Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13

Choose the correct or the most suitable questions.

Question 1.
If |x + 2| ≤ 9, then x belongs to
(a) (-∞, -7)
(b) [-11, 7]
(c) (-∞, -7) ∪ [11, ∞)
(d)(-11, 7)
Solution:
(b) [-11, 7]
Hint:
Given |x + 2| ≤ 9
– 9 ≤ (x + 2) ≤ 9
– 9 – 2 ≤ x ≤ 9 – 2
– 11 ≤ x ≤ 7
∴ x ∈ [-11, 7]

Question 2.
Given that x, y and b are real numbers x < y, b ≥ 0, then ……..
(a) xb < yb (b) xb > yb
(c) xb ≤ vb
(d) xlb ≥ ylb
Solution:
(a) xb < yb
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 1

Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 3
(a) [2, ∞]
(b) (2, ∞)
(c) (-∞, 2)
(d) (-2, ∞)
Solution:
(b) (2, ∞)
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 4

Question 4.
The solution of 5x – 1 < 24 and 5x + 1 > -24 is …….
(a) (4, 5)
(b) (-5, -4)
(c) (-5, 5)
(d) (-5, 4)
Solution:
(c) (-5, 5)
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 5

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13

Question 5.
The solution set of the following inequality |x – 1| ≥ |x – 3| is …….
(a) [0, 2]
(b) (2, ∞)
(c) (0, 2)
(d) (-∞, 2)
Solution:
(b) (2, ∞)

Question 6.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 6
(a) 16
(b) 18
(c) 9
(d) 12
Solution:
(b) 18
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 7

Question 7.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 8
(a) -2
(b) -8
(c) -4
(d) -9
Solution:
(c) -4
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 9

Question 8.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 10
(a) 0.5
(b) 2.5
(c) 1.5
(d) 1.25
Solution:
(a) 0.5
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 11

Question 9.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 12
(a) 2
(b) 1
(c) 3
(d) 4
Solution:
(b) 1
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 13

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13

Question 10.
If 3 is the logarithm of 343, then the base is ……
(a) 5
(b) 7
(c) 6
(d) 9
Solution:
(b) 7
Hint.
Given logx 343 = 3
343 = x3
7 × 7 × 7 = x3
73 = x3
x = 7
Base of the logarithm x = 7

Question 11.
Find a so that the sum and product of the roots of the equation 2x2 + (a – 3)x + 3a – 5 = 0 are equal is ……..
(a) 1
(b) 2
(c) 0
(d) 4
Solution:
(b) 2
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 14

Question 12.
If a and b are the roots of the equation x2 – kx + 16 = 0 and satisfy a2 + b2 = 32, then the value of k is ……
(a) 10
(b) -8
(c) (-8, 8)
(d) 6
Solution:
(c) -8, 8
Hint:
Given a and b are the roots of x2 – kx + 16 = 0 satisfying a2 + b2 = 32
Sum of the roots a + b = \(\frac{-(-k)}{1}\)
a + b = k
Product of the roots ab = \(\frac{16}{1}\)
ab = 16
a2 + b2 = (a + b)2 – 2ab
32 = k2 – 2 × 16
32 = k2 – 32
k2 = 32 + 32 = 64
k = ± 8

Question 13.
The number of solutions of x2 + |x – 1| = 1 is ………
(a) 1
(b) 0
(c) 2
(d) 3
Solution:
(c) 2
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 15
We have two solutions 0, 1

Question 14.
The equations whose roots are numerically equal but opposite in sign to the roots of 3x2 – 5x – 7 = 0 is ……
(a) 3x2 – 5x – 7 = 0
(b) 3x2 + 5x – 7 = 0
(c) 3x2 – 5x + 7 = 0
(d) 3x2 + x – 7 = 0
Solution:
(b) 3x2 + 5x – 7 = 0
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 16

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13

Question 15.
If 8 and 2 are the roots of x2 + ax + c = 0 and 3, 3 are the roots of x2 + ax + b = 0, then the roots of the equation x2 + ax + b = 0 are …….
(a) 1, 2
(b) -1, 1
(c) 9, 1
(d) -1, 2
Solution:
(c) 9, 1
Hint:
Sum = 8 + 2 = 10 = -a ⇒ a = -10
Product = 3 × 3 = 9 = b ⇒ b = 9
Now the equation x2 + ax + b = 0
⇒ x2 – 10x + 9 = 0
⇒ (x- 9) (x – 1) = 0
x = 1 or 9

Question 16.
If a and b are the real roots of the equation x2 – kx + c = 0, then the distance
between the points (a, 0) and (b, 0) is ……..
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 17
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 18
Hint:
a + b = k, ab = c
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 19

Question 17.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 20
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 21

Question 18.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 22
(a) -1/2
(b) -2/3
(c) 1/2
(d) 2/3
Solution:
(a) -1/2
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 23

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13

Question 19.
The number of real roots of (x + 3)4 + (x + 5)4 = 16 is ……
(a) 4
(b) 2
(c) 3
(d) 0
Solution:
(a) 4
Hint:
The given equation is (x + 3)4 + (x + 5)4 = 16
Since it is a fourth degree equation it has four roots.
∴ Number of roots = 4

Question 20.
The value of log3 11 . log11 13 . log13 15 . log15 27 . log27 81 is …….
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(d) 4
Solution:
(d) 4
Hint.
log3 11 . log11 13 . log 13 15 . log 15 27 . log 27 81
= log3 13 . log 13 15 . log 15 27 . log 27 81
= log 3 15 . log 15 27 . log 27 81
= log 3 27 . log 27 81
= log 3 81
= log 334
= 4 log 33
= 4 × 1
= 4
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13 50

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