Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.13 Text Book Back Questions and Answers, Notes.

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

Question 1.

If |x + 2| ≤ 9 then x belongs to

(1) (- ∞,- 7)

(2) [- 11, 7]

(3) (-∞, – 7) ∪ [11, ∞]

(4) (-11, 7)

Answer:

(2) [- 11, 7]

Explanation:

-x – 2 ≤ 9 x + 2 ≤ 9

-x < 9 + 2 = 11 x ≤ 9 – 2 = 7

⇒ x ≥ -11

so x ∈ [-11, 7]

Question 2.

Given that x, y and b are real numbers x < y, b > 0 then

(1) xb < yb

(2) xb > yb

(3) xb < yb

(4) \(\frac{x}{b}\) ≥ \(\frac{y}{b}\)

Answer:

(1) xb < yb

Explanation:

Given x, y and b are real numbers and b ≥ 0. x > b

Multiplying by positive real number the inequality is not affected.

∴ xb < yb

Question 3.

If \(\frac{|x-2|}{x-2}\) ≥ 0 , then x belongs to

(1) [2, ∞)

(2) (2, ∞)

(3) (-∞, 2)

(4) (-2, ∞)

Answer:

(1) [2, ∞)

Explanation:

Given \(\frac{|x-2|}{x-2}\) ≥ 0

By the definition of mod function

|x – 2 | = – (x – 2) if x – 2 < 0 | x – 2 | = x – 2 if x – 2 > 0

Suppose x – 2 < 0 then

Question 4.

The solution of 5x – 1 < 24 and 5x + 1 > – 24 is

(1) (4, 5)

(2) (- 5, – 4)

(3) (- 5, 5)

(4) (- 5, 4)

Answer:

(3) (- 5, 5)

Explanation:

The given inequalities are

5x – 1 < 24 ——— (1)

5x + 1 > – 24 ——– (2)

(1) ⇒ 5x – 1 < 24

⇒ 5x < 24 + 1

⇒ 5x < 25 ⇒ x < 5 ——— (3)

(2) ⇒ 5x + 1 > – 24

⇒ 5x > – 24 – 1

⇒ 5x > – 25

⇒ x > – 5 ——– (4)

Combining (3) and (4), we have

-5 < x < 5

∴ x ∈ (- 5, 5)

Question 5.

The solution set of the following inequality

|x – 1| ≥ |x – 3| is

(1) [0, 2]

(2) [2, ∞)

(3) (0, 2)

(4) (-∞, 2)

Answer:

(2) [2, ∞)

Explanation:

The given inequality is |x – 1| ≥ |x – 3|

(x – 1)^{2} ≥ (x – 3)^{2}

x^{2} – 2x + 1 ≥ x^{2} – 6x + 9

– 2x + 1 ≥ – 6x + 9

6x – 2x + 1 – 9 ≥ 0

4x – 8 ≥ 0 ⇒ 4x ≥ 8

⇒ x ≥ 2

∴ The solution set of the given inequality lies in the interval (2, ∞)

Question 6.

The value of \(\log _{\sqrt{2}} 512\) is

(1) 16

(2) 18

(3) 9

(4) 12

Answer:

(2) 18

Explanation:

Question 7.

The value of \(\log _{3} \frac{1}{81}\) is

(1) – 2

(2) – 8

(3) – 4

(4) – 9

Answer:

(3) – 4

Explanation:

\(\log _{3} \frac{1}{81}\) = log_{3}1 – log_{3}81

= o – log_{3} 3^{4}

= – 4 log_{3}3

= – 4 × 1

= – 4

Question 8.

If log_{√x}0.25 = 4, then the value of x is

(1) 0.5

(2) 2.5

(3) 1.5

(4) 1.25

Answer:

(1) 0.5

Explanation:

log_{√x}0.25 = 4

By the definition of logarithm

0.25 = \((\sqrt{x})^{4}\)

(0.5 )2 = \(\left(x^{\frac{1}{2}}\right)^{4}\)

(0.5 )2 = x2

x = 0.5

Question 9.

The value of log_{a}b . log_{b}c . log_{c}a is

(1) 2

(2) 1

(3) 3

(4) 4

Answer:

(2) 1

Explanation:

log_{a}b . log_{b}c . log_{c}a = log_{a}c . log_{c}a

= log_{a}a = 1

Question 10.

If 3 is the logarithm of 343 then, the base is

(1) 5

(2) 7

(3) 6

(4) 9

Answer:

(2) 7

Explanation:

⇒ log_{x}343 = 3 ⇒ 343 = x^{3}

(.i.e.,) 7^{3} = x^{3} ⇒ x = 7

⇒ x = 7

Question 11.

Find a so that the sum and product of the roots of the equation 2x^{2} + (a – 3 ) x + 3a – 5 = 0 are equal is

(1) 1

(2) 2

(3) 0

(4) 4

Answer:

(2) 2

Explanation:

Given quadratic equation is

2x^{2} + (a – 3) x + 3a – 5 = 0

Let the roots be α, β

Sum of the roots α + β = \(-\frac{(a-3)}{2}\)

Product of the roots α β = \(\frac{3 a-5}{2}\)

Given α + β = α β

∴ \(-\frac{(a-3)}{2}=\frac{3 a-5}{2}\)

– a + 3 = 3a – 5

3a + a = 5 + 3

4a = 8 ⇒ a = 2

Question 12.

If a and b are the roots of the equation x^{2} – kx +16 = 0 and satisfy a^{2} + b^{2} = 3^{2}, then the value of k is

(1) 10

(2) – 8

(3) – 8, 8

(4) 6

Answer:

(3) – 8, 8

Explanation:

a + b = k ….(1) ab = 16 ….(2)

a^{2} + b^{2} = (a + b)^{2} – 2ab = 32 .

k^{2} – 32 = 32 ⇒ k^{2} = 64 ⇒ k = ±8

Question 13.

The number of solutions of x^{2} + |x – 1| = 1 is

(1) 1

(2) 0

(3) 2

(4) 3

Answer:

(3) 2

Explanation:

The given quadratic equatiuon is

x^{2} + |x – 1| = 1 ——– (1)

Case(i)

By the definition of mod function if x – 1 ≥ 0, then

|x – 1| = x – 1

∴ (1) ⇒ x^{2} + x – 1 = 1

x^{2} + x – 2 = 0

x^{2} + 2x – x – 2 = 0

x(x + 2) – 1 (x + 2) = 0

(x – 1)(x + 2) = 0

x – 1 = 0 or x + 2 = 0

x = 1 or x = – 2

Since x – 1 ≥ 0

x = – 2 is not possible. ∴ x = 1

Case (ii)

Again by the definition of mod function if x – 1 < 0 then

|x – 1| = – (x – 1)

∴ (1) ⇒ x^{2} – (x – 1) = 1

x^{2} – x + 1 = 1

x^{2} – x = 0

x (x – 1 ) = 0

x = 0 or x – 1 = 0

x = 0 or x = 1

Since x < 1, x = 1 is not possible

∴ x = 0

∴ The required solution set is {0, 1}

Number of solutions = 2

Question 14.

The equation whose roots are numerically equal but opposite in sign to the roots of 3x^{2} – 5x – 7 = 0 is

(1) 3x^{2} – 5x – 7 = 0

(2) 3x^{2} + 5x – 7 = 0

(3) 3x^{2} – 5x + 7 = 0

(4) 3x^{2} + x – 7 = 0

Answer:

(2) 3x^{2} + 5x – 7 = 0

Explanation:

The given quadratic equation is

3x^{2} – 5x – 7 = 0 ——— (1)

Let α and β be the roots of eqn (1)

Sum of the roots α + β = \(-\left(\frac{-5}{3}\right)\)

α + β = \(\frac{5}{3}\)

Product of the roots α β = – \(\frac{7}{3}\)

The quadratic equation whose roots are – α and – β is

x^{2} – (sum of the roots) x + Product of the roots = 0

x^{2} – (- α – β)x + (- α)(- β) = 0

x^{2} + (α + β)x + αβ = 0

x^{2} + \(\frac{5}{3}\) x – \(\frac{7}{3}\) = 0

3x^{2} + 5x – 7 = 0 is the required equation.

Question 15.

If 8 and 2 are the roots of x^{2} + ax + c = 0 and 3, 3 are the roots of x^{2} + dx + b = 0 , then the roots of the equation x^{2} + ax + b = 0 are

(1) 1, 2

(2) -1, 1

(3) 9, 1

(4) -1, 2

Answer:

(3) 9, 1

Explanation:

Given that 8 and 2 are the roots of the equation

x^{2} + ax + c = 0 ———- (1)

Sum of the roots 8 + 2 = \(-\frac{a}{1}\) ⇒ a = -10

Product of the roots 8 × 2 = \(\frac{c}{1}\) ⇒ c = 16

Also given 3 , 3 are the roots of

x^{2} + dx + b = 0 ——— (2)

Sum of the roots 3 + 3 = – \(\frac{d}{1}\) ⇒ d = – 6

Product of the roots 3 × 3 = \(\frac{b}{1}\) ⇒ b = 9

Let a and p be roots of the equation x^{2} + ax + b = 0

Sum of the roots α + β = – y

α + β = -(-10)

α + β = 10 ——– (3)

Product of the roots α β = y

α β = 9 ———- (4)

(α – β) = (α + β)^{2} – 4αβ

= 10^{2} – 4 × 9

= 100 – 36 = 64

α – β = 8 ——— (5)

Solving equations (3) and (5)

Substituting in equation (3),

9 + β = 10

⇒ β = 10 – 9 = 1

∴ The required roots are 9, 1

Question 16.

If a and b are the real roots of the equation x^{2} – kx + c = 0 , then the distance between the

points(a, 0) and (b, 0)is

(1) \(\sqrt{\mathbf{k}^{2}-4 \mathbf{c}}\)

(2) \(\sqrt{4 k^{2}-c}\)

(3) \(\sqrt{4 \mathbf{c}-\mathbf{k}^{2}}\)

(4) \(\sqrt{\mathbf{k}-8 \mathbf{c}}\)

Answer:

(1) \(\sqrt{\mathbf{k}^{2}-4 \mathbf{c}}\)

Explanation:

Given that a and b are the roots of the equation

x^{2} – kx + c = 0 ——— (1)

Sum of the roots a + b = \(-\frac{(-k)}{1}\)

a + b = k ———- (2)

Product of the roots ab = \(\frac{c}{1}\)

ab = c ——– (3)

Distance between the points ( a, 0) and (b, 0) is

Question 17.

If then the value of k is

(1) 1

(2) 2

(3) 3

(4) 4

Answer:

(3) 3

Explanation:

kx = 2x – 2 + x + 2

kx = 3x ⇒ k = 3

Question 18.

If , then the value of A + B is

(1) – \(\frac{1}{2}\)

(2) – \(\frac{2}{3}\)

(3) \(\frac{1}{2}\)

(4) \(\frac{2}{3}\)

Answer:

(1) – \(\frac{1}{2}\)

Explanation:

1 – 2x = A (x + 1) + B(3 – x) ——— (1)

Put x = – 1 in equation (1)

1 – 2 (- 1) = A (- 1 + 1) + B (3 + 1)

1 + 2 = 0 + 4B ⇒ B = \(\frac{3}{4}\)

Put x = 3 in equation (1)

1 – 2 × 3 = A(3 + 1) + B(3 – 3)

1 – 6 = 4A + 0

– 5 = 4A

Question 19.

The number of roots of (x + 3)^{4} + (x + 5)^{4} = 16 is

(1) 4

(2) 2

(3) 3

(4) 0

Answer:

(1) 4

Explanation:

The equation is (x + 3)^{4} + (x + 5)^{4} = 16

(x + 3)^{4} + (x + 5)^{4} = 2^{4}

This is biquadratic equation. It has 4 roots.

Question 20.

The value of

log_{3} 11 . log_{11} 13 . log _{13} 15 . log _{15} 27 . log _{27} 81 is

(1) 1

(2) 2

(3) 3

(4) 4

Answer:

(4) 4

Explanation:

log_{3} 11 . log_{11} 13 . log _{13} 15 . log _{15} 27 . log _{27} 81

= log_{3} 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 _{3}3^{4}

= 4 log _{3}3

= 4 × 1

= 4