BOUDDHA MERIDIAN SCHOOL
FIRST TERMINAL MODEL QUESTION – 2016/17
Class: IX Subject: Opt. Maths F.M.: 100
Time: 3 hours P.M.: 40
Attempt ALL the questions.
Group ‘A’ [8 (2 + 2) = 32]
1. a) If (x + 5, 5) = (3, – 2 – y), then find the value of x and y.
b) If f'(x) = 3x – 10, find the value of f'(–3) and f'.
2. a) Find the pre-image of 8 in the function f'(x) = . b) If x = 7 + 4, find the value of .
3. a) Simplify: b) Solve: = 2
4. a) Construct a 2 3 matrix where aij = j – i.
b) Define identity matrix with example.
5. a) If AT = and B = , find (A – B)T.
b) Find a, b and c if 2 = .
6. a) Define polynomial. Give an example of trinomial of degree 3.
b) If p(x) = x2 + x + 1 and r(x) = 2x3 – 7x – 3, then find the degree
of r(x) – p(x).
7. a) If the end points of a diameter of a circle are (1, –6) and (3, 4) find the centre of the circle.
b) Find the ratio in which the line segment joining the points (–5, 2) and (2, 3) is divided by the Y-axis.
8. a) If the slope of the line passing through the points (4, 5) and (6, k) is 5, find the value of k.
b) If the slope of a line is 1, find its inclination.
Group ‘B’ [17 4 = 68]
9. If g = {(1, a), (2, b), (3, c), (4, d)} is a relation find its inverse. Represent g–1 in arrow diagram and table.
10.Write down the set of ordered pair of relation r defined as:
r = {(x, y) : y = 4x – 2, x whole number less than 3}
11.If f'(x) = ax + b, f'(2) = 2 and f'(–3) = 12, then find the value of a and
b.
12.If f'(3x + 1) = f'(x) + f'(1), xR, prove that f'(0) = 0 and
f'= – f'(1). 13.If A = , find:
a) The order of matrix A
b) The elements for a12 and a23
c) i, j for aij = 0 and aij = 6
14.If A = , B = and C = , find a matrix X such that 4A + 2B – 3X = C.
15. If AT = and B = , prove that (A + B)T = AT + BT.
16.Arrange in ascending order: , and
17.Simplify: + –
18.If x = 3 + , find x2 + .
19.Solve for x: = 2 +
20. Solve for x: − = 2
21.If p(x) = 3x3 + 4x2 + 2, q(x) = 5x3 – 4x + 7 and r(x) = 6x2 – 4x + 7, then
find p(x) – [q(x) + r(x)] and write its degree also.
22.Prove that the points (2, 3), (3, –1) and (4, –5) are collinear.
[1] 24.If the three vertices of a parallelogram ABCD are A(2, 3), B(4, –1),
and C(0, 5) in order. Find the coordinates of the 4th vertex D.
25.If the line AB is trisected by the points C(1, 2) and D(2, 4), find the coordinates of A and B.
***THE END***
Marking Schemes
Group ‘A’ [8 (2 + 2) = 32] 1. a) (x + 5, 5) = (3, – 2 – y)
Equating corresponding elements of equal order pairs,
x + 5 = 3
5 = – 2 – y [1]
x = – 2
y = – 7
b) f'(–3) = 3 (– 3) – 10 = – 19 [1]
f'= 3 – 10 = – [1] 2. a) f'(x) =
8 =
or, 512 = x + 1 [Cubing both sides] [1]
x = 511 [1]
b) x = 7 + 4
x= 4 + 4+ 3
= (2)2 + 2 2 + [1]
or, x =
x = 2 + [1]
3. a)
=
= [1]
= =
= 2 [1]
b) = 2
Raising power '4' ... 3x + 1 = 24
3x + 1 = 16
x = 5 [1]
[1] [1]
[3] 4. a) Let A =
aij = j – i [1]
For correct value of each with matrix [1] b) For correct definition and example [1 + 1] 5. a) A =
For correct value A– B [1]
For correct value of (A– B)T [1]
b) 2 = or, = Equating,
2a + 2 = 0 2b – 4 = 1 10 – 2c = 4
a = – 1
b =
c = 3
6. a) For correct definition and example [1 + 1]
b) r(x) – p(x) = x2 + x + 1 – (2x3 – 7x – 3)
= x2 + x + 1 – 2x3 + 7x + 3
= – 3x3 + x2 + 8x + 2 [1]
Degree =3 [1]
7. a) Let (x, y) be the co-ordinates of centre then by using midpoint formula,
x = & y = [1] x = 2 & y = – 1
(2, –1) is the centre of circle. [1] b) Let the point on y-axis be P(0, y) and divides the line segment
joining the points (–5, 2) & (2, 3) in the ratio m1 : m2, then By internal section formula,
0 = &
or, 0 = [1]
or, 0 = 2m1 – 5m2
or, 5m2 = 2m1
or, =
Required ratio = 5 : 2 [1] 8. a) Slope =
or, 5 = [1]
or, 10 = k – 5 or, 15 = k
k = 15 [1]
b) The slope of a line (m) = 1
tan = 1 [m = tan] [1] = tan–1 (1)
Inclination () = 45 [1] Group ‘B’ [17 4 = 68]
9. g = {(1, a), (2, b), (3, c), (4, d)}
g–1 = {(a, 1), (b, 2), (c, 3), (d, 4)} [1]
For representing in arrow diagram For representing in table
10. r = {(x, y) : y = 4x – 2, xw, and x < 3}
When x = 0, y = 4 0 – 2 = – 2 [1]
When x = 1, y = 4 1 – 2 = 2 [1]
When x = 2, y = 4 2 – 2 = 4 [1] r = {(0, –2), (1, 2), (2, 4)} [1]
11. f'(x) = ax + b
f'(2) = 2 and f'(–3) = 12 When x = 2,
f'(2) = 2 x + b
[1 + 1]
f'(–3) = – 3a + b
or, 12 = – 3a + b ...(ii) [1]
Solving equation (i) and (ii),
a = – 2
b = 6
12. f'(3x + 1) = f'(x) + f'(1), xR, When x = 0,
f'(3 0 + 1) = f'(0) + f'(1) or, f'(1) – f'(1) = f'(0)
or, 0 = f'(0) [2] Again, when x = f'
f'= f'+ f'(1) or, f'(0) = f'+ f'(1) or, 0 – f'(1) = f' f'(0) = 0
& f'= – f'(1) [2]
13.A =
a) Order of matrix A is 2 3. [1] b) a12 = 0 and a23 = 4 [1] c) When aij = 0, i = 1, j = 2 [1]
aij = 6, when i = 2, j = 1 [1]
14. 4A + 2B – 3X = C
or, 4 + 2 – 3X =
or, + – = 3X [1]
or, = 3X [1]
or, = 3X [1]
or, = X
X = [1]
15.AT =
A = (AT)T = [1]
B =
BT = [1]
(A + B) = + =
(A + B)T = [1]
AT + BT = +
=
(A + B)T = AT + BTproved [1]
16.L.H.S. =
=– [1]
=cosec4A – cot4A
=(cosec2A + cot2A) (cosec2A – cot2A) [1]
=(cosec2A + cosec2A – 1) 1
= 2cosec2A – 1 [1]
= R.H.S.
L.H.S. = R.H.S. proved [1] 17.L.H.S. =
= [1]
= [2]
=
= R.H.S. proved [1] 18.L.H.S. = (sinA + cosecA)2 + (cosA + secA)2
= sin2A + 2sinA cosA + cosec2A + cos2A + 2cosA secA + sec2A [1]
= sin2A + cos2A + 2 1 + cosec2A + sec2A + 2 1 [1]
= 1 + 2 + 2 + 1 + cot2A + 1 + tan2A [1]
= 1 + tan2A + cot2A
[2] L.H.S. = R.H.S. proved [1]
19.L.H.S. = +
= + [1]
= +
= [1]
=
= [1]
= +
= cosecB secB + 1
= R.H.S. proved [1]
20. Solu: Given equation is − = 2...(i) or, = 2 +
Squaring on both sides
2x + 3 = 4 + 4+ x – 2 [1]
or, x + 1 = 4
Again, squaring on both sides
x2 −14 x −33 = 0 [1]
or, (x −11) (x − 3) =0
Either (x −11) = 0 or (x − 3) =0
x = 11 or 3 [1]
For checking [1]
21.p(x) – [q(x) + r(x)]= 3x3 + 9x2 + 2 – [5x3 – 9x + 7 + 6x2 – 9x + 7] [1]
= 3x3 + 9x2 + 2 – 5x3 – 6x2 + 8x – 14 [1]
= – 2x3 – 2x2 + 8x – 12 [1]
Degree = 3 [1]
22.Slope of AB = = = = [1]
Slope of BC = = = = [1]
Here, slope of AB = slope of BC and B is common point. [1]
Hence, A, B and C are collinear. [1]
23.In the given figure,
ABC is an isosceles triangle in which A = 90 and side BC parallel to X-axis.
Inclination of BC = 0
Slope of BC = tan 0 = 0 [1]
Inclination of AB = 45
Slope of AB = tan 45 = 1 [1]
and Inclination of AC = 180 – 45 = 135
So, Slope of AC = tan 135 = – 1
24.Let D(x, y) be 4th vertex.
Join the diagonals AC & BD. Midpoint of AC =
= (1, 4) [1]
(1, 4) is also the midpoint of BD. Midpoint of BD =
or, (1, 4) = or, 1 = and 4 =
x = – 2 and 9 = y [1 + 1] 4th vertex of ABCD is the point D(–2, 9). [1]
25.Let A(x1, y1) and B(x2, y2) be the points and C(1, 2) and D(2, 4) trisects the line segments AB.
i.e. AC = CD = BD
C is midpoint of AD and D is midpoint of CB. By midpoint formula,
(1, 2) = Equating,
x1 = 0 and y1 = 0 [1 + 1]
Again, (2, 4) = Equating,
Co-ordinates of A and B are (0, 0) and (3, 6). [1 + 1]