# CBSE Solutions for Class 11 Maths

#### Select CBSE Solutions for class 10 Subject & Chapters Wise :

Define whether the statement is true or false: The relation defined as {(3, 1), (2, 1), (8, 1), (10, 1), (4, 1), (7, 1)} is a function

True

Define whether the statement is true or false: The relation defined as {(3, 1), (4, 2), (6, 4), (8, 3), (10, 5), (12, 7), (14, 6)} is not a function

False

If f (a) = a2 – 3a + 4, then find the values of a satisfying the equation f (a) = f (2a + 1).

Given:
f (a) = a2 – 3a + 4
Therefore,
f (2a + 1) = (2a + 1)– 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

Hence, a=−1, 23

 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 ∞

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, ∞)

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

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

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}