……. (i)
Now, 2
2 2x 0 x 0 x 0
1 cos 4x 2sin 2xlim f x lim limx x
2
2
x 0
sin 2xLHL lim 8 8 1 82x
x 0 x 0
xRHL lim f x lim16 x 4
x 0
x 16 x 4lim
16 x 4 16 x 4
x 0
x 16 x 4lim
16 x 16
x 0lim 16 x 4 8
From Eq (i), 8 8 a a 8 24. (D)
Given, x ,if x 1
f x cx k, , if1 x 42x ,ifx 4
Since, f x is continuous everywhere.
x 1lim f x f 1
and x 4limf x f 4
x 1lim cx k 1
and x 4lim cx k 2 4
c k 1 and 4c k 8 On solving, we get c 3,k 4 25. (A) Let y tan x On differentiating both sides w.r.t. x, we get
dy 1 d tan xdx dx2 tan x
21 1sec x2 x2 tan x
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2sec x
4 x tan x
26. (B)
Give, 1 1sin x sin y2
1 1sin x sin y2
1 1sin x cos y
1 2 1cos 1 x cos y On differentiating both sides w.r.t. x, we get
2
dy 1 x2xdx y2 1 x
27. (C) 1 1 1 1y tan x cot x sec x cos ec x
dy 0
2 2 dx
28. (C) Let u sin x3 and v cos x3 On differentiating both functions w.r.t. x, we get
3 2du cos x .3xdx
and 3 2dv sin x .3xdx
2 3
32 3
du du dx 3x cos x cot xdv dv dx 3x sin x
29. (B) Given, y xy log y ylog x
1 dy 1 dy. y. .log xy dx x dx
2dy 1 y dy ylog xdx y x dx x 1 y log x
30. (B)
Given, sin yx
sin a y
On differentiating both sides w.r.t. y, we get
2
sin a y cos y sin ycos a ydxdy sin a y
2
sin a y ysin a y
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2sin a ydydx sin a
31. (A) Draw BP AC , the longer diagonal. In right angled triangle APB, we have
oAP cos30AB
o 3AP ABsin 30 10 5 3cm2
Also, oBP sin 30AB
o 1BP ABcos30 10 5cm
2
In 2 2 2BPC, BP PC BC 2 2 25 PC 6 2 2 2PC 6 5 11 AC AP PC 5 3 11 cm
option (a) is correct 32. (C) They will encounter if 2 210 6t 3 t t 6t 7 0 t 7 At t 7s, speed of first point 1v
1d 10 6t 6cmsdt
At t 7s , speed of second point 2v
2 1d 3 t 2t 2 7 14cmsdt
Resultant velocity 12 1v v 14 6 8cms
33. (B)
2
1 dxdx1 cos ax 2cos ax 2
2 tan ax 21 ax 1sec dx .2 2 2 a 2
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1 axtan Ca 2
34. (A)
Let
4
3 24
1 xI dx1 x
33 3
3 2 3 23 2 2
2 2
1 1x x xx xdx dx
1 1x x xx x
Put 22
1 x zx
3
1 dzx dxx 2
1 23 2 4
1 dz xI z C C2 z 1 x
35. (A)
Put 12
dxtan x t dt1 x
12
tan x t 22
1 x xe dx e tan t sec t dt1 x
1t tan xe tan t C xe C
36. (C)
4 3 4 3 3 3 2
x 1 xdx dx 1 1dx dxx x x x x x 1 x x x 1
3 2
1 1 1 dxx x x x 1
3 2
1 1 1 1 dxx x x x 1
2
1 1 log x log x 1 C2x x
2
1 1 xlog C2x x x 1
1A and B 12
37. (B)
2
2
1 1 1dx dx4 99x 4x x x
4
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2 2
2
1 1 dx2 9 9 9x x
4 8 8
2 2
1 1 dx2 9 9x
8 8
1 1
9x1 1 8x 98sin C sin C92 2 98
38. (C) Let 17 4f x x cos x f x is an odd function.
1 17 4
1x cos x dx 0
39. (A)
Let a xf x loga x
a xf x loga x
a xloga x
f x is an odd function.
10
10f x dx 0
40. (A)
Given pair of lines 2 2x y 2xy 0
p q h
OR 2 21 1 1x y 2 xy 0P q h
1 2 1 2
22q qhm m m m1 h p
q
2 1 12qm 2m given 3mh
21 2
q 4q q2m 2p 9h p
2
pq 9h 8
2pq : h 9 :8
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41. (A) Given xy 2x 2y 4 0 And x y 2 0 From eq. (i) and (ii) we get xy 0 x 0 y 0 Vertices of triangle 2,0 0,0 and 0, 2 In a right angled triangle circumcentre is mid point of hypotenuse. 42. (B)
Intersection points of both the curves are (0, 0) and 2
16 16,m m
. Therefore, required area
216 m
0
216x mx dx3
216 m2
3 2
0
2 mx 24, x3 2 3
3 2 2
2 2
8 16 m 16 23 m 2 m 3
3 3 3
8 64 16 16 1 512 256 2.3 m 2m m 3 2 3
3 128 3m 643 2
m 4 43. (B) Given, 2 2x y 2ax 0 On differentiating w.r.t. x, we get 2x 2yy' 2a 0 a x yy' On putting the value of a in Eq. (i), we get 2 2x y 2x x yy ' 0
2 2y x 2xyy ' 44. (D) Given differential equation can be rewritten as
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x x
x x
ydy e dx 1 e1 dy dxy 1 e 1 y 1 e 1
xy log y 1 log e 1 log C [Integrating both sides]
xe 1 y 1
y logC
x ye 1 y 1 Ce 45. (C)
PQ lines x 2y 1 Equation of PQ
1y 5 x 32
2y 10 x 3 x 2y 7 …….. (1) Solving 2x 3y 14 and x 2y 7 0 We get B 1,4
2 2PQ 3 1 5 4 5 46. (B) Equation of family of parabolas with focus at (0, 0) and X – axis as axis is 2y 4a x a ….. (i) On differentiating Eq. (i), we get
12yy 4a 1dyHere, ydx
On putting the value of a in Eq. (i),
2 11
yyy 2yy x2
21 1y 2xy yy
2dy dyy 2x y
dx dx
47. (A) We know that sum of pdf is one 2 2 20 k 2k 2k 3k k 2k 7k k 1 210k 9k 1 0 10k 1 k 1 0
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1k k 1
10
P x 3 P x 0 P x 1 P x 2
30 k 2k 3k
10
P x 6 P x 7
2 7 1 177k k100 10 100
P 0 x 3 P x 1 P x 2
3k 2k 3k
10
48. (B) The equation of line through 1 2A L L 0 2x 3y 1 ax by 1 0
It passes through 0 0,0 1 So 2 a x 3 b y 0 Side AD BC
2 a 1 13 b 2
a 2b 8 …… (2) Similarly BE AC a b 0 ……… (3) b 8 and a 8 Hence a, b is 8,8 49. (B)
Given, n = 4 and 16P X 081
Let p be the probability of success and q be the probability of failure in a trial.
Then 16P X 081
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4 o 40
16C p q81
4
4 2 2q q3 3
1p3
4
4 4 o 44
1 1P X 4 C p q p3 81
50. (B) The probability of suffering of a disease is 10%.
10 1p
100 10 and
9q10
Total number of patients, n = 6 Required probability
3 3
63
1 9C10 10
6.5.4 1 9 9 93.2.1 1000 1000
55
2 729 1458 1010