જવાબ : True
જવાબ : False
જવાબ : False
જવાબ : False
જવાબ : False
જવાબ : False
જવાબ : Now this function is defined for p where 11−p>0 as
or p<11
so Domain is (−∞, 11) (−∞, 11)
since square root gives positive values only. Also this function cannot have zero value; So Range is (0, ∞)
જવાબ : Now this function is defined for x where 2−p≥0
or p≤2
So Domain is (−∞, 2] (−∞, 2]
since square root gives positive values only, Range is [0, ∞) [0, ∞)
જવાબ : False
જવાબ : False
જવાબ : False
જવાબ : Domain is all the Real number as function is defined for all values
Domain =R
Now it can be written as
q=3−p4
Now p4 will always be positive for all real values of pesos Range will be (−∞,3](−∞,3]
જવાબ : False
જવાબ : True
જવાબ : True
જવાબ : Domain is all the Real number as function is defined for all values
Domain =R
The is a linear function. Range is also R
જવાબ : Domain is all the Real number as function is defined for all values
Domain =R
The function always provides negative value. So range is (−∞, 0] (−∞, 0]
જવાબ : Domain is all the Real number as function is defined for all values
Domain =R
The function always provides positive value. So range is [0, ∞) [0, ∞)
જવાબ : The ordered pair of B/A
(1, 1/4), (2, 4/3), (3, 9/2), (4, 16)
જવાબ : The domain for B/A is {1, 2, 3, and 4} as on 5 functions p is zero
જવાબ : (1,-3), (2, 1), (3, 7), (4, 15), (25)
જવાબ : The domain of function (B-A) is the intersection of domain of A and B
So, Domain of (B-A) = {1, 2, 3, 4, 5}
જવાબ : Domain of A ={0,1,2,3,4,5}
Range of A ={5,4,3,2,1,0}
જવાબ : Domain of B ={1,2,3,4,5,6}
Range of A ={1,4,9,16,25,36}
જવાબ : Number of relations from A to B is 21 = 2.
જવાબ : Number of relations from A to B is 22 = 4.
જવાબ : Number of relations from A to B is 29 = 512.
જવાબ : Number of relations from A to B is 26 = 64.
જવાબ : Number of relations from A to B is 26 = 64.
જવાબ : R = {(2, 4), (3, 9), (5, 25)}
જવાબ : R = {(2, 4)}
જવાબ : R = {(2, 4), (3, 9), (5, 25), (7, 49)}
જવાબ : R = {(2, 8)}
જવાબ : F (x) = (x – a) 2(x – b) 2
=> f (a + b) = (a + b – a) 2(a + b – b) 2
= b2a2
Hence, f (a + b) = a2b2
જવાબ : Given:
f (x) = | x |, x ∈ R
∴ the range of f is [0, ∞)
જવાબ : Given:
f (x) = -| x |, x ∈ R
∴ the range of f is [0, -∞)
જવાબ : Given:
f (x) = | -x |, x ∈ R
∴ the range of f is [0, ∞)
જવાબ : Given:
f (x) = | x |-1, x ∈ R
∴ the range of f is [-1, ∞)
જવાબ : Given:
f (x) = 1+ | x |, x ∈ R
∴ the range of f is [0, ∞)
જવાબ : Given:
f (x) = | x |, x ∈ R
∴ the range of f is [0, ∞)
જવાબ : Given:
f (x) = | x |, x ∈ R
∴ the range of f is [0, ∞)
જવાબ : Total number of functions from A to B = ax
જવાબ : Total number of functions from A to B = 9
જવાબ : Total number of functions from A to B = 8
જવાબ : Total number of functions from A to B = 1
જવાબ : Total number of functions from A to B = 0
જવાબ : Total number of functions from A to B = 625
જવાબ : Total number of functions from A to B = 64
જવાબ : R = {(2, 8), (3, 27), (5, 125), (7, 343)}
જવાબ : R = {(2, 8)}
જવાબ : It is given that the functions f(x) = x2 + 16 and g(x) = -8x are equal.
∴f(x) =g(x)
જવાબ : It is given that the functions f(x) = x2 + 16 and g(x) = -8x are equal.
∴f(x) =g(x)
જવાબ : It is given that the functions f(x) = x2 + 18 and g(x) = -11x are equal.
∴f(x) =g(x)
જવાબ : It is given that the functions f(x) = x2 + 14 and g(x) = -9x are equal.
∴f(x) =g(x)
જવાબ : It is given that the functions f(x) = x2 and g(x) = -12 - 8x are equal.
∴f(x) =g(x)
જવાબ : It is given that the functions f(x) = 3x2 and g(x) = 6 + x are equal.
∴f(x) =g(x)
જવાબ : It is given that the functions f(x) = x2 − 1 and g(x) = 1 - x are equal.
∴f(x) =g(x)
જવાબ : It is given that the functions f(x) = 3x2 − 1 and g(x) = 3 + x are equal.
∴f(x) =g(x)
જવાબ : Given:
f (a) = a2 – 3a + 4
Therefore,
f (2a + 1) = (2a + 1)2 – 3(2a + 1) + 4
= 4a2 + 1 + 4a – 6a – 3 + 4
= 4a2 – 2a + 2
Now,
f (a) = f (2a + 1)
⇒ a2 – 3a + 4 = 4a2 – 2a + 2
⇒ 4a2 – a2 – 2a + 3a + 2 – 4 = 0
⇒ 3a2 + a – 2 = 0
⇒ 3a2 + 3a – 2a – 2 = 0
⇒ 3x(a + 1) – 2(a +1) = 0
⇒ (3a – 2)(a +1) = 0
⇒ (a + 1) = 0 or ( 3a – 2) = 0
⇒a=−1 or a=23
જવાબ : Given:
X = {2, 3}, Y = {4, 5} and Z ={5, 6}
Also,
(Y ∪ Z) = {4, 5, 6}
Thus, we have:
X × (Y ∪ Z) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3,6)}
And,
(Y ∩ Z) = {5}
Thus, we have:
X × (Y ∩ Z) = {(2, 5), (3, 5)}
Now,
(X × Y) = {(2, 4), (2, 5), (3, 4), (3, 5)}
(X × Z) = {(2, 5), (2, 6), (3, 5), (3, 6)}
∴ (X × Y) ∪ (X × Z) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
જવાબ : Given:
A = {1, 2, 3}, B = {4} and C = {5}
A × (B ∪ C) = (A × B) ∪ (A × C)
We have:
(B ∪ C) = {4, 5}
LHS: A × (B ∪ C) = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
Now,
(A × B) = {(1, 4), (2, 4), (3, 4)}
And,
(A × C) = {(1, 5), (2, 5), (3, 5)}
RHS: (A × B) ∪ (A × C) = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}
∴ LHS = RHS
જવાબ : Given:
A = {1, 2, 3}, B = {4} and C = {5}
A × (B ∩ C) = (A × B) ∩ (A × C)
We have:
(B ∩ C) = ϕ
LHS: A × (B ∩ C) = ϕ
And,
(A × B) = {(1, 4), (2, 4), (3, 4)}
(A × C) = {(1, 5), (2, 5), (3, 5)}
RHS: (A × B) ∩ (A × C) = ϕ
∴ LHS = RHS
જવાબ : Given:
A = {1, 2, 3}, B = {4} and C = {5}
A × (B − C) = (A × B) − (A × C)
We have:
(B − C) = ϕ
LHS: A × (B − C) = ϕ
Now,
(A × B) = {(1, 4), (2, 4), (3, 4)}
And,
(A × C) = {(1, 5), (2, 5), (3, 5)}
RHS: (A × B) − (A × C) = ϕ
∴ LHS = RHS
જવાબ : Given:
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
A × C ⊂ B × D
LHS: A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
RHS: B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
∴ A × C ⊂ B × D
જવાબ : Given:
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
A × (B ∩ C) = (A × B) ∩ (A × C)
We have:
(B ∩ C) = ϕ
LHS: A × (B ∩ C) = ϕ
Now,
(A × B) = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
(A × C) = {(1, 5), (1, 6), (2, 5), (2, 6)}
RHS: (A × B) ∩ (A × C) = ϕ
∴ LHS = RHS
જવાબ : R = {(x, y): x, y ∈ W, 2x + y = 7}
As y = 8−2x
જવાબ : R = {(x, y): x, y ∈ W, x + y = 3}
As y = 3−x
જવાબ : R = {(x, y): x, y ∈ W, -x + y = 3}
As y = 3 - x
જવાબ : R = {(x, y): x, y ∈ W, x + 2y = 7}
As y = (7-x)/2
જવાબ : R = {(x, y): x, y ∈ W, 2x + y = 5}
As y = -2x+5
જવાબ : Given:
f(x)=(x-2)/(2-x)
Domain ( f ) :
Clearly, f (x) is defined for all x satisfying: if 2 -x ≠ 0 ⇒ x ≠ 2.
Hence, domain ( f ) = R - {2}
Range of f :
Let f (x) = y
⇒ (x-2)/(2-x)=y
⇒ x - 2 = y (2 - x)
⇒ x - 2 = - y (x - 2)
⇒ y = - 1
Hence, range ( f ) = {- 1}
જવાબ : Given:
f(x) = 4x − x2, x ∈ R
Now,
f(p + 1) = 4(p + 1) - (p + 1)2
= 4p + 4 - (p2 + 1 + 2p)
= 4p + 4 - p2 - 1 – 2p
= 2p - p2 + 3
f(p - 1) = 4(p - 1) - (p - 1)2
= 4p - 4 - (p2 + 1 – 2p)
= 4p - 4 - p2 - 1 + 2p
= 6p - p2 - 5
Thus,
f(p + 1) − f(p − 1) = ( 2p - p2 + 3) - (6p - p2 - 5)
= 2p - p2 + 3 – 6p + p2 + 5
= 8 – 4p
= 4(2 - p)
write the number of functions from B to A
1 |
n(A) = 2 and n(B) = 3 |
A |
9 |
2 |
n(A) = 3 and n(B) = 2 |
B |
1 |
3 |
n(A) = 0 and n(B) = 1 |
C |
8 |
જવાબ :
1-C, 2-A, 3-B
write the number of functions from A to B
1 |
n(A) = 2 and n(B) = 3 |
A |
1 |
2 |
n(A) = 3 and n(B) = 2 |
B |
9 |
3 |
n(A) = 0 and n(B) = 1 |
C |
8 |
જવાબ :
1-B, 2-C, 3-A
the number of functions that can be defined from A into B
1 |
If A = {1, 2, 3, 4 } and B = {x, y} |
A |
3 |
2 |
If A = {1, 2, 3} and B = {x, y} |
B |
16 |
3 |
If A = {1, 2 } and B = {x, y} |
C |
9 |
4 |
If A = {1, 2, 3} and B = { y} |
D |
4 |
જવાબ :
1-B, 2-C, 3-D, 4-A
1 |
If [x]2−5[x]+4=0, where [.] denotes the greatest integer function, then |
A |
x ∈ [-10, -4) |
2 |
If [x]2−8[x]+16=0, where [.] denotes the greatest integer function |
B |
x ∈ [-2, -2) |
3 |
If [x]2+5[x]+6=0, where [.] denotes the greatest integer function |
C |
x ∈ [4, 5) |
4 |
If [x]2+15[x]+50=0, where [.] denotes the greatest integer function |
D |
x ∈ [1, 5) |
જવાબ :
1-D, 2-C, 3-B, 4-A
1 |
Real No. |
A |
{0,1,2,3…..} |
2 |
Natural No. |
B |
{…..-2, -1, 0, 1, 2…..} |
3 |
Whole No. |
C |
{1,2,3,4……} |
4 |
Integers |
D |
[-∞,∞] |
જવાબ :
1-D, 2-C, 3-A, 4-B
1 |
The domain of definition of the function f(x) = 1/ |x-1| |
A |
R − {2} |
2 |
The domain of definition of the function f(x) = x - 21 |
B |
R − {1} |
3 |
The domain of definition of the function f(x) = 1/ |x-2| |
C |
∞ |
જવાબ :
1-B, 2-C, 3-A
1 |
The domain of definition of the function f(x) = log |x| |
A |
R-{0} |
2 |
The range of the function f(y) = |y − 1| |
B |
(−∞, 0) |
3 |
The range of the function f(y) = - |y| |
C |
[0, ∞) |
જવાબ :
1-A, 2-C, 3-B
1 |
N |
A |
Real No. |
2 |
W |
B |
Natural No. |
3 |
Z |
C |
Whole No. |
4 |
R |
D |
Integers |
જવાબ :
1-B, 2-C, 3-D, 4-A
If X = {1, 2, 4}, Y= {2, 4, 5}, Z = {2, 5}
1 |
Y+Z |
A |
X-Y |
2 |
X+Z |
B |
Y |
3 |
X-(Y+Z) |
C |
X |
4 |
X+Y-Z |
D |
X+Y |
જવાબ :
1-B, 2-D, 3-A, 4-C
If X = {1, 2, 4}, Y= {2, 4, 5}, Z = {2, 5}
1 |
X-Y |
A |
{4} |
2 |
Y-Z |
B |
{1,4} |
3 |
X-Z |
C |
{1} |
4 |
X+Y |
D |
{1,2,4,5} |
જવાબ :
1-C, 2-A, 3-B, 4-D
Math
The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.
The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.
For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.