mht-cet 2016 : mathematics - actual test...
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MHT-CET 2016 : Mathematics - Actual Test Paper
(Solution at the end)
This is a Simulator Test. Do not attempt this Test Paper until you are ready to take this Test under strict
examination conditions.
IMPORTANT INSTRUCTIONS
1. The test is of 1½ hours duration 2. The Test Booklet consists of 50 questions. The maximum marks are 100. 3. Each question is allotted 2 (two) marks for each correct response. 4. No deduction from the total score will be made if incorrect response or no answer is marked. 5. There is only one correct answer for each question.
6. All calculations / writing work are to be done on the, space provided for this purpose in the Test Booklet itself, marked ‘Space for Rough Work’. 7. Use of Electronic / Manual Calculator is prohibited.
8. Use Black Ball Point Pen only for writing particulars / marking responses on the Answer Sheet.
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1. Let X ~ B (n, p). If E(X) = 5, Var(X) = 2.5, then P(X < 1) = ________
(A) 11
1
2
(B) 10
1
2
(C) 6
1
2
(D) 9
1
2
2. Derivative of 1
2
xtan
1 x
with respect to sin1(3x 4x3) is _______
(A) 2
1
1 x (B)
2
3
1 x (C) 3 (D)
1
3
3. The differential equation of the family of circles touching yaxis at the origin is
(A) 2 2 dyx y 2xy 0
dx (B) 2 2 dy
x y 2xy 0dx
(C) 2 2 dyx y 2xy 0
dx (D) 2 2 dy
x y 2xy 0dx
4. If
1 1 0
A 2 1 5
1 2 1
, then a11A21 + a12A22 + a13A23 = __________
(A) 1 (B) 0 (C) 1 (D) 2
5. If Rolle‟s theorem for f(x) = ex (sinx cosx) is verified on 5
,4 4
, then the value of c is
(A) 3
(B)
2
(C)
3
4
(D)
6. The joint equation of lines passing through the origin and trisecting the first quadrant is
(A) 2 2x 3xy y 0 (B) 2 2x 3xy y 0 (C) 2 23x 4xy 3y 0 (D) 3x2 y2 = 0
7. If 2 tan1
(cosx) = tan1
(2cosecx), then sinx + cosx =
(A) 2 2 (B) 2 (C) 1
2 (D)
1
2
8. Direction cosines of the line x 2 2y 5
, z 12 3
are
(A) 4 3
, ,05 5 (B)
3 4 1, ,
5 5 5 (C)
3 4, ,0
5 5 (D)
4 2 1, ,
5 5 5
9. 2
1dx
8 2x x
(A) 11 x 1sin c
3 3
(B) 1 x 1
sin c3
(C) 11 x 1
sin c3 3
(D) 1 x 1
sin c3
10. The approximate value of f(x) = x3 + 5x2 7x + 9 at x = 1.1 is
(A) 8.6 (B) 8.5 (C) 8.4 (D) 8.3
11. If r.v. X : waiting time in minutes for bus and p.d.f. of X is given by
1, 0 x 5
f(x) 5
0, otherwise,
then probability of waiting time not more than 4 minutes is =
(A) 0.3 (B) 0.8 (C) 0.2 (D) 0.5
12. In ABC : (a b)2cos2 2 2C C(a b) sin
2 2 =
(A) b2 (B) c2 (C) a2 (D) a2 + b2 + c2
MHT-CET 2016 : Mathematics Paper (3)
13. Derivative of log(sec + tan ) with respect to sec at = 4
is
(A) 0 (B) 1 (C) 1
2 (D) 2
14. The joint equation of bisectors of angles between lines x = 5 and y = 3 is
(A) (x 5)(y 3) = 0 (B) x2 y
2 10x + 6y +16 = 0
(C) xy = 0 (D) xy 5x 3y +15 = 0
15. The point on the curve 6y = x3 + 2 at which ycoordinate is changing 8 times as fast as xcoordinate is
(A) (4, 11) (B) (4, 11) (C) (4, 11) (D) (4, 11)
16. If the function f(x) defined by
1
f(x) x sin ,x
for x 0
= k, for x = 0
is continuous at x = 0, then k =
(A) 0 (B) 1 (C) 1 (D) 1
2
17. If y = 1m sin xe and
2
2 2dy(1 x ) Ay
dx
, then A =
(A) m (B) m (C) m2 (D) m2
18. If x
x
x
4e 25dx Ax B log | 2e 5 | c
2e 5
, then
(A) A = 5, B = 3 (B) A = 5, B = 3 (C) A = 5, B = 3 (D) A = 5, B = 3
19. 1 1
1 1
tan 3 sec ( 2)
1cosec ( 2) cos
2
(A) 4
5 (B)
4
5 (C)
3
5 (D) 0
20. For what value of k, the function defined by
0
2
log(1 2x) sinxf(x) for x 0
x
k for x 0
is continuous at x = 0 ?
(A) 2 (B) 1
2 (C)
90
(D)
90
21. If 2 2
10 2 2
x y dylog 2, then
dxx y
_______ .
(A) 99x
101y (B)
99x
101y (C)
99y
101x (D)
99y
101x
22. /2
/2
2 sinxlog dx
2 sinx
(A) 1 (B) 3 (C) 2 (D) 0
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23.
12 (x tan x)
2
(x 2)adx
x 1
=
(A) 1x tan xloga a c
(B) 1(x tan x)
cloga
(C)
1x tan xac
loga
(D) log a . (x + tan1x) + c
24. The degree and order of the differential equation
7/33 2
2
dy d y1 7
dx dx
respectively are
(A) 3 and 7 (B) 3 and 2 (C) 7 and 3 (D) 2 and 3
25. The acute angle between the line ˆ ˆˆ ˆ ˆ ˆr (i 2 j k) (i j k) and the plane ˆˆ ˆr (2i j k) 5
(A) 1 2cos
3
(B) 1 2sin
3
(C) 1 2tan
3
(D) 1 2sin
3
26. The area of the region bounded by the curve y = 2x x2 and xaxis is
(A) 2
3 sq.units (B)
4
3sq. units (C)
5
3sq.units (D)
8
3sq.units
27. If f(x)
dx log logsinx c,log(sinx)
then f(x) =
(A) cot x (B) tan x (C) sec x (D) cosec x
28. If A and B are foot of perpendicular drawn from point Q (a, b, c) to the planes YZ and ZX respectively,
then the equation of plane through the points A, B and O is
(A) x y z
0a b c (B)
x y z0
a b c (C)
x y z0
a b c (D)
x y z0
a b c
29. If ˆ ˆ ˆˆ ˆ ˆ ˆ ˆa i j 2k, b 2i j k and c 3i k and c ma nb then m + n = _____
(A) 0 (B) 1 (C) 2 (D) 1
30. /2 n
n n0
sec xdx
sec x cosec x
(A) 2
(B)
3
(C)
4
(D)
6
31. If the p.d.f. of r.v. X is given as
xi 2 1 0 1 2
P(X = xi) 0.2 0.3 0.15 0.25 0.1
then F(0) =
(A) P(X < 0) (B) P(X > 0) (C) 1 P(X > 0) (D) 1 P(X < 0)
32. The particular solution of the differential equation dx
y(1 logx) xlogx 0dy
when x = e, y = e2 is
(A) y = ex log x (B) ey = x logx (C) xy = e logx (D) y logx = ex
33. M and N are the midpoints of the diagonals AC and BD respectively of quadrilateral ABCD, then
AB D CB CD =
(A) 2MN (B) 2NM (C) 4MN (D) 4NM
34. If sin x is the integrating factor (I.F.) of the linear differential equation dy
Py Qdx
, then P is
(A) log sin x (B) cos x (C) tan x (D) cot x
MHT-CET 2016 : Mathematics Paper (5)
35. Which of the following equation does not represent a pair of lines?
(A) x2 x = 0 (B) xy x = 0 (C) y2 x + 1 = 0 (D) xy + x + y + 1 =0
36. Probability of guessing correctly atleast 7 out of 10 answers in a “True” or “False” test is = ______
(A) 11
64 (B)
11
32 (C)
11
16 (D)
27
32
37. Principal solutions of the equation sin 2x + cos 2x = 0, where < x < 2 are
(A) ,8 8
(B) ,
8 8
(C)
15,
8 8
(D)
15 19,
8 8
38. If line joining points A and B having position vectors 6a 4b 4c and 4c respectively, and the line
joining the points C and D having position vectors a 2b 3c and a 2b 5c intersect, then their
point of intersection is
(A) B (B) C (C) D (D) A
39. If A = 2 2 0 1
, B ,3 2 1 0
then
11 1B A
____
(A) 2 2
2 3
(B) 2 2
2 3
(C) 2 3
2 2
(D) 1
2 3
40. If p : Every square is a rectangle
q : Every rhombus is a kite,
then truth values of p q and p q are _________ and _______ respectively.
(A) F, F (B) T, F (C) F, T (D) T, T
41. If G(g), H(h) and C(c) are centroid, orthocenter and circumcenter of a triangle and xc yh zg 0 ,
then (x, y, z) = _____
(A) 1, 1, 2 (B) 2, 1, 3 (C) 1, 3, 4 (D) 2, 3, 5
42. Which of the following quantified statements is true?
(A) The square of every real number is positive
(B) There exists a real number whose square is negative
(C) There exists a real number whose square is not positive
(D) Every real number is rational
43. The general solution of the equation tan2x = 1 is
(A) n4
(B) n
4
(C) n
4
(D) 2n
4
44. The shaded part of given figure indicates the feasible region
then the constraints are
(A) x, y 0, x + y 0, x 5, y 3 (B) x, y 0, x y 0, x 5, y 3
(C) x, y 0, x y 0, x 5, y 3 (D) x, y 0, x y 0, x 5, y 3
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45. Direction ratios of the line which is perpendicular to the lines with direction ratios 1, 2, 2 and 0, 2, 1 are
(A) 1, 1, 2 (B) 2, 1, 2 (C) 2, 1, 2 (D) 2, 1, 2
46. If Matrix 1 2
A4 3
such that Ax = 1, then x = ________
(A) 1 31
2 15
(B)
4 21
4 15
(C)
3 21
4 15
(D)
1 21
1 45
47. If ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆa i j k, b 2i j k and c i j 4k and a (b c) 10 , then is equal to
(A) 6 (B) 7 (C) 9 (D) 10
48. If r.v. X ~ B1
n 5,p ,3
then P(2 < x < 4) = _______
(A) 80
243 (B)
40
243 (C)
40
343 (D)
80
343
49. The objective function Z = x1 + x2 subject to x1 + x2 10, 2x1 + 3x2 15, x1 6, x1, x2 0 has
maximum value _______ of the feasible region.
(A) at only one point
(B) at only two points
(C) at every point of the segment joining two points
(D) at every point of the line joining two points
50.
Symbolic form of the given switching circuit is equivalent to______ (assume S1 = p and S2 = q)
(A) p ~q (B) p ~ q (C) p q (D) ~(p q)
Solution to MHT-CET 2016 : Mathematics - Actual Test Paper
ANSWER KEY
Q. No. Ans. Q. No. Ans. Q. No. Ans. Q. No. Ans. Q. No. Ans.
1. (B) 2. (D) 3. (B) 4. (B) 5. (D)
6. (C) 7. (B) 8. (A) 9. (D) 10. (A)
11. (B) 12. (B) 13. (B) 14. (B) 15. (A)
16. (A) 17. (C) 18. (B) 19. (B) 20 (C)
21. (A) 22. (D) 23. (C) 24. (B) 25. (B)
26. (B) 27. (A) 28. (A) 29. (C) 30. (C)
31. (C) 32. (A) 33. (C) 34. (D) 35. (C)
36. (A) 37. (C) 38. (A) 39. (A) 40. (D)
41. (B) 42. (A) 43. (C) 44. (B) 45. (B)
46. (C) 47. (A) 48. (B) 49. (C) 50. (D)
S1 2S
S2 1S
L
MHT-CET 2016 : Mathematics Paper (7)
DETAILED SOLUTIONS
1. (B)
Here, E(x) = np = 5 and Var(x) = npq = 2.5
2.5 1
q5 2
1
p2
and np = 5 5 5
n 101p
2
Now, P(x < 1) = P(x = 0)
=
0 10 0 10 1010
0
1 1 1 1C 1
2 2 2 2
2. (D)
Let u = 1
2
xtan
1 x
and v = sin1(3x 4x3)
Put x = sin = sin1x
u = 1
2
sintan
1 sin
and v = sin1(3sin 4sin3)
= 1 sin
tancos
and v = sin1(sin3)
= 1tan tan and v = sin1(sin3)
u = = sin1x and v = 3 = 3sin1x
2
du 1
dx 1 x
and
2
dv 3
dx 1 x
2
2
1du
du 11 xdxdvdv 33dx
1 x
3. (B)
As required circle touches yaxis at the origin, let centre of the circle be (a, 0) and radius is a.
Equation of circle will be,
(x a)2 + (y 0)2 = a2
x2 2ax + a2 + y2 = a2
x2 + y2 2ax = 0 …(i)
By differentiating above equation w.r.t. x, we get
dy
2x 2y 2a 0dx
2a = dy
2x 2ydx
…(ii)
From (i) and (ii),
x2 + y2
dy2x 2y x 0
dx
x2 + y2 2x2 2xydy
0dx
2 2 dy
x y 2xy 0dx
x2 y2 + 2xydy
dx = 0
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4. (B)
We know that, aij stands for element in matrix A in ith row and jth column, and
Aij stands for cofactor of element aij of matrix A.
a11 = 1, a12 = 1 and a13 = 0
2 1
21
1 0A ( 1) 1
2 1
2 2
22
1 0A ( 1) 1
1 1
2 3
23
1 1A ( 1) 1
1 2
a11A21 + a12A22 + a13A23 = 1 (1) + 1(1) + 0 (1) = 0
5. (D)
Given, f(x) = ex (sin x cos x)
x x xf (x) e cosx sin x sin x cosx e e cosx sin x sin x cosx
xf (x) 2e sin x
To verify Rolle‟s Theorem,
f (c) 0 c2e sin c 0 sin c 0 c 2 , 3 , ...... [ as ec 0]
c = 5
as ,4 4
6. (C)
As given, both lines pass through (0, 0), and 1 2,6 3
Equation of first line is :
y 0 = tan x 06
1y x
3 3y x
x 3 y 0 …(i)
Equation of second line is :
y 0 = tan x 03
y 3x
3 x y 0 …(ii)
Hence, joint equation of these lines is
(x 3y) ( 3x y) 0
2 23x xy 3xy 3y 0
2 23x 3y 4xy 0
7. (B)
As given, 2tan1(cos x) = tan1 (2cosec x)
tan1(cos x) + tan1(cos x) = tan1(2cosec x)
tan11
2
2cos xtan (2cosecx)
1 cos x
… 1 1 1 A B
tan A tan B tan1 AB
2
2cos x2cosec x
sin x
cos x 1 1cot x 1
sin x sin x sin x
x4
1 1 2sin x cos x sin cos 2
4 4 2 2 2
2nd line
1st line
1 2
y
x
MHT-CET 2016 : Mathematics Paper (9)
8. (A)
The equation of line is
x 2 2y 5
, (z 1) i.e.2 3
52 y
x ( 2) 2, z 1
2 3
, i.e.
5y
x ( 2) 2, z 1
32
2
Hence the line is passing through the point 5
2, , 12
and
direction ratios of the line are 3
2, , 02
i.e. 4, 3, 0
direction cosines of the line are
2 2 2 2 2 2
4 3 0 4 3, , i.e. , , 0
5 5(4) (3) 0 (4) (3) 0 (4) (3) 0
9. (D)
Let 2
dxI
8 2x x
2 2 2 2
dx dx dx
9 2x x 1 9 (x 2x 1) (3) (x 1)
1 x 1
I sin c3
1
2 2
dx xsin c
aa x
10. (A)
f(x) = x3 + 5x2 7x + 9
2f (x) 3x 10x 7
Let a = 1 and h = 0.1
f (a) = f(1) = (1)3 + 5(1)2 7(1) + 9 = 15 7 = 8
2f (a) f (1) 3(1) 10(1) 7 13 7 6
f(1.1) = f(1) + (0.1) f (1)
= 8 + (0.1)(6) = 8 + (0.6) = 8.6
11. (B)
Required probability = P(X 4)
=
4 4
0 0
1 1dx dx
5 5 =
4
0
1x
5 =
14 0.8
5
12. (B)
2 2 2 2C C
(a b) cos (a b) sin2 2
= 2 2 2 2 2 2C Ca 2ab b cos (a 2ab b )sin
2 2
= 2 2 2 2 2 2 2 2C C C Ca b cos 2abcos (a b )sin 2absin
2 2 2 2
= 2 2 2 2 2 2C C C Ca b cos sin 2ab cos sin
2 2 2 2
= 2 2a b 1 2ab cosC
= 2 2 2a b 2abcosC c …[By cosine rule]
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13. (B)
Let y1 = log(sec + tan )
21dy 1sec tan sec
d sec tan
1dy sec [sec tan ]
d [sec tan
1dysec
d
…(i)
Let y2 = sec
2dysec tan
d
…(ii)
1
1
22
dy
dy d
dydy
d
= sec
cotsec tan
1
24
dycot 1
dy 4
14. (B)
We can say both line passes through point (5, 3) and makes angle 45° and 135° with x axis
Equation of 1st line is,
y 3 = tan 45°(x 5) y 3 = x 5
y x + 2 = 0 …(i)
Equation of 2nd line is
y 3 = tan 135°(x 5) y 3 = 1(x 5)
y + x 8 = 0 …(ii)
Required joint equation of line is
(y x + 2) (y + x 8) = 0
y2 xy + 2y + xy x2 + 2x 8y + 8x 16 = 0
x2 + y2 + 10x 6y 16 = 0
x2 y2 10x + 6y + 16 = 0
15. (A)
As required point is on the curve : 6y = x3 + 2, we will go by options.
We find that only point (4, 11) satisfies the given equation.
This problem can be alternatively solved as follows :
We have 6y = x3 + 2 and dy dx
8dt dt
2dy dx
6 3xdt dt
2 2dx dx
6 8 3x 48 3xdt dt
x2 = 16 x = 4
3x 2y
6
3(4) 2y
6
or
3( 4) 2y
6
= 64 2
6
or
64 2
6
y = 11 or y = 62
6
Hence required points are (4, 11) and 62
4,6
x = 5
y = 3
x
(5, 3)
y
45° 135°
2nd line 1st line
MHT-CET 2016 : Mathematics Paper (11)
16. (A)
Since f(x) is continuous at x = 0, we write
f(0) = x 0lim f (x)
x 0
1k lim x sin 0 (finite number) 0
x
k = 0
17. (C)
Given, 1msin xy e
1msin x
2
dy 1e m
dx 1 x
1
22 2msin x 2
2
dy 1e (m)
dx 1 x
= 1
2 22msin x 2
2 2
1 y me m
1 x 1 x
2
2 2 2dy1 x m y
dx
Comparing, we write A = m2
18. (B)
Let I =
x
x
4e 25dx
2e 5
=
x x
x
10e 25 6edx
2e 5
x x
x x
5(2e 5) 6 edx dx
2e 5 2e 5
=
x
x
2e5 dx dx
2e 5
I = 5x 3log(2ex 5 ) + c
Comparing we get, A = 5 and B = 3
19. (B)
Let y = 1 1
1 1
tan 3 sec ( 2)
1cos ec ( 2) cos
2
= 1 1
1 1
tan 3 sec (2)
1cos ec ( 2) cos
2
=
1 1
1 1
1tan 3 cos
2
1 1sin cos
22
=
2
3 3 3 37 5
4 3 12 12
= 12 4
3 5 5
20. (C)
Since f(x) is continuous at x = 0, we write
0
2x 0
log(1 2x) sin xlim k
x
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0
x 0
log(1 2x) sin xlim
x x
= k
c
x 0
xsin
log(1 2x) 180lim k
2x x
x 0 x 0
sin x2 log(1 2x) 180
lim lim k2x 180
x180
2 k180
k
90
21. (A)
Given,
2 2
10 2 2
x ylog 2
x y
2 22
2 2
x y10 100
x y
…[By definition of logarithm]
x2 y2 = 100x2 + 100y2 101y2 = 99 x2
99x2 + 101 y2 = 0
Differentiating, we get
(2 99)x + (2 101)y dy
0dx
dy
99x 101y 0dx
dy
101y 99xdx
dy 99x
dx 101y
22. (D)
Let I =
/2
/2
2 sin xlog dx
2 sin x
Let f(x) = log2 sin x
2 sin x
log(2 sinx) log(2 + sinx)
f(x) = log2 sin( x) 2 sin x
log2 sin( x) 2 sin x
= log(2 + sinx) log(2 sinx)
= [log(2 sinx) log(2 + sinx)]
= f(x)
Hence f(x) is odd function.
I = 0
23. (C)
Let I =
12 x tan x
2
(x 2) adx
1 x
Put 1x tan xa t
1x tan x
2
1a 1 (log a)dx dt
1 x
1
2x tan x
2
1 x 1a (log a)dx dt
1 x
1
2x tan x
2
2 x dta dx
log a1 x
MHT-CET 2016 : Mathematics Paper (13)
dt 1 t
I dt clog a log a log a
1x tan xaI c
log a
24. (B)
We have
7/33 2
2
dy d y1 7
dx dx
Raising both sides to power 3, we get
7 33 23
2
dy d y1 (7)
dx dx
Hence order = 2, degree = 3 degree and order are respectively 3 and 2.
25. (B)
We have line ˆ ˆ ˆ ˆ ˆ ˆr (i 2 j k) (i j k) and plane ˆ ˆ ˆr (2i j k) 5
Standard equation of line and plane are r a b and r n p respectively
ˆ ˆ ˆ ˆ ˆ ˆb i j k and n 2i j k
Let be the required angle.
ˆ ˆ ˆ ˆ ˆ ˆi j k 2i j kb n
sinˆ ˆ ˆ ˆ ˆ ˆ| b | | n | i j k 2i j k
2 2 2 2 2 2
(1)(2) (1)( 1) (1)(1) 2 1 1
3 6(1) (1) (1) (2) ( 1) (1)
= 2
18 =
2 2
33 2
1 2sin
3
26. (B)
Point of intersection of y = 2x x2 and x axis i.e. y = 0 is
0 = 2x x2 x(2x) = 0 x = 0 or x = 2
When x = 0, y = 0 and when x = 2, y = 0
Hence point of intersection are (0, 0) and (2, 0)
Hence required area is
22
0
A (2x x )dx
2 22 2 2 32
0 0 0 0
x xA 2 x dx x dx
2 3
=
23
2
0
xx
3
=
84
3
=
4
3 sq. unit
27. (A)
Given f (x)
dx log logsin x clog(sin x)
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Differentiating both sides w.r.t. x, we get
f (x) d
log(log(sin x)) clog(sin x) dx
f (x) 1 1
cos x 0log(sin x) log(sin x) sin x
f (x) 1
cot xlog(sin x) log(sin x)
f(x) = cot x
28. (A)
A and B are foot of perpendicular from point (a, b, c) on planes YZ and ZX respectively.
A ≡ (0, b, c) and B ≡ (a, 0, c)
We have to find equation of plane passing through A(0, b, c), B(a, 0, c) and O(0, 0, 0)
ˆ ˆ ˆ ˆa bj ck , b ai ck , o 0
Hence required vector equation of plane is
r (AB AO ) a (AB AO )
ˆ ˆ ˆ ˆ ˆ ˆAB b a (ai ck bj ck) ai bj
ˆ ˆ ˆ ˆAO a bj ck) bj ck
ˆ ˆ ˆi j k
AB AO a b 0
0 b c
= ˆ ˆ ˆi(bc) j( ac) k( ab) = ˆ ˆ ˆbc i ac j ab k
a (AB AO ) ˆ ˆ ˆ ˆ ˆ(bj ck) (bc i ac j ab k) abc abc 0
Required equation is ˆ ˆ ˆr (bc i ac j ab k) 0
If ˆ ˆ ˆr x i y j z k , then above equation becomes ˆ ˆ ˆ ˆ ˆ ˆ(x i y j z k) (bc i ac j ab k) 0
bc x + ac y ab z = 0
Dividing throughout by abc, we get
x y z
0a b c
29. (C)
From given data, we write
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ3i k m( i j 2k) n( 2i j k)
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ3i k m i mj 2mk 2ni nj nk
ˆ ˆ ˆ ˆ ˆ3i k m 2n) i (m n) j (n 2m)k
Comparing, we get
m+2n = 3 …(i)
m n = 0 …(ii)
n 2m = 1 …(iii)
Solving, we get m = 1 and n = 1 m + n = 2
30. (C)
Let I =
/2 n
n n0
sec xdx
sec x cosec x
…(i)
I =
n/2
0 n n
sec x2
dx
sec x cosec x2 2
…(ii)
MHT-CET 2016 : Mathematics Paper (15)
=
/2 n
n n0
cosecxdx
cosecx sec x
Adding equation (i) and (ii), we get
2I =
/2 n n
n n0
sec x cosec xdx
sec x cosec x
2I =
/2
0
dx
2I = /2
0x
2I
2
I =
4
31. (C)
F(0) = P(X 0)
= P(X = 2) + P(X = 1) + P(X = 0)
= (0.2) + (0.3) + (0.15) = 0.65
P(X>0) = P(1) + P(2)
= 0.25 + 0.1 = 0.35
F(0) = 1 P(X > 0)
32. (A)
We have dx
y(1 log x) x log x 0dy
dx
y(1 log x) x log xdy
1 log x dy
dxx log x y
Integrating both sides, we get
1 log x dy
dxx log x y
In LHS, put xlogx = t
x
log x dx dtx
(1 log x) dx dt
Given equation becomes
dt dy
t y
log t = log y + log c
log(xlog x) = log y + log c log(xlog x) = log(y . c)
xlog x = y . c …(i)
As x = e, y = e2
e log e = e2 . c e(1) = e2 . c c = 1
e
Putting 1
ce
in eq.(i) we get
y
x log xe
i.e. y = ex log x
33. (C)
AB AD CB CD
= (AM MN NB) (AM MN ND)
(CM MN NB) (CM MN ND)
A B
D
N M
C
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= 2AM 4MN 2NB 2ND 2CM
= 4MN 2(AM CM NB ND)
= 4MN 2(AM MC NB DN)
= 4MN ...[AM MC and NB DN , as M & N are mid points of AC & BD resp. given]
34. (D)
It is given that Integrating factor of dy
Py Qdx
is sin x
Pdx
e sin x P dx ln(sin x)
Differentiating both sides w.r.t. x, we get
d
P ln(sin x)dx
P = 1
cos xsin x
P = cotx
35. (C)
We will go by options
Option A :
x2 x = 0 x(x1) = 0 x = 0 and x = 1 are two lines
Option B :
xy x = 0 x(y 1) = 0 x = 0 and y = 1 are two lines
Option D :
xy + x + y + 1 x(y + 1) + 1(y + 1) = 0 x + 1 = 0 and y + 1 = 0 are two lines
Option C :
y2 x + 1 = 0 y2 = x 1 This is equation of parabola
Alternatively this problem can be solved as follows :
When ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of straight lines, then
abc + 2fgh af2 bg2 ch2 = 0
Students may check all options by using this formula.
36. (A)
Let „p‟ be the probability of guessing correctly in a „True‟ or „False‟ test.
1
p2
1 1
q 12 2
There are 10 questions n = 10
From given condition, we write
P(x 7) = P(x = 7) + P(x = 8) + P(x = 9) + P(x = 10)
=
7 3 8 2 9 1 1010 10 10 10
7 8 9 10
1 1 1 1 1 1 1C C C C
2 2 2 2 2 2 2
=
1010 10 10 10
7 8 9 10
1C C C C
2
= 1
120 45 10 11024
176 22 11
1024 128 64
37. (C)
We have, sin2x + cos2x = 0
Multiplying by 1
2 on both sides, we get
1 1
sin 2x cos 2x 02 2
sin 2x cos cos 2x sin 04 4
MHT-CET 2016 : Mathematics Paper (17)
sin 2x 04
sin 2x sin(0)
4
2x n4
2x = n
1 4n 1n
4 4 4
(4n 1)
x8
11
x8
and
15
8
….[ < x < 2] (Putting n = 3, 4 gives required answer)
38. (A)
Vector equation of line passing through A and B is
r = (6a 4b 4c) [( 4c) (6a 4b 4c)]
= (6a 4b 4c) ( 6a 4b 8c)
xa yb zc = (6 6 )a ( 4 4 )b (4 8 )c
Comparing, we get
x = (6 6), y = (4 + 4), z = (4 8)
Thus, coordinates of any point on line AB are
[(6 6), (4 + 4), (4 8)] …(i)
Similarly vector equation of line passing through C and D is
r = ( a 2b 3c) [(a 2b 5c) ( a 2b 3c) ]
( a 2b 3c) 2a 4b 2c]
xa yb zc = (1 + 2) a ( 2 4 ) b ( 3 2 )c
Comparing, we get
x = (1 + 2), y = ( 2 4 ), z ( 3 2 )
Thus coordinates of any point on line CD are
[(1 + 2), ( 2 4 ), ( 3 2 )] …(ii)
It is given that lines AB and CD intersect.
From equation (i) and (ii), we write
6 6 = 1 + 2 6 + 2 = 7 …(1)
4 + 4 = 2 + 4 4 4 = 2 …(2)
4 8 = 3 2 8 2 = 7 …(3)
Solving equation (1) and (2), we get = 1 and 1
2
These values of and satisfy equation (3)
Hence point of intersection is
[(66), (4 + 4), (4 8)] = (6 6, 4 + 4, 4 8) i.e. (0, 0, 4) which is point B.
Students may get the answer by using value of .
39. (A)
1 1 1 1 1 1 1(B A ) (A ) (B ) AB
1 1 1 2 2 0 1
(B A )3 2 1 0
0 2 2 0 2 2
0 2 3 0 2 3
40. (D)
p : Every square is a rectangle : Truth value of statement p is T
q : Every rhombus is a kite : Truth value of statement q is T
p q ≡ T T ≡ T
p q ≡ T T ≡ T
(18) CET Score Booster : Mathematics
41. (B)
We know that centroid, orthocenter and circumcentre of a triangle are collinear and distance between
centroid and orthocenter is twice the distance between centroid and circumcentre.
H : Orthocentre
G : Centroid
C : Circumcentre
GH = 2 GC
Thus G divides segment HC in the ratio 2 : 1
2c h
g 3g 2c h i.e. 2c h 3g 02 1
We have xc yh zg 0 …(given)
Comparing, we get x = 2, y = 1, z = 3
42. (A)
By fundamental concepts about real numbers, we find that only option (A) is correct.
43. (C)
tan2x = 1
tan x = 1 tan x = 1 or tan x = 1
tan x = 4
or tan x =
3
4
x = n4
44. (B)
By fundamental concepts of LPP, the inequalities of feasible region are as follows :
Line OC : x = y x y 0
Line AB : x = 5 x 5
Line BC : y = 3 y 3
Feasible region is in 1st quadrant x 0, y 0
45. (B)
Let direction ratios of the required line be a, b, c
1a + 2b + 2c = 0
0a + 2b + 1c = 0
a b c
2 2 1 2 1 2
2 1 0 1 0 2
a b c
2 4 1 0 2 0
a b c
2 1 2
a b c
2 1 2
Hence direction ratios of the required line are 2, 1, 2
46. (C)
Given : Ax = I A1Ax = A1 I Ix = A1
x = A1
We have 1 2
A4 3
| A | = 3 8 = 5
and adj A = 3 2
4 1
1 3 21
A x4 15
3 21x
4 15
H G C
D
F E
MHT-CET 2016 : Mathematics Paper (19)
47. (A)
1 1 1
a (b c) 2 1 10
1 1 4
1(4 + 1) 1(8 1) + 1(2) = 10
4 + 1 7 2 = 10 3 9 + 1 = 10 3 = 18 = 6
48. (B)
We have, n = 5, p = 1
3 q = 1 p =
1 21
3 3
P(2 < x <4) = P(x = 3)
=
3 25
3
1 2C
3 3
= 5 4 1 4
2 27 9
=
40
243
49. (C)
First we draw lines AB, CD, EF whose equations are x1 + x2 = 10, 2x1 + 3x2 = 15, x1 = 6 respectively
Line Equation Point on X axis Point on Y axis Same/Opposite side of origin
AB 1 2x x 10 A(10, 0) B(0, 10) Same
CD 2x1 + 3x2 = 15 C(7.5, 0) D(0, 5) Same
EF x1 = 6 E(6, 0) Same
Point of intersection of AB and CD is P(3, 7)
Point of intersection of AB and EF is Q(6, 4)
Point of intersection of CD and EF is R(6, 9)
The feasible region is shown shaded in figure.
The vertices of feasible region are O(0, 0), D(0, 5), P(3, 7), Q(6, 4) and E(6, 0).
The values of objective function Z = x1 + x2 at these vertices are as follows.
Z(O) = 0 + 0 = 0
Z(D) = 0 + 5 = 5
Z(P) = 3 + 7 = 10
Z(Q) = 6 +4 = 10
Z(E) = 6 + 0 = 6
Z has maximum value at P and Q. Hence Z has maximum value at every point of the segment joining P
and Q.
50. (D)
We have, S1 ≡ p and S2 ≡ q
The symbolic form of given switching circuit is
(p ~q) (~p q) ≡ ~(p q) …[By fundamental concept]
7 8
C
O E 6 10
A
B
R
F
D
y
10
Q
x
5
P