10th Class, MATHS - Paper-I

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Guess Paper-II
10th Class, MATHS - Paper-I
SECTION -I
Group - A
1)
Prove that “If x is even the x2 is even”.
2)
Prove that A ∩ B1 = A – B.
3)
If f ( x ) =
4)
Define Remainder theorem and prove it.
5)
Group - B
The values of an objective function at (2, 0) and (0, 5) are 15 and 12. Then find the objective
function.
6)
If ax = by = cz = dw; ab = cd show that
7)
Evaluate Lt
8)
If g1, g2, g3 are three Geometric means between m and n. Show that g1g3 = g22 = mn.
9)
Section -II
If n(A ∪ B) = 51, n(A) = 20, n(B) = 44. Then find the value n(A ∩ B).
S
1 1 1 1
+ = + .
x y w z
K
x →0
1+ x −1
.
x
H
I
1− x
( x ≠ −1) then find f(0), f(1), f(2), & f(3)?
1+ x
x +1
⎛1⎞
, x ≠ 1, then find the value of f ⎜ ⎟ .
x −1
⎝ 3⎠
A
10) If f : R − {1} → R , f ( x ) =
S
11) State the principle of mathematical Induction.
12) Define the term Isoprofit Line.
13) Solve |12 – 3x| = 9.
14) Find the values of x. So that
−2
−7
, x,
are three consecutive terms of a G.P.
7
2
SECTION -III
Group - A
15) Prove that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) for any three sets A, B, C.
16) Let f : R – {2} → R be defined by f ( x ) =
⎛ 2x + 1 ⎞
2x + 1
. Show that f ⎜
⎟ = x.
⎝ x−2 ⎠
x−2
17) Given f(x) = x – 2, g(x) = x2 – 2; h(x) = x3 – 3 ∀ x ∈ R. Then find (i) (fog)oh (ii) fo(goh).
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8
6 ⎞
⎛
18) Find the constant term of ⎜ 5 x 2 − 2 ⎟ .
x ⎠
⎝
Group - B
19) A sweet shop makes gift packet of sweets combines two special types of sweets A and B
which weighs 7 kg of A and no more than 5 kg of B should be used. The shop makes a profit
of Rs. 15 on A and Rs. 20 on B per kg. Determine the product of mixer so as to obtain
maximum profit?
x
y
z
20) If a = b = c ;
y
2z
b c
.
= . Show that =
x x+z
a b
21) The sum of the first ‘n’ natural numbers is s1 and that of their squares s2 and cubes s3. Show
that 9s22 = s3(1 + 8s1).
1
a
b
,
2
,
2
1
I
1
c2
will also be in H.P.
H
22) If (b + c), (c + a), (a + b) are in H.P. Show that
SECTION-IV
S
23) Using graph of y = x2 solve the equation x2 – 3x + 2 = 0.
24) Maximize f : 5x + 7y subject to the constraints 2x + 3y ≤ 12, 3x + y ≥ 12, x ≥ 0 and y ≥ 0.
K
ANSWERS
SECTION -I
Group - A
x2 = (2k)2 = 4k2
S
3.
A
1) Prove that “If x is even the x2 is even.”
A: S.No. Steps
Reasons
1.
x is even
Given
2.
x = 2k
Definition of even numbers
L is an integer
= 2(2k2)
= 2L
4.
2)
x2 is even
Definition of even numbers.
Prove that A ∩ B1 = A – B.
A: Let x ∈ A ∩ B1
⇔ x ∈ A and x ∈ B1
⇔ x ∈ A and x ∉ B (∵ x ∈ B1 ⇔ x ∉ B)
⇔ x ∈ A – B.
∴ A ∩ B1 = A – B
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A:
If f ( x ) =
1− x
( x ≠ −1) then find f(0), f(1), f(2), & f(3)?
1+ x
f ( x) =
1− x
1+ x
f (0) =
1− 0 1
= =1
1+ 0 1
f (1) =
1−1 0
= =0
1+1 2
f (2) =
1 − 2 −1
=
1+ 2 3
f (3) =
1 − 3 −2 −1
=
=
.
2
1+ 3 4
I
3)
A
K
S
H
4) Define Remainder theorem and prove it?
A: Remainder Theorem: If a rational integral function of x, say f(x) is divided by (x – a) then
the remainder is f(a).
Proof: Let Q(x), the Quotient and R, the remainder when f(x) is divided by (x – a). Then we
have f(x) = Q(x) (x – a) + R.
Where in the degree of Q is one less than that of f(x) and R is a constant.
Since the equation is true for all real values of x putting x = a in the above, we get
⇒ f(a) = Q(a) (a – a) + R
⇒ f(a) = R
∴ f(x) = (x – a) Q(x) + f(a).
S
Group - B
5) The values of an objective function at (2, 0) and (0, 5) are 15 and 12. Then find the objective
function.
A: Let the objective function be f = ax + by –––– (1)
The value of ‘f ’ at (2, 0) is 15.
15 = a(2) + b(a)
⇒ 2a = 15
⇒a=
15
.
2
The value of f at (0, 5) is 12.
12 = a(0) + b(5) ⇒ 5b = 12
⇒b=
12
.
5
∴ From (1) objective function f =
15
12
x + y.
2
5
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6)
If ax = by = cz = dw; ab = cd show that
1 1 1 1
+ = + .
x y w z
A: ax = by = cz = dw = k
ax = k ⇒ a = k1/x;
by = k ⇒ b = k1/y
cz = k ⇒ c = k1/z;
ab = cd (Given)
dw = k ⇒ d = k1/w
1
k x. k
k
1
1 1
+
x y
y
= k z.k
1
1 1
+
=kz w
1
w
(∵ As the bases of equal powers are equal their exponents must be equal)
x →0
Lt
x →0
H
A:
1+ x −1
.
x
Evaluate Lt
1+ x −1
x
rationalising the numerator we get
x →0
1+ x −1 1+ x + 1
×
x
1+ x +1
K
Lt
S
7)
I
1 1 1 1
+ = +
x y z w
A
( 1 + x ) 2 − (1) 2
= Lt
x → 0 x ( 1 + x + 1)
1 + x −1
x
= Lt
x → 0 x ( 1 + x + 1) x →0 x ( 1 + x + 1)
= Lt
1
1+ x +1
S
= Lt
x →0
=
∴ Lt
x →0
8)
1
1
1
1
=
=
=
1+ 0 +1
1 +1 1+1 2
1+ x −1 1
=
x
2
If g1, g2, g3 are three Geometric means between m and n. Show that g1g3 = g22 = mn.
A: g1, g2, g3 are the three Geometric means between m and n.
∴ m, g1, g2, g3, n are in G.P.
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r=
⇒
t2 t3 t 4 t5
= = =
t1 t2 t3 t 4
g 2 g 2 g3 n
=
=
=
––––– (1)
m g1 g 2 g 3
g1 n
=
m g3
From (1)
⇒ g1 g3 = mn ––––– (2)
From (1) we get
g 2 g3
=
g1 g 2
⇒ g1 g3 = g22 ––––– (3)
I
From (2) and (3) we get
H
g1g3 = mn = g22
K
S
SECTION -II
9) If n(A ∪ B) = 51, n(A) = 20, n(B) = 44. Then find the value n(A ∩ B).
A: n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
51 = 20 + 44 – n(A ∩ B)
51 = 64 – n(A ∩ B)
∴ n(A ∩ B) = 64 – 51 = 13
10) If f : R − {1} → R , f ( x ) =
f ( x) =
x +1
x −1
A
A:
x +1
⎛1⎞
, x ≠ 1, then find the value of f ⎜ ⎟ .
x −1
⎝ 3⎠
S
1+ 3
1
+1
4
⎛1⎞
3
=
f ⎜ ⎟= 3
=
= −2.
⎝ 3 ⎠ 1 − 1 1 − 3 −2
3
3
11) State the principle of mathematical Induction.
A: If p(n) is a statement which true for n = 1, and is true for n + 1, whenever it is true for 1, 2,
.......n. Then the statement p(n) is true for all natural numbers n.
12) Define the term Isoprofit Line.
A: Isoprofit Line: Any line belonging to the system of parallel lines given by the objective
function for various values of the objective function is called Isoprofit Line.
13) Solve |12 – 3x| = 9.
A:
|12 – 3x| = 9
12 – 3x = 9
or
12 – 3x = –9
–3x = 9 – 12
–3x = – 9 – 12
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–3x = –3
x=1
∴ x = 1 or x = 7
–3x = –21
x=7
14) Find the values of x. So that
−2
−7
, x,
are three consecutive terms of a G.P..
7
2
A: If a, b, c are three consecutive terms of a G.P. then b = ac .
Given that
−2
−7
, x,
are three consecutive terms of a G.P.
7
2
⎛ −2 ⎞⎛ −7 ⎞
∴ x = ⎜ ⎟⎜ ⎟ = 1 = ± 1
⎝ 7 ⎠⎝ 2 ⎠
I
∴ x = ±1
S
A
K
S
H
SECTION -III
Group - A
15) Prove that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) for any three sets A, B, C.
A: It is enough to prove
i) A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C)
ii) (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C)
i) To show that A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C)
Let x ∈ A ∩ (B ∪ C)
⇒ x ∈ A and x ∈ (B ∪ C)
⇒ x ∈ A and (x ∈ B or x ∈ C)
⇒ (x ∈ A and x ∈ B) or (x ∈ A and x ∈ C)
⇒ x ∈ (A ∩ B) or x ∈ (A ∩ C)
⇒ x ∈ (A ∩ B) ∪ (A ∩ C)
∴ A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C) ––––– (1)
ii) To show that (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C)
Let x ∈ (A ∩ B) ∪ (A ∩ C)
⇒ x ∈ (A ∩ B) or x ∈ (A ∩ C)
⇒ x ∈ A and x ∈ B or x ∈ A and x ∈ C
⇒ x ∈ A and (x ∈ B or x ∈ C)
⇒ x ∈ A and (x ∈ B ∪ C
⇒ x ∈ A ∩ (B ∪ C)
∴ (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C) ––––– (2)
From (1) and (2) we get
A ∩ (B ∪ C) = (A ∩ B) ∪ A ∩ C
(∵ A ⊂ B, B ⊂ A then A = B.)
16) Let f : R – {2} → R be defined by f ( x ) =
⎛ 2x + 1 ⎞
2x + 1
. Show that f ⎜
⎟ = x.
⎝ x−2 ⎠
x−2
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A:
f ( x) =
2x + 1
x−2
⎛ 2x + 1 ⎞
2⎜
⎟ +1
x−2 ⎠
⎛ 2x + 1 ⎞
⎝
L.H.S f ⎜
⎟ = 2x + 1
⎝ x−2 ⎠
−2
x−2
2(2 x + 1) + 1( x − 2)
x−2
=
2 x + 1 − 2( x − 2)
x−2
I
4 x + 2 + x − 2 5x
=
=x
2 x + 1 −2x + 4 5
=
L.H.S = R.H.S proved.
= f(x2 – 2)
(∵ g(x) = x2 – 2)
= x2 – 2 – 1
(∵ f(x) = x – 1)
K
= x2 – 3
= fog[h(x)]
= fog(x3 – 3)
(∵ h(x) = x3 – 3)
= (x3 – 3)2 – 3
(∵ fog(x) = x2 – 3)
A
(fog)oh(x)
S
A: f(x) = x – 2, g(x) = x2 – 2; h(x) = x3 – 3
i) (fog)oh
(fog)oh
= f[g(x)]
H
17) Given f(x) = x – 2, g(x) = x2 – 2; h(x) = x3 – 3 ∀ x ∈ R. Then find (i) (fog)oh (ii) fo(goh).
= (x3)6 – 2 · x3 · 3 + (3)2 – 3
= x6 – 6x3 + 9 – 3
= x6 – 6x3 + 6
S
(fog)oh
ii) fo(goh)
goh(x)
= g[h(x)]
= g(x3 – 3)
(∵ h(x) = x3 – 3)
= (x3 – 3)2 – 2
(∵ g(x) = x2 = 2)
= (x3)2 – 2 · x3 · 3 + (3)2 – 2
= x6 – 6x3 + 9 – 2
= x6 – 6x3 + 7
∴ fo(goh)(x) = f[goh(x)]
= f(x6 – 6x3 + 7) (∵ goh(x) = x6 – 6x3 + 7)
= x6 – 6x3 + 7 – 1 (∵ f(x) = x – 1)
= x6 – 6x3 + 6
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∴ fo(goh)(x) = –
∴ (fog)oh = fo(goh)
x6
6x3
+6
8
6 ⎞
⎛
18) Find the constant term of ⎜ 5 x 2 − 2 ⎟ .
x ⎠
⎝
8
A:
−6
⎛ 2 6 ⎞
2
⎜ 5 x − 2 ⎟ k = 5x , y = 2 , n = 8
x ⎠
x
⎝
T r + 1 = ncr · xn–r · yr
8
2 8− r
= cr ·(5 x )
⎛ −6 ⎞
·⎜ 2 ⎟
⎝x ⎠
r
= 8cr · (5)8–r · (–6)r · x16 – 2r – 2r
––––– (1)
H
= 8cr · (5)8–r · (–6)r · x16 – 4r
I
= 8cr · (5)8–r · x16 – 2r · (–6)r · x–2r
∴ Constant term = 8c4 · 54 · 64
= 8c4 · 54 · 64
S
∴ To find the constant term the exponent of x should be zero.
∴ 16 – 14r = 0
⇒ 4r = 16
⇒r=4
(∵ from (1))
S
A
K
Group - B
19) A sweet shop makes gift packet of sweets combines two special types of sweets A and B
which weight 7 kg of A and no more than 5 kg of B should be used. The shop makes a profit
of Rs. 15 on A and Rs. 20 on B per kg. Determine the product of mixer so as to obtain
maximum profit.
A: Let the no.of kgs of A type sweets be x, the number of kgs of B type sweets by y. The no.
of kgs of sweets cannot be negative.
∴ x ≥ 0, y ≥ 0
––––– (1)
The gift packet of sweets contains A and B type sweets which weight 7 kgs.
∴x+y=7
––––– (2)
At least 3 kgs of A should be used in each packet.
⇒x≥3
––––– (3)
Not more than 5 kgs of B should be used in each packet.
⇒y≤5
––––– (4)
The shop makes profit of Rs. 15 on A and Rs. 20 on B per kg.
f = 15x + 20y
––––– (5)
‘f ’ is to be maximized.
x
y
z
20) If a = b = c ;
y
2z
b c
.
= . Show that =
x x+z
a b
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A: Let
ax
=
by
=
cz
= k.
ax = k
⇒ a = k1/x
by = k
⇒ b = k1/y
cz = k
⇒ c = k1/z
Given that
b c
= ⇒ b 2 = ac
a b
⇒ (k1/y)2 = k1/x · k1/z
2
y
⇒k =
1 1
+
kx z
As the bases of equal powers are equal, their exponents must be equal.
2 z+x
=
y
xz
H
⇒
I
2 1 1
= +
y x z
⇒ 2(xz) = y(z + x)
y
2z
.
=
x z+x
S
∴
21) The sum of the first ‘n’ natural numbers is s1 and that of their squares s2 and cubes s3. Show
K
that 9s22 = s3(1 + 8s1).
A: Sum of the first ‘n’ natural numbers s1 =
n(n + 1)
––––– (1)
2
A
Sum of the squares of first ‘n’ natural numbers s2 =
n(n + 1)(2n + 1)
6
S
n 2 (n + 1)2
Sum of the cubes of first ‘n’ natural numbers s3 =
4
R.H.S. s3 (1 + 8s1)
n 2 (n + 1)2 ⎛
n (n + 1) ⎞
=
⎜ 1 + 8·
⎟
4
2 ⎠
⎝
=
n 2 (n + 1)2
(1 + 4n(n + 1) )
4
n 2 (n + 1)2
=
(1 + 4n 2 + 4n)
4
=
n 2 (n + 1) 2 (2n + 1) 2
4
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n 2 (n + 1)2 (2n + 1) 2
n 2 (n + 1) 2 (2n + 1) 2
⎛ n (n + 1)(2n + 1) ⎞
=9
= 9·
= 9· ⎜
⎟
4· 9
36
6
⎝
⎠
2
= 9s22 L.H.S
∴ 9s22 = s3(1 + 8s1) is proved.
22) If (b + c), (c + a), (a + b) are in H.P. Show that
1
1
1
a
b
c2
,
2
,
2
will also be in H.P.
A: (b + c), (c + a), (a + b) are in H.P.
1
1
1
,
,
are in A.P.
b +c c +a a +b
In A.P t2 – t1 = t3 – t2
b + c − (c + a ) (c + a ) − (a + b )
=
(c + a ) (b + c ) (a + b ) (c + a )
⇒
b + c − c − 9 c + a − a −b
=
b+c
a +b
⇒
b−a c −b
=
b+c a+b
K
S
⇒
H
I
1
1
1
1
−
=
−
c + a b +c a +b c +a
⇒ (b + a)(b – a) = (c + b)(c – b)
⇒ b2 – a2 = c2 – b2
1
a
2
,
1
b
2
,
1
c2
are in H.P.
S
⇒
A
∴ a2, b2, c2 are in A.P. (∵ If t1, t2, t3 are in A.P. then t2 – t1 = t3 – t2)
SECTION-IV
23) Using graph of y = x2 solve the equation x2 – 3x + 2 = 0.
A: Given quadratic equation x2 – 3x + 2 = 0
⇒ x2 = 3x – 2
If y = x2 then y = 3x – 2
∴ The roots of x2 – 3x + 2 = 0 are the x-coordinates of the points of intersection of parabola
y = x2 and the straight line y = 3x – 2.
y = x2:
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x
0
1
2
3
–1
–2
–3
y
0
1
4
9
1
4
9
y = 3x – 2:
0
4
y
–2
10
A
K
S
H
I
x
S
Observe that the line y = 3x – 2 meets the parabola y = x2 in two points (1, 1) and (2, 4)
whose x-coordinates are 1, 2.
∴ The set of solution of the given equation is {1, 2}
24) Maximize f : 5x + 7y subject to the constraints 2x + 3y ≤ 12, 3x + y ≥ 12, x ≥ 0 and y ≥ 0.
A: i) x ≥ 0, y ≥ 0 represent first Quadrant.
ii) Given inequation 2x + 3y ≤ 12 ––––– (1)
line eq 2x + 3y = 12
x
0
6
y
4
0
Mark the points (0, 4), (6, 0) on a graph paper and join them by means of a line.
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Substituting (0, 0) in equation (1)
⇒ 2(0) + 3(0) ≤ 12
0 ≤ 12 (True)
Shade the region on the origin side of the line.
iii) 3x + y ≤ 12 ––––– (2)
Corresponding equation is 3x + y = 12
x
0
4
y
12
0
H
I
Mark the points (0, 12), (4, 0) on a graph and join them by means of a line.
Substituting (0, 0) in equation (2).
3(0) + 0 ≤ 12
0 ≤ 12 (True)
12
11
Scale:
x-axis: 1 cm = 1 unit
y-axis: 1 cm = 1 unit
S
10
9
8
K
7
6
5
A
4
C
3
2
B2
x+
S
1
0
1
2
3
A4
3y
=
12
x
5
6
7
8
y
Shade the region on the origin side of the line.
The closed convex polygonal region OABC is the feasible region. The vertices of
the feasible region are O(0, 0), A(4, 0), B(3.4, 1.7) and C(0, 4).
Objective function f = 5x + 7y
At O(0, 0)
f = 5(0) + 7(0) = 0
At A(4, 0)
f = 5(4) + 7(0) = 20
At B(3.4, 1.7)
f = 5(3.4) + 7(1.7) = 28.9
at C(0, 4)
f = 5(0) + 7(4) = 28
∴ f is maximized at B(3.4, 1.7) and maximum value is 28.9
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PART - B
3.
4.
5.
(x + 1) is a factor to ax4 + bx3 + cs2 + dx + e. Then which of the following
is true?
A) a + c + e = b + d
B) a + b + c = 0
C) a + b + c + d + e = 0
D) a + b + c = d + e
2⎞
⎛
Last term in the expansion of ⎜ x + ⎟
x⎠
⎝
A)
8.
9.
x5
10
C)
x5
32
x5
[D]
[A]
[C]
D)
32
x
S
Which of the following inequations represents the region containing the points (1, 2) and
(2, 1) ?
[A]
A) x + y < 2
B) x + y > 5
C) 2x + y < 6
D) x + 2y < 7
| x | = 3 then x = ____.
[A]
A) 3 or –3
B) 1/3
C) –1/3
D) 0
The modulus of real number is ____.
[B]
A) Positive
B) Never Negative
C) Not Negative, Not Positive
D) Always zero
In an A.P., 1st term is –1, d = –3 then 12th term = ____.
[D]
A) 34
B) 32
C) –32
D) –34
S
10. nth term in the series 1.3 + 3.5 + 5.7 + ____.
A) n (n + 2)
B) n(n + 5)
C) 2n(2n – 1)
II.
11.
12.
13.
[C]
K
7.
B)
is ____.
A
6.
2
5
[C]
I
2.
In the following which is True?
A) 3 + 7 = 10 ⇔ 1 + 2 = 2
B) 3 × 7 = 10 ⇔ 1 × 3 = 3
C) 3 + 7 = 10 ⇔ 1 + 2 = 3
D) 3 × 7 = 8 ⇔ 1 × 2 = 2
If A ⊂ B, n(A) = 12 and n(B) = 20; then n(B – A) = _____.
A) 32
B) –8
C) 8
D) –32
If f(x) = 2 – x; g(x) = 3x – 2 then fog(2) = _____.
A) 4 + 2x
B) –4x
C) 2
D) –6
H
1.
[D]
D) (2n – 1) (2n + 1)
Fill in the blanks.
If Truth value of p ⇒ q, then the truth value of ~q ⇒ ~p = _____.
If A ∪ B = A ∩ B, then_____.
The zero of the function is ___________.
y
x 1 (–2, 0) (0, 0)
(2, 0)
x
y1
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(F)
(A = B)
(0, ±2)
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14. If two functions f, g are equal, when their _____ are equal.
(Domains)
15. If α, β are the roots of quadratic equation x2 + ax + b = 0. Then α2 + β2 = _____.
16. If x ≥ 0, y ≥ 0 then, the minimum value of f = x + y = ____.
(a2 – 2b)
(0)
x
17. If (64) = 2 2 then x = _____.
(1/4)
18.
Lt
x→4
x + 12
= ______
4
.
(±1)
19. The Tangent values of π/3, π/4, π/6 are in ______.
(G.P.)
10
.
(0)
Group-B
A) –
B) 0
C) 2 and 4
[E]
[D]
[A]
S
III. Match the following.
i)
Group-A
21) p ∧ (~p) = ____
22) Co-domain of the function f : A → B
23) If f(x) = x+2, then f(–3) = _____
I
n =1
H
20. 55 − ∑ n = _____
[B]
D) B
25) If 10c2n = 10cn+4, then n = _____
[C]
E) Tautology
K
F) A
G) Contradiction
H) 4
[L]
Group-B
I) 0.1
[M]
J) 10
28) (0.001)1/3
[I]
K)
29) 13+23+33+______+m3 = 3025 then m = ___
30) Geometric mean of a, b
[J]
[K]
L) (2, 1)
M) 2
N) 0.01
O) (1, –10)
27)
A
Group-A
26) The point which belongs to 2x – y ≤ 4
4x + 3
= ___
x →α 2 x + 3
Lt
S
ii)
24) Sum of the roots of 6x2 – 5 = 0
P)
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ab
a+b
2

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