Samacheer Class 11 Maths Basic Algebra Book Back Answers & Solutions

Q: Classify each element of {√7, -\frac{1}{4}, 0, 3.14, 4, \frac{22}{7}} as a member of N, Q, R – Q or Z.

√7 is an irrational number. ∴ √7 ∈ R – Q -\frac{1}{4} is a negative rational number. ∴ -\frac{1}{4} ∈ Q 0 is an integer. ∴ 0 ∈ Z, Q 3.14 is a rational number. ∴ 3.14 ∈ Q 4 is a positive integers. ∴ 4 ∈ Z, N, Q \frac{22}{7} is an rational number. ∴ \frac{22}{7} ∈ Q

Q: Prove that √3 is an irrational number.

Suppose that √3 is rational. Let √3 = \frac{\mathrm{m}}{\mathrm{n}} where m and n are positive integers with no common factors greater than 1. √3 = \frac{\mathrm{m}}{\mathrm{n}} ⇒ √3n = m ⇒ 3n 2 = m 2 ——– (1) By assumption n is an integer ∴ n 2 is an integer. Hence 3n 2 is an integral multiple of 3. ∴ From equation (1) m 2 is an integral multiple of 3 ⇒ m is an intergral multiple of 3 [Here m is a

Q: Are there two distinct irrational numbers such that their difference is a rational number? Justify.

Taking two irrational numbers as 3 + \sqrt{2} and 1 + \sqrt{2} Their difference is a rational number. But if we take two irrational numbers as 2 – \sqrt{3} and 4 + \sqrt{7}. Their difference is again an irrational number. So unless we know the two irrational numbers we cannot say that their difference is a rational number or irrational number.

Q: Find two irrational numbers such that their sum is a rational number. Can you find two irrational numbers whose product is a rational number?

(i) Let the two irrational numbers as 2 + \sqrt{3} and 3 – \sqrt{3} Their sum is 2 + \sqrt{3} + 3 – 3\sqrt{3} which is a rational number. But the sum of 3 + \sqrt{5} and 4 – \sqrt{7} is not a rational number. So the sum of two irrational numbers is either rational or irrational. (ii) Again taking two irrational numbers as π and \frac{3}{\pi} their product is \sqrt{3} and \sqrt{2} = \sqrt{3} × \sqr

Q: Find a positive number smaller than \frac{1}{2^{1000}} Justify.

The given number is \frac{1}{2^{1000}} We have 1000 \frac{1}{2^{1001}} ∴ A positive number smaller than \frac{1}{2^{1000}} is \frac{1}{2^{1001}}

Q: Solve for x (i) |3 – x| < 7

-7 -7 -x > -7 -3 (= -10) -x > -10 ⇒ x -4 ….(2) From (1) and (2) ⇒ x > -4 and x < 10 ⇒ -4 < x < 10 (ii) |4x – 5| ≥ – 2 |4x – 5| ≥ -2 (4x – 5) ≤ -(-2) or (4x – 5) ≥ -2 (4x – 5) ≤ 2 or (4x – 5) ≥ -2 4x ≤ 2 + 5 or 4x ≥ -2 + 5 ∴ x ∈ (-∞, ∞) = R (iii) |3 – \frac{3}{4}x| ≤ \frac{1}{4} Multiplying by 4, we have – 13 ≤ – 3x ≤ – 11 ——– (1) We k

Q: Solve – 3 |x| + 5 ≤ – 2 and graph the solution set in a number line.

-3|x| + 5 ≤ – 2 ⇒ -3 |x| ≤ – 2 – 5 (= -7) -3|x| ≤ – 7 ⇒ 3 |x| ≥ 7

Q: SoIve 2|x + 1| – 6 ≤ 7 and graph the solution set in a number line.

Given 2|x + 1| – 6 ≤ 7 2|x + 1| ≤ 7 + 6 2|x + 1| ≤ 13 |x + 1| ≤ \frac{13}{2}

Q: Solve \frac{1}{5} |10x – 2| < 1

Given \frac{1}{5} |10x – 2| < 1 |10x – 2| < 5 -5 < (10x – 2) < 5 – 5 + 2 < 10x < 5 + 2 – 3 < 10x < 7 -\frac{3}{10} < x < \frac{7}{10} ∴ The solution set is x ∈ \left(-\frac{3}{10}, \frac{7}{10}\right)

Q: Solve |5x – 12| < – 2

By the definition of modulus function. |5x – 12| always positive. ∴ |5x – 12| < -2 is not possible. ∴ Solution does not exist.

Your Progress - Chapter 20% complete

1Exercise 2.15 questions

Q.2.1.1Classify each element of {√7, \(-\frac{1}{4}\), 0, 3.14, 4, \(\frac{22}{7}\)} as a member of N, Q, R – Q or Z.v

Solution

√7 is an irrational number. ∴ √7 ∈ R – Q
\(-\frac{1}{4}\) is a negative rational number. ∴ \(-\frac{1}{4}\) ∈ Q
0 is an integer. ∴ 0 ∈ Z, Q
3.14 is a rational number. ∴ 3.14 ∈ Q
4 is a positive integers. ∴ 4 ∈ Z, N, Q
\(\frac{22}{7}\) is an rational number. ∴ \(\frac{22}{7}\) ∈ Q

Answer:

Q.2.1.2Prove that √3 is an irrational number.v

Solution

Suppose that √3 is rational. Let √3 = \(\frac{\mathrm{m}}{\mathrm{n}}\) where m and n are positive integers with no common factors greater than 1.
√3 = \(\frac{\mathrm{m}}{\mathrm{n}}\)
⇒ √3n = m
⇒ 3n 2 = m 2 ——– (1)
By assumption n is an integer
∴ n 2 is an integer. Hence 3n 2 is an integral multiple of 3.
∴ From equation (1) m 2 is an integral multiple of 3
⇒ m is an intergral multiple of 3
[Here m is an integer and m 2 is an integral multiple of 3. That m 2 is cannot take all integral multiples of 3. For example suppose m 2 = 3 = 1 × 3 which is an integral multiple of 3. In this case m = √3 which is not an integer. Suppose m 2 = 6 = 2 × 3 which is an integer multiple of 3, but m = √2 √3 which is an integer. Hence m 2 is an integral multiple of 3. Such that m is an integer.
Examples: m 2 = 4 × 9,
m 2 = 9,
m 2 = 9 × 9 etc.]
Let m = 3k
where k is an integer
Using equation (1) we have
3n 2 = (3k) 2
⇒ 3n 2 = 9k 2
⇒ n 2 = 3k 2
∴ n 2 is an integral multiple of 3. Since, n is an integer, we have n is also an integral multiple of 3.
Thus we have proved both m and n are integral multiple of 3. Hence both m and n have common factor 3, which is a contradiction to our assumption that m and n are integers with no common factors greater than 1.
Hence our assumption that √3 is a rational number is wrong.
∴ √3 is an irrational number.

Answer:

Q.2.1.3Are there two distinct irrational numbers such that their difference is a rational number? Justify.v

Solution

Taking two irrational numbers as 3 + \(\sqrt{2}\) and 1 + \(\sqrt{2}\)
Their difference is a rational number. But if we take two irrational numbers as 2 – \(\sqrt{3}\) and 4 + \(\sqrt{7}\).
Their difference is again an irrational number. So unless we know the two irrational numbers we cannot say that their difference is a rational number or irrational number.

Answer:

Q.2.1.4Find two irrational numbers such that their sum is a rational number. Can you find two irrational numbers whose product is a rational number?v

Solution

(i) Let the two irrational numbers as 2 + \(\sqrt{3}\) and 3 – \(\sqrt{3}\)
Their sum is 2 + \(\sqrt{3}\) + 3 – 3\(\sqrt{3}\) which is a rational number.
But the sum of 3 + \(\sqrt{5}\) and 4 – \(\sqrt{7}\) is not a rational number. So the sum of two irrational numbers is either rational or irrational.
(ii) Again taking two irrational numbers as π and \(\frac{3}{\pi}\) their product is \(\sqrt{3}\) and \(\sqrt{2}\) = \(\sqrt{3}\) × \(\sqrt{2}\) which is irrational, So the product of two irrational numbers is either rational or irrational.

Answer:

Q.2.1.5Find a positive number smaller than \(\frac{1}{2^{1000}}\) Justify.v

Solution

The given number is \(\frac{1}{2^{1000}}\)
We have 1000 < 1001
⇒ 2 1000 < 2 1001
⇒ \(\frac{1}{2^{1000}}\) > \(\frac{1}{2^{1001}}\)
∴ A positive number smaller than \(\frac{1}{2^{1000}}\) is \(\frac{1}{2^{1001}}\)

Answer:

2Exercise 2.25 questions

Q.2.2.1Solve for x (i) |3 – x| < 7v

Solution

-7 < 3 – x < 7 3 – x > -7
-x > -7 -3 (= -10)
-x > -10 ⇒ x < 10
3 – x < 7
– x < 7 – 3 (= 4)
– x < 4x > -4 ….(2)
From (1) and (2)
⇒ x > -4 and x < 10
⇒ -4 < x < 10
(ii) |4x – 5| ≥ – 2
|4x – 5| ≥ -2
(4x – 5) ≤ -(-2) or (4x – 5) ≥ -2
(4x – 5) ≤ 2 or (4x – 5) ≥ -2
4x ≤ 2 + 5 or 4x ≥ -2 + 5
∴ x ∈ (-∞, ∞) = R
(iii) |3 – \(\frac{3}{4}\)x| ≤ \(\frac{1}{4}\)
Multiplying by 4, we have
– 13 ≤ – 3x ≤ – 11 ——– (1)
We know that a < b ⇒ \(\frac{\mathrm{a}}{\mathrm{y}}\) > \(\frac{\mathrm{b}}{\mathrm{y}}\) when y < 0
Divide equation (1) by – 3, we have
(iv) |x| – 10 < – 3
|x| – 10 < – 3
|x| < – 3 + 10
|x| < 7
– 7 < x < 7
∴ The solution set is (-7, 7)

Answer:

Q.2.2.3Solve – 3 |x| + 5 ≤ – 2 and graph the solution set in a number line.v

Solution

-3|x| + 5 ≤ – 2
⇒ -3 |x| ≤ – 2 – 5 (= -7)
-3|x| ≤ – 7 ⇒ 3 |x| ≥ 7

Answer:

-3|x| + 5 ≤ – 2
⇒ -3 |x| ≤ – 2 – 5 (= -7)
-3|x| ≤ – 7 ⇒ 3 |x| ≥ 7

Q.2.2.4SoIve 2|x + 1| – 6 ≤ 7 and graph the solution set in a number line.v

Solution

Given 2|x + 1| – 6 ≤ 7
2|x + 1| ≤ 7 + 6
2|x + 1| ≤ 13
|x + 1| ≤ \(\frac{13}{2}\)

Answer:

Given 2|x + 1| – 6 ≤ 7
2|x + 1| ≤ 7 + 6
2|x + 1| ≤ 13
|x + 1| ≤ \(\frac{13}{2}\)

Q.2.2.5Solve \(\frac{1}{5}\) |10x – 2| < 1v

Solution

Given \(\frac{1}{5}\) |10x – 2| < 1
|10x – 2| < 5
-5 < (10x – 2) < 5
– 5 + 2 < 10x < 5 + 2
– 3 < 10x < 7
\(-\frac{3}{10}\) < x < \(\frac{7}{10}\)
∴ The solution set is x ∈ \(\left(-\frac{3}{10}, \frac{7}{10}\right)\)

Answer:

Q.2.2.6Solve |5x – 12| < – 2v

Solution

By the definition of modulus function. |5x – 12| always positive.
∴ |5x – 12| < -2 is not possible.
∴ Solution does not exist.

Answer:

By the definition of modulus function. |5x – 12| always positive.
∴ |5x – 12| < -2 is not possible.
∴ Solution does not exist.

3Exercise 2.39 questions

Q.2.3.1Represent the following inequalities in the interval notation: (i) x ≥ – 1 and x < 4v

Solution

x ≥ – 1 and x < 4
x ∈ [- 1, 4)
(ii) x ≤ 5 and x ≥ – 3
x ≤ 5 and x ≥ – 3
– 3 ≤ x ≤ 5
∴ x ∈ [- 3, 5 ]
(iii) x < – 1 or x < 3
x < – 1 or x < 3
x ∈ (-∞, 3)
(iv) -2x > 0 or 3x – 4 < 11
– 2x > 0 or 3x – 4 < 11
2x < 0 or 3x < 11 + 4
x < 0 or x < \(\frac{15}{3}\)
x < 0 or x < 5
x ∈ (- ∞, 5)

Answer:

Q.2.3.2Solve 23x < 100 when (i) x is a natural number, (ii) x is an integer.v

Solution

Given 23x < 100
(i) When x is a natural number
23x < 100
⇒ x < \(\frac{100}{23}\)
⇒ x < 4.347
⇒ x = 1, 2, 3, 4
∴ The solution set is { 1, 2, 3, 4 }
(ii) When x is an integer
23x < 100
⇒ x < \(\frac{100}{23}\)
⇒ x < 4.347
⇒ x = ……, – 3, – 2, – 1, 0, 1, 2, 3, 4
Hence the solution set is
{ ………, – 3, – 2, – 1, 0, 1, 2, 3, 4 }

Answer:

Q.2.3.3Solve – 2x ≥ 9 when (i) x is a real number, (ii) x is an integer (iii) x is a natural number.v

Solution

Given – 2x ≥ 9
⇒ – x ≥ \(\frac{9}{2}\)
⇒ x ≤ \(-\frac{9}{2}\)
(i) When x is a real number
x ≤ \(-\frac{9}{2}\)
The solution set is \(\left(-\infty,-\frac{9}{2}\right]\)
(ii) x is an integer
x ≤ –\(-\frac{9}{2}\)
x ≤ – 4.5
x = …………., -7, – 6, -5
(iii) x is a natural number
x ≤ – \(\frac{9}{2}\)
x ≤ – 4.5
Since there exists no natural number less than – \(\frac{9}{2}\)
∴ No solution

Answer:

Q.2.3.5To secure an A grade, one must obtain an average of 90 marks or more in 5 subjects each of a maximum of 100 marks. If one scored 84, 87, 95, 91 in the first four subjects, what is the minimum mark one scored in the fifth subject to get an A grade in the course?v

Solution

Required marks = 5 × 90 = 450
Total marks obtained in 4 subjects = 84 + 87 + 95 + 91 = 357
So required marks in the fifth subject = 450 – 357 = 93

Answer:

Required marks = 5 × 90 = 450
Total marks obtained in 4 subjects = 84 + 87 + 95 + 91 = 357
So required marks in the fifth subject = 450 – 357 = 93

Q.2.3.6A manufacturer has 600 litres of a 12 percent solution of acid. How many litres of a 30 percent acid solution must be added to it so that the acid content in the resulting mixture will be more than 15 percent but less than 18 percent?v

Solution

Amount of 12% solution of acid = 600 litres
Let x be the required number litres of 30 % acid solution to be added to the given 600 litres of 12 % acid solution to make the resulting mixture will be more than 15 % but less than 18 %.
∴ Total amount of mixture = (600 + x) litres
30% acid solution of x litres + 12% acid solution of 600 litres > 15% acid solution of (600 + x) litres
\(\frac{30}{100}\) × x + \(\frac{12}{100}\) × 600 > \(\frac{15}{100}\) × (600 + x)
30x + 7200 > 9000 + 15x
30x – 15x > 9000 – 7200
15x > 1800
x > \(\frac{1800}{15}\) = 120
x > 120 ——– (1)
Also 30% acid solution of x litres + 12% acid solution of 600 litres < 18% acid solution of ( 600 + x) litres.
\(\frac{30}{100}\) × x + \(\frac{12}{100}\) × 600 < \(\frac{15}{100}\) × (600 + x)
30x + 7200 < 18 (600 + x)
30x + 7200 < 10800 + 18x
30x – 18x < 10,800 – 7200
12x < 3600
x < \(\frac{3600}{12}\) = 300
x < 300 ——- (2)
From equations (1) and (2), we get ) 120 < x < 300
∴ The numbers of litres of the 30% acid solution to be added is greater than 120 litres and less than 300 litres.

Answer:

Q.2.3.7Find all pairs of consecutive odd natural numbers both of which are larger than 10 and their sum is less than 40.v

Solution

Let x and x + 2 be the two pair of consecutive odd natural numbers.
Given x > 10 ——— (1)
and x + 2 > 10 ——— (2)
Also x + (x + 2) < 40 ——— (3)
From equations (1), we have
x = 11, 13, 15, 17, 19, 21 …………
Using equation (3), the required pairs are
(11, 13), (13, 15), (15, 17), ( 17, 19), (19,21 ) is not possible since 19 + 21 = 40

Answer:

Q.2.3.8A model rocket is launched from the ground. The height h reached by the rocket after t seconds from lift off is given by h(t) = – 5t 2 + 100t, 0 ≤ t ≤ 20. At what times, the rocket is 495 feet above the ground?v

Solution

h(t) = -5t 2 + 100t
at t = 0, h(0) = 0
at t = 1, h(1) = -5 + 100 = 95
at t = 2, h(2) = -20 + 200 = 180
at t =3, h(3) = -45 + 300 = 255
at t = 4, h(4) = -80 + 400 = 320
at t = 5, h(5) = -125 + 500 = 375
at t = 6, h(6) = – 180 + 600 = 420
at t = 7, h(7) = -245 + 700 = 455
at t = 8, h(8) = – 320 + 800 = 480
at t = 9, h(9) = -405 + 900 = 495
So, at 9 secs, the rocket is 495 feet above the ground.

Answer:

Q.2.3.9A plumber can be paid according to the following schemes: In the first scheme he will be paid rupees 500 plus rupees 70 per hour and in the second scheme, he will pay rupees 120 per hour. If he works x hours, then for what value of x does the first scheme give better wages?v

Solution

I scheme with x hr
500 + (x- 1) 70 = 500 + 70x – 70 = 430 + 70x
II scheme with x hours
120x
Here I > II
⇒ 430 + 70x > 120x
⇒ 120x – 70x < 430
50x < 430
\(\frac{50 x}{50}<\frac{430}{50}\)
x < 8.6 (i.e.) when x is less than 9 hrs the first scheme gives better wages.

Answer:

Q.2.3.10A and B are working on similar jobs but their monthly salaries differ by more than Rs. 6000. If B earns rupees 27,000 per month, then what are the possibilities of A’s salary per month?v

Solution

A’s monthly salary = ₹ x
B’s monthly salary = ₹ 27000
Their annual salaries differ by ₹ 6000
A’s salary – 27000 > 6000
A’s salary > ₹ 33000
B’s salary – A’s salary > 6000
27000 – A’s salary > 6000
A’s salary < ₹ 21000
A’s monthly salary will be lesser than ₹ 21,000 or greater than ₹ 33,000

Answer:

4Exercise 2.410 questions

Q.2.4.1Construct a quadratic equation with roots 7 and – 3.v

Solution

The given roots are 7 and -3
Let α = 7 and β = -3
α + β = 7 – 3 = 4
αβ = (7)(-3) = -21
The quadratic equation with roots α and β is x 2 – (α + β) x + αβ = 0
So the required quadratic equation is
x 2 – (4) x + (-21) = 0
(i.e.,) x 2 – 4x – 21 = 0

Answer:

Q.2.4.2A quadratic polynomial has one of its zero 1 + √5 and it satisfies p(1) = 2. Find the quadratic polynomial.v

Solution

Let p(x) = ax 2 + bx + c be the required quadratic polynomial.
Given p (1) = 2, we have
a × 1 2 + b × 1 + c = 2
a + b + c = 2 ——— (1)
Also given 1 + √5 is a zero of p(x)
∴ a(1 + √5) 2 + b (1 + √5) + c = 0
a( 1 + 5 + 2√5) + b (1 + √5) + c = 0
6a + 2a√5 + b + b√5 + c = 0 ——— (2)
If 1 + √5 is zero then 1 – √5 is also a zero of p (x).
∴ a(1 – √5) 2 + b (1 – √5) + c = 0
a( 1 – 2√5 + 5) + b (1 – √5) + c = 0
6a – 2a√5 + b – b√5 + c = 0 ——— (3)
Substituting the value of a in equation (4)
5 × – \(\frac{2}{5}\) + 2 × – \(\frac{2}{5}\) × √5 + b√5 = – 2
– 2 – \(\frac{4}{5}\)√5 + b√5 = – 2
b√5 = – 2 + 2 + \(\frac{4}{5}\). √5
b√5 = \(\frac{4}{5}\). √5
b = \(\frac{4}{5}\)
Substituting the value of a and b in equation (1), we have
∴ The required quadratic polynomial is
p(x) = \(-\frac{2}{5}\)x 2 + \(\frac{4}{5}\)x + \(\frac{8}{5}\)
p(x) = \(-\frac{2}{5}\)(x 2 – 2x – 4)

Answer:

Q.2.4.3If α and β are the roots of the quadratic equation x 2 + √2x + 3 = 0 form a quadratic polynomial with zeros \(\frac{1}{\alpha}, \frac{1}{\beta}\).v

Solution

Given α and β are the roots of the quadratic polynomial
x 2 + √2x + 3 = 0 ——— (1)
∴ The required quadratic equation whose roots are \(\frac{1}{\alpha}, \frac{1}{\beta}\) is
x 2 – (sum of the roots)x + product of the roots = 0

Answer:

Q.2.4.4If one root of k (x – 1) 2 = 5x – 7 is double the other root, show that k = 2 or – 25v

Solution

The given quadratic equation is
k(x – 1) 2 = 5x – 7
k(x 2 – 2x + 1) – 5x + 7 = 0
kx 2 – 2kx + k – 5x + 7 = 0
kx 2 – (2k + 5)x + k + 7 = 0 ———- (1)
Let the roots be α and 2α
Sum of the roots α + 2α = –\(\frac{b}{a}\)
Product of te roots α(2α) = \(\frac{c}{a}\)
Using equation (2) and (3) we have
2(4k 2 + 20k + 25) = 9k(k + 7)
8k 2 + 40k + 50 = 9k 2 + 63k
9k 2 + 63k – 8k 2 – 40k – 50 = 0
k 2 + 23k – 50 = 0
k 2 + 25k – 2k – 50 = 0
k(k + 25) – 2(k + 25) = 0
(k – 2) (k + 25) = 0
k – 2 = 0 or k + 25 = 0
k = 2 or k = – 25

Answer:

Q.2.4.5If the difference of the roots of the equation 2x 2 – (a + 1)x + a – 1 = 0 is equal to their product then prove that a = 2.v

Solution

The given quadratic equation is
2x 2 – (a + 1) x + a – 1 = 0 ———– (1)
Let α and β be the roots of the given equation
Given that α – β = αβ —— (2)
Sum of the roots α + β = – \(\frac{b}{a}\)
Substituting the values of α and β in equation (2)
2(a – 1) = a
2a – 2 – a = 0
a – 2 = 0
⇒ a = 2

Answer:

Q.2.4.6Find the condition that one of the roots of ax 2 + bx + c may be (a) negative of the other (b) thrice the other (c) reciprocal of the other.v

Solution

The given quadratic equation is
ax 2 + bx + c = 0 ——- (1)
Let α and β be the roots of the equation (1) then
Sum of the roots α + β = ——- (2)
Product of the roots αβ = ——- (3)
(a) Given one root is the negative of the other
β = – α
(2) ⇒ α + (-α) = – \(\frac{b}{a}\)
0 = – \(\frac{b}{a}\)
⇒ b = 0
(3) ⇒ α(-α) = \(\frac{c}{a}\)
– α 2 = \(\frac{c}{a}\)
Hence the required condition is b = 0
(b) Given that one root is thrice the other
β = 3α
When is the required condition?
(c) One root is reciprocal of the other
When is the required condition?

Answer:

Q.2.4.7If the equations x 2 – ax + b = 0 and x 2 – ex + f = 0 have one root in common and if the second equation has equal roots then prove that ae = 2(b + f).v

Solution

The given quadratic equations are
x 2 – ax + b = 0 ———- (1)
x 2 – ex + f = 0 ——— (2)
Let α be the common root of the given quadratic equations (1) and (2)
Let α, β be the roots of x 2 – ax + b = 0
Sum of the roots α + β = \(-\left(-\frac{a}{1}\right)\)
α + β = a ———- (3)
Product of the roots αβ = \(\frac{b}{1}\)
αβ = b ——– (4)
Given that the second equation has equal roots.
∴ The roots of the second equation are a, a
Sum of the roots α + α = \(-\left(-\frac{e}{1}\right)\)
2α = e ——— (5)
Product of the roots α.α = \(\frac{f}{1}\)
α 2 = f ———- (6)
ae = (α + β)2α (Multiplying equations (3) and (5))
ae = 2α2 + 2αβ
ae= 2 (f) + 2b From equations (4) and (6)
ae= 2(f + b) Hence proved.

Answer:

Q.2.4.8Discuss the nature of roots of (i) – x 2 + 3x + 1 = 0 (ii) 4x 2 – x – 2 = 0 (iii) 9x 2 + 5x = 0.v

Solution

(i) -x 2 + 3x + 1 = 0
⇒ comparing with ax 2 + bx + c = 0
∆ = b 2 – 4ac = (3) 2 – 4(1)(-1) = 9 + 4 = 13 > 0
⇒ The roots are real and distinct
(ii) 4x 2 – x – 2 = 0
a = 4, b = -1, c = -2
∆ = b 2 – 4ac = (-1) 2 – 4(4)(-2) = 1 + 32 = 33 >0
⇒ The roots are real and distinct
(iii) 9x 2 + 5x = 0
a = 9, b = 5, c = 0
∆ = b 2 – 4ac = 5 2 – 4(9)(0) = 25 > 0
⇒ The roots are real and distinct

Answer:

Q.2.4.9Without sketching the graphs, find whether the graphs of the following functions will intersect the x-axis and if so in how many points. (i) y = x 2 + x + 2 (ii) y = x 2 – 3x – 7 (iii) y = x 2 + 6x + 9v

Solution

(i) y = x 2 + x + 2
y = x 2 + x + 2 ——— (1)
Compare this equation with the equation
ax 2 + bx + c = 0
we have a = 1, b = 1, c = 2
b 2 – 4ac = 1 2 – 4 × 1 × 2 = 1 – 8
b 2 – 4ac = – 7 < 0
Since the discriminant is negative the quadratic equation has no real roots and therfore the graph does not meet x-axis.
(ii) y = x 2 – 3x – 7
y = x 2 – 3x – 7 ——— (2)
Compare this equation with the equation ax 2 + bx + c = 0
we have a = 1, b = – 3, c = – 1
b 2 – 4ac = (-3) 2 – 4(1)(-1)
= 9 + 4
b 2 – 4ac = 13 > 0
Since the discriminant is positive the quadratic equation has real and distinct roots and therefore the graph intersect the x – axis at two different points,
(iii) y = x 2 + 6x + 9
y = x 2 + 6x + 9 ——— (3)
Compare this equation with the equation
ax 2 + bx + c = 0
we have a = 1, b = 6, c = 9
b 2 – 4ac = 6 2 – 4 × 1 × 9
= 36 – 36 =0
Since the discriminant is zero the quadratic equation has real and equal roots and therefore the graph touches the x-axis at one point.

Answer:

Q.2.4.10Write f(x) = x 2 + 5x + 4 in completed square form.v

Solution

The given quadratic equation is
f(x) = x 2 + 5x + 4
Let y = x 2 + 5x + 4
y – 4 = x 2 + 5x

Answer:

The given quadratic equation is
f(x) = x 2 + 5x + 4
Let y = x 2 + 5x + 4
y – 4 = x 2 + 5x

5Exercise 2.52 questions

Q.2.5.1Solve 2x 2 + x – 15 ≤ 0v

Solution

The given inequality is
2x 2 + x – 15 ≤ 0 ——— (1)
2x 2 + x – 15 = 2x 2 + 6x – 5x – 15
= 2x (x + 3) – 5 (x + 3)
= (2x – 5)(x + 3)
2x 2 + x – 15 = 2\(\left(x-\frac{5}{2}\right)\))(x + 3) ——— (2)
The critical numbers are x – \(\frac{5}{2}\) = 0 or x + 3 = 0
The critical numbers are x = \(\frac{5}{2}\) or x = – 3
Divide the number line into three intervals
(i) (- ∞, – 3)
When x < – 3 say x = – 4
The factor x – \(\frac{5}{2}\) = – 4 – \(\frac{5}{2}\) < 0 and
x + 3 = – 4 + 3 = – 1 < 0
x – \(\frac{5}{2}\) < 0 and x + 3 < 0
⇒ \(\left(x-\frac{5}{2}\right)\) (x + 3) > 0
Using equation (2) 2x 2 + x – 15 > 0
∴ 2x 2 + x – 15 ≤ 0 is not true in (- ∞, – 3)
(ii) \(\left(-3, \frac{5}{2}\right)\)
When – 3 < x < \(\frac{5}{2}\) say x = 0
The factor x – \(\frac{5}{2}\) = 0 – \(\frac{5}{2}\) = – \(\frac{5}{2}\) < 0 and
x + 3 = 0 + 3 = 3 > 0
x – \(\frac{5}{2}\) < 0 and x + 3 > 0
⇒ \(\left(x-\frac{5}{2}\right)\) (x + 3) < 0
using equation (2) 2x 2 + x – 15 < 0
∴ 2x 2 + x – 15 ≤ 0 is true in \(\left(-3, \frac{5}{2}\right)\)
(iii) \(\left(\frac{5}{2}, \infty\right)\)
When x > \(\frac{5}{2}\) say x = 3
The factor x – \(\frac{5}{2}\) = 3 – \(\frac{5}{2}\) > 0 and
x + 3 = 3 + 3 > 0
x – \(\frac{5}{2}\) > 0 and x + 3 > 0
= \(\left(x-\frac{5}{2}\right)\) (x + 3) > 0
Using equation (2) 2x 2 + x – 15 > 0
∴ 2x 2 + x – 15 ≤ 0 is not true in \(\left(\frac{5}{2}, \infty\right)\)
We have proved the inequality 2x 2 + x – 15 ≤ 0 is true in the interval \(\left(-3, \frac{5}{2}\right)\)
But it is not true in the interval
Inequality solver that solves an inequality with the details of the calculation: linear inequality, quadratic inequality.

Answer:

Q.2.5.2Solve x 2 + 3x – 2 ≥ 0v

Solution

The given inequality is
– x 2 + 3x – 2 ≥ 0
x 2 – 3x + 2 < 0 ——– (1)
x 2 – 3x + 2 = x 2 – 2x – x + 2
= x(x – 2) – 1(x – 2)
x 2 – 3x + 2 = (x – 1) (x – 2) ——— (2)
The critical numbers are
x – 1 = 0 or x – 2 = 0
The critical numbers are
x = 1 or x = 2
Divide the number line into three intervals
(- ∞, 1), (1, 2) and (2, ∞).
(i) (- ∞, 1)
When x < 1 say x = 0
The factor x – 1 = 0 – 1 = – 1 < 0 and
x – 2 = 0 – 2 = – 2 < 0
x – 1 < 0 and x – 2 < 0
⇒ (x – 1)(x – 2) > 0
Using equation (2) x 2 – 3x + 2 > 0
∴ The inequality x 2 – 3x + 2 ≤ 0 is not true in the interval (- ∞, 1 )
(ii) (1, 2)
When x lies between 1 and 2 say x = \(\frac{3}{2}\)
The factor x – 1 = \(\frac{3}{2}\) – 1 = \(\frac{1}{2}\) > 0 and
x – 2 = \(\frac{3}{2}\) – 2 = – \(\frac{1}{2}\) – < 0
x – 1 > 0 and x – 2 < 0
⇒ (x – 1)(x – 2) < 0
Using equation (2) x 2 – 3x + 2 < 0
∴ The inequality x 2 – 3x + 2 ≤ 0 is true in the interval (1, 2 )
(iii) (2, ∞)
When x > 2 say x = 3
The factor x – 1 = 3 – 1 = 2 > 0 and
x – 2 = 3 – 2 = 1 > 0
x – 1 > 0 and x – 2 > 0
= (x – 1)(x – 2) > 0
Using equation (2) x 2 – 3x + 2 > 0
∴ The inequality x 2 – 3x + 2 ≤ 0 is not true in the interval (2, ∞)
We have proved the inequality x 2 – 3x + 2 ≤ 0 is true in the interval [ 1, 2 ].
But it is not true in the interval
(- ∞, 1) and (2, ∞)
∴ The solution set is [1, 2]

Answer:

6Exercise 2.64 questions

Q.2.6.1Find the zeros of the polynomial function f(x) = 4x 2 – 25v

Solution

Given f(x) = 4x 2 – 25
To find the zeors of f(x), put f(x) = 0
∴ 4x 2 – 25 = 0
⇒ 4x 2 = 25
⇒ x 2 = \(\frac{25}{4}\)
⇒ x = ±\(\sqrt{\frac{25}{4}}\) = ±\(\frac{5}{2}\)
Hence the zeros of f(x) are \(-\frac{5}{2}, \frac{5}{2}\)

Answer:

Q.2.6.2If x = – 2 is one root of x 3 – x 2 – 17x = 22, then find the other roots of equation.v

Solution

Let f(x) = x 3 – x 2 – 17x – 22 = 0 —– (1)
Given that x = – 2 is a root of f(x).
∴ x + 2 is a factor of f (x)
Using synthetic division
Comparing equation (1) with the equation ax 2 + bx + c = 0 we have
a = 1, b = – 3, c = – 11

Answer:

Q.2.6.3Find the real roots of x 4 = 16.v

Solution

x 4 = 16
⇒ x 4 – 16 = 0
(i.e.,) x 4 – 4 2 = 0
⇒ (x 2 + 4)(x 2 – 4) = 0
x 2 + 4 = 0 will have no real roots
so solving x 2 – 4 = 0
x 2 = 4

Answer:

x 4 = 16
⇒ x 4 – 16 = 0
(i.e.,) x 4 – 4 2 = 0
⇒ (x 2 + 4)(x 2 – 4) = 0
x 2 + 4 = 0 will have no real roots
so solving x 2 – 4 = 0
x 2 = 4

Q.2.6.4Solve (2x + 1) 2 – (3x + 2) 2 = 0v

Solution

The given equation is (2x + 1) 2 (3x + 2) 2 = 0
(2x + 1 + 3x + 2) [(2x + 1) – (3x + 2)] = 0
[a 2 – b 2 = (a + b) (a – b)]
(5x + 3) (2x + 1 – 3x – 2) = 0
(5x + 3)(- x – 1) = 0
– (5x + 3)(x + 1) = 0
5x + 3 = 0 or x + 1 = 0
x = – \(\frac{3}{5}\) or x = – 1
∴ Solution set is { – 1, \(\frac{3}{5}\)}

Answer:

7Exercise 2.72 questions

Q.2.7.1Factorize x 4 + 1v

Solution

The given equation is x 4 + 1
x 4 + 1 = (x 2 ) 2 + 1 2
= (x 2 + 1) 2 – 2 (x 2 ) (1)
[ a 2 + b 2 = (a + b) 2 – 2ab]
= (x 2 + 1) 2 – (√2x) 2
= (x 2 + 1 + √2x) (x 2 + 1 – √2x)
x 4 + 1 = (x 2 + 2x + 1) (x 2 – √2x + 1)

Answer:

Q.2.7.2If x 4 + x + 1 is a factor of the polynomial 3x 3 + 8x 2 + 8x + a, then find the value of a.v

Solution

Given that x 2 + x + 1 is a factor of the polynomial 3x 3 + 8x 2 + 8x + a.
∴ 3x 3 + 8x 2 + 8x + a is divisible by x 2 + x + 1
Since 3x 3 + 8x 2 + 8x + a is divisible by x 2 + x + 1, the remainder must be zero.
a – 5 = 0
⇒ a = 5

Answer:

8Exercise 2.83 questions

Q.2.8.1Find all values of x for which \(\frac{x^{3}(x-1)}{x-2}\) > 0v

Solution

The given inequality is f(x) = \(\frac{x^{3}(x-1)}{x-2}\) > 0
[The critical numbers of f(x) are those values of x for which f(x) = 0, and those values of x for which f(x) is not defined.
When x = 2, f(x) = ∞ ⇒ f(x) is not defined.]
The critical numbers are x = 0, 1, 2
Divide the number line into 4 intervals
(- ∞, 0), (0, 1), (1, 2) and (2, ∞)
(1) (- ∞, 0)
When x < 0 say x = – 1
The factor x 3 = (- 1) 3 = – 1 < 0
The factor x – 1 = – 1 – 1 = – 2 < 0
The factor x – 2 = – 1 – 2 = – 3 < 0
∴ \(\frac{x^{3}(x-1)}{x-2}\) < 0
Thus \(\frac{x^{3}(x-1)}{x-2}\) > 0 is not true in the interval (- ∞, 0)
Therefore, it has no solution in the interval (- ∞, 0)
(2) (0, 1)
When 0 < x < 1 say x = 0.5
The factor x 3 = (0.5 ) 3 > 0
The factor x – 1 = 0.5 – 1 = – 0.5 < 0
The factor x – 2 = 0.5 – 2 = – 1.5 < 0
Thus x 3 > 0, x – 1 < 0 and x – 2 < 0
∴ \(\frac{x^{3}(x-1)}{x-2}\) > 0
Thus \(\frac{x^{3}(x-1)}{x-2}\) > 0 is true in the interval (0, 1)
Therefore it has solution in (0,1)
(3) (1, 2)
When 1 < x < 2 say x = 1.5
The factor x 3 = 0
The factor x – 1 = 1.5 – 1 = 0.5 > 0
The factor x – 2 = 1.5 – 2 = – 0.5 < 0
Thus x 3 > 0, x – 1 > 0 and x – 2 < 0
∴ \(\frac{x^{3}(x-1)}{x-2}\) < 0
Thus \(\frac{x^{3}(x-1)}{x-2}\) > 0 is not true in the interval (1, 2).
Therefore it has no solution in (1, 2).
(4) (2, ∞)
When x > 2 say x = 3
The factor x 3 = 3 3 > 0
The factor x – 1 = 3 – 1 = 2 > 0
The factor x – 2 = 3 – 2 = 1 > 0
Thus x 3 > 0, x – 1 > 0 and x – 2 > 0
∴ \(\frac{x^{3}(x-1)}{x-2}\) > 0
Thus \(\frac{x^{3}(x-1)}{x-2}\) > 0 is true in the interval (2, ∞).
Therefore it has a solution in (2, ∞)
∴ \(\frac{x^{3}(x-1)}{x-2}\) > 0 has solution in the intervals (0, 1) and (2, ∞)
∴ The solution set is given by (0, 1) ∪ (2, ∞)

Answer:

Q.2.8.2Find all values of x that satisfies the inequality \(\frac{2 x-3}{(x-2)(x-4)}\) < 0.v

Solution

The given inequality is
[The critical numbers of f(x) are those values of x for which f(x) = 0, and those values of x for which f(x) is not defined. When x = 2, f(x) = ∞ ⇒ f(x) is not defined.]
The critical numbers are x = \(\frac{3}{2}\), x = 2, x = 4
Divide the number into 4 intervals
(1) \(\left(-\infty, \frac{3}{2}\right)\)
When x < \(\frac{3}{2}\) say x = 0
The factor x – \(\frac{3}{2}\) = 0 – \(\frac{3}{2}\) < 0
The factor x – 2 = 0 – 2 < 0
The factor x – 4 = 0 – 4 < 0
Thus x – \(\frac{3}{2}\) < 0, x – 2 < 0 and x – 4 < 0
(2) \(\left(\frac{3}{2}, 2\right)\)
(3) (2, 4)
When 2 < x < 4 say x = 3 The factor x – \(\frac{3}{2}\) = 3 – \(\frac{3}{2}\) = \(\frac{3}{2}\) > 0
The factor x – 2 = 3 – 2 = 1 > 0
The factor x – 4 = 3 – 4 = – 1 < 0 Thus x – \(\frac{3}{2}\) > 0, x – 2 > 0 and x – 4 < 0
Thus \(\frac{2 x-3}{(x-2)(x-4)}\) < 0 is true in the interval (2, 4) ∴ It has solution in (2, 4). (4) (4, ∞) When x > 4 say x = 5
The factor x – \(\frac{3}{2}\) = 5 – \(\frac{3}{2}\) = \(\frac{7}{2}\) > 0
The factor x – 2 = 5 – 2 = 3 >0
The factor x – 4 = 5 – 4 = 1 > 0
Thus x – \(\frac{3}{2}\) > 0, x – 2 > 0 and x – 4 > 0
Thus \(\frac{2 x-3}{(x-2)(x-4)}\) < 0 is not true in the interval (4, ∞)
∴ It has a solution in (4, ∞)

Answer:

Q.2.8.3Solve: \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0v

Solution

[The critical numbers of f(x) are those values of x for which f(x) = 0, and those values of x for which f(x) is not defined. When x = – 3, 5. f(x) = ∞ ⇒ f(x) is not defined.]
The critical numbers are x = – 2, 2, – 3, 5
Divide the number line into five intervals
(- ∞, – 3), (- 3, – 2), (- 2, 2), (2, 5),(5, ∞)
(a) (- ∞, – 3)
When x <- 3 say x = – 4
The factor x + 2 = – 4 + 2 = – 2 < 0
The factor x – 2 = – 4 – 2 = 6 < 0
The factor x + 3 = – 4 + 3 = – 1 < 0
The factor x – 5 = – 4 – 5 = – 9 < 0
Thus x + 2 < 0, x + 3 < 0, x – 2 < 0, x – 5 < 0
∴ \(\frac{x^{2}-4}{x^{2}+4 x-15}\) > 0
Thus \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0 is not true in the interval (- ∞, – 3).
∴ It has no solution in (- ∞, – 3)
(b) (- 3, – 2)
When – 3 < x ≤ – 2 say x = – 2.5
The factor x + 2 = – 2.5 + 2 = – 0.5 < 0
The factor x – 2 = – 2.5 – 2 = – 4.5 < 0 The factor x + 3 = – 2.5 + 3 = 0.5 > 0
The factor x – 5 = – 2.5 – 5 = – 7.5 < 0
Thus x + 2 < 0, x + 3 > 0
and
x – 2 < 0
x – 5 < 0
∴ \(\frac{x^{2}-4}{x^{2}+4 x-15}\) < 0
Thus \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0 is not true in the interval (- 3, – 2).
∴ It has no solution in (- 3, – 2)
(c) (-2, 2)
When – 2 ≤ x ≤ 2 say x = 0
The factor x + 2 = 0 + 2 = 2 > 0
The factor x – 2 = 0 – 2 = – 2 < 0
The factor x + 3 = 0 + 3 = 3 > 0
The factor x – 5 = 0 – 5 = – 5 < 0
Thus x + 2 > 0,
x + 3 > 0
and
x – 2 < 0
x – 5 < 0
∴ \(\frac{x^{2}-4}{x^{2}+4 x-15}\) > 0
Thus \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0 is not true in the interval (- 2, – 2).
∴ It has no solution in (- 2, – 2)
(d) (2, 5)
When 2 ≤ x < 5 say x = 3 The factor x + 2 = 3 = 3 + 2 = 5 > 0
The factor x – 2 = 3 – 2 = 1 > 0
The factor x + 3 = 3 + 3 = 6 > 0
The factor x – 5 = 3 – 5 = – 2 < 0 Thus x + 2 > 0,
x + 3 > 0
and
x – 2 > 0
x – 5 < 0
∴ \(\frac{x^{2}-4}{x^{2}+4 x-15}\) < 0
Thus \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0 is not true in the interval (2, 5).
∴ It has no solution in (2, 5)
(e) (5, ∞)
When 5 < x < ∞ say x = 6 The factor x + 2 = 6 + 2 = 8 > 0
The factor x – 2 = 6 – 2 = 4 > 0
The factor x + 3 = 6 + 3 = 9 > 0
The factor x – 5 = 6 – 5 = 1 > 0
Thus
x + 2 > 0,
x + 3 > 0
and
x – 2 > 0,
x – 5 > 0
∴ \(\frac{x^{2}-4}{x^{2}+4 x-15}\) > 0
Thus \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0 is not true in the interval (5, ∞).
∴ It has no solution in (5, ∞)
The given inequality f(x) = \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0 has solution in the intervals (-3, – 2]
∴ The solution set is (-3, 2] ∪ [2, 5)

Answer:

9Exercise 2.912 questions

Q.2.9.1Resolve the following rational expressions into partial fractions: \(\frac{1}{x^{2}-a^{2}}\)v

Solution

1 = A (x – a) + B (x + a) ——– (1)
Put x = a in equation (1)
1 = A (0) + B (a + a)
1 = B(2a)
⇒ B = \(\frac{1}{2 a}\)
Put x = – a in equation (1)
1 = A(- a – a) + B(- a + a)
1 = – 2a A + 0
⇒ A = \(-\frac{1}{2 a}\)
∴ The required partial fraction is

Answer:

Q.2.9.2\(\frac{3 x+1}{(x-2)(x+1)}\)v

Solution

3x + 1 = A(x + 1) + B (x – 2) ——— (1)
Put x = 2 in equation (1)
3(2) + 1 = A (2 + 1) + B (2 – 2)
6 + 1 = 3A + 0
⇒ A = \(\frac{7}{3}\)
Put x = – 1 in equation (1)
3(-1) + 1 = A (-1 + 1 ) + B (- 1 – 2)
– 3 + 1 = A × 0 – 3B
– 2 = 0 – 3B
⇒ B = \(\frac{2}{3}\)
∴ The required partial fractions are

Answer:

Q.2.9.3\(\frac{x}{\left(x^{2}+1\right)(x-1)(x+2)}\)v

Solution

x = Ax (x + 1) (x + 2) + B(x – 1)(x + 2) + C(x 2 + 1)(x + 2) + D(x 2 + 1)(x – 1) ——— (1)
Put x = 1 in equation (1)
1 = A(1)(1 – 1) (1 + 2) + B ( ( 1 – 1 ) (1 + 2) + C(1 2 + 1 ) ( 1 + 2) + D(1 2 + 1) (1 – 1)
1 = A × 0 + B × 0 + C(2)(3) + D × 0
1 = 6C
⇒ C = \(\frac{1}{6}\)
Put x = – 2 in equation (1)
-2 = A(- 2)(- 2 – 1)(- 2 + 2) + B(- 2 – 1)(- 2 + 2) + C ((- 2) 2 + 1) (- 2 + 2) + D((-2) 2 + 1 ) (- 2 – 1)
– 2 = A × 0 + B × 0 + C × 0 + D(4 + 1)(-3)
-2 = D(5)(-3)
⇒ – 2 = – 15 D
⇒ D = \(\frac{2}{15}\)
Put x = 0 in equation (1)
0 = A(0) (0 – 1) (0 + 2) + B(0 – 1)(0 + 2)+ C(0 2 + 1) (0 + 2) + D(0 2 + 1) (0 – 1)
0 = 0 + B (- 2 ) + C (2) + D (- 1)
0 = – 2B + 2C – D
In equation (1), equate the coefficient of x 3 on both sides
0 = A + C + D

Answer:

Q.2.9.4\(\frac{x}{(x-1)^{3}}\)v

Solution

X = A(x – 1) 2 + B(x – 1) + C ——— (1)
Put x = 1 in equation (1)
⇒ 1 = A(1 – 1) 2 + B(1 – 1) + C
1 = 0 + 0 + C
⇒ C = 1
In equation (1), equating the coefficient of x 2 on both sides
0 = A ⇒ A = 0
Put x = 0 in equation (1) ⇒ 0 = A(0 – 1) 2 + B(0 – 1) + C
⇒ 0 = A – B + C
0 = 0 – B + 1
⇒ B = 1

Answer:

Q.2.9.5\(\frac{1}{x^{4}-1}\)v

Solution

1 = Ax (x + 1)(x – 1) + B (x + 1) (x – 1) + C (x<sup2 + 1)(x – 1) + D(x + 1)(x 2 + 1) —— (1)
Put x = 1 in equation (1)
1 = A(1 ) (1 + 1) (1 – 1) + B(1 + 1) (1 – 1) + C(1<sup2 + 1) (1 – 1) + D(1 + 1) (1<sup2 + 1)
1 = A × 0 + B × 0 + C × 0 + D(2)(2)
⇒ 1 = – 4D
⇒ D = \(\frac{1}{4}\)
Put x = -1 in equation (1)
1 = A(- 1)(- 1 + 1)(- 1 – 1) + B(- 1 + 1)(- 1 – 1) + C ((- 1) 2 + 1)(- 1 + 1) + D(- 1 + 1)((-1) 2 + 1)
1 = A × 0 + B × 0 + C (2) (-2) + D × 0
⇒ 1 = -4C
⇒ C = \(-\frac{1}{4}\)
Put x = 0 in equation (1)
I = A(0) (0 + 1 )(0 – 1) + B(0 + 1 )(0 – 1) + C(0 2 + 1 )(0 – 1) + D(0 + 1)(0 2 + 1)
1 = A × O + B(-1) + C(- 1) + D(1)
⇒ 1 = – B – C + D
1 = – B + \(\frac{1}{4}\) + \(\frac{1}{4}\)
⇒ B = \(\frac{1}{2}\) – 1 = – \(\frac{1}{2}\)
⇒ B = – \(\frac{1}{4}\)
In equation (1), equate the coefficient of x3 on both sides
0 = A + C + D
⇒ 0 = A – \(\frac{1}{4}+\frac{1}{4}\)
⇒ A = 0

Answer:

Q.2.9.6\(\frac{(x-1)^{2}}{x^{3}+x}\)v

Solution

(x – 1) 2 = A(x 2 + 1) + Bx 2 + Cx ——- (1)
Put x = 0 in equation (I)
(0 – 1) 2 = A(0 2 + 1) + B × 0 + C × 0
1 = A + 0 + 0
⇒ A = 1
Equating the coefficient of x 2 on both sides
1 = A + B
1 = 1 + B
⇒ B = 0
Put x = 1 in equation (1)
(1 – 1) 2 = A(1 2 + 1) + B × 1 2 + C × 1
0 = 2A + B + C
0 = 2 × 1 + 0 + C
⇒ C = – 2
∴ The required partial fraction is

Answer:

Q.2.9.7\(\frac{x^{2}+x+1}{x^{2}-5 x+6}\)v

Solution

\(\frac{x^{2}+x+1}{x^{2}-5 x+6}\)
Here the degree of the numerator is equal to the degree of the denominator. Let us divide the numerator by the
6x – 5 = A(x – 3) + B(x – 2) ——- (3)
Put x = 3 in equation (3)
6(3) – 5 = A(3 – 3) + B(3 – 2)
18 – 5 = 0 + B
⇒ B = 13
Put x = 2 in equation (3)
6(2) – 5 = A(2 – 3) + B(2 -2)
12 – 5 = – A + 0
7 = -A
⇒ A = – 7
Substituting the values of A and B in equation (2)

Answer:

Q.2.9.8\(\frac{x^{3}+2 x+1}{x^{2}+5 x+6}\)v

Solution

\(\frac{x^{3}+2 x+1}{x^{2}+5 x+6}\)
Since the numerator is of degree greater than that of the denominator divide the numerator by the denominator.
21x + 31 = A(x + 3) + B(x + 2) ——- (3)
Put x = – 3. in equation (3)
21(- 3) + 31 = A(- 3 + 3) + B(- 3 + 2)
– 63 + 31 = 0 – B
– 32 = – B
⇒ B = 32
Put x = – 2, in equation (3)
21(- 2) + 31 = A(- 2 + 3) + B(- 2 + 2)
– 42 + 31 = A + 0
⇒ A = – 11
Substituting the values of A and B in equation (2)

Answer:

Q.2.9.9\(\frac{x+12}{(x+1)^{2}(x-2)}\)v

Solution

x + 12 = A(x + 1 ) (x – 2) + B(x – 2) + C(x + 1) 2 ——– (1)
Put x = 2 in equation (l)
2 + 12 = A(2 + 1 ) (2 – 2) + B(2 – 2) + C(2 + 1) 2
14 = A(3) (0) + B × 0 + C (3 ) 2
4 = 0 + 0 + 9C
⇒ C = \(\frac{14}{9}\)
Put x = – 1 in equation (1)
-1 + 12 = A(- 1 + 1)(- 1 – 2) + B(- 1 – 2) + C(- 1 + 1) 2
11 = A × 0 + B (- 3 ) + C × 0
11 = -3 B
⇒ B = \(-\frac{11}{3}\)
Put x = 0 in equation (1)
0 + 12 = A (0 + 1)(0 – 2) + B(0 – 2) + C(0 + 1) 2
12 = – 2A – 2B + C

Answer:

Q.2.9.10\(\frac{6 x^{2}-x+1}{x^{3}+x^{2}+x+1}\)v

Solution

6x 2 – x + 1 = Ax (x + 1) + B (x + 1) + C(x 2 + 1) ——— (1)
Put x = – 1 in equation (1)
6 x (- 1) 2 – (- 1) + 1 = A (- 1 ) (- 1 + 1 ) + B(- 1 + 1) + C ( (- 1) 2 + 1 )
6 + 1 + 1 = A × 0 + B × 0 + C (2)
8 = 2C
⇒ C = 4
Put x = 0 in equation (1)
6 × 0 2 – 0 + 1 = A(0)(0 + 1 ) + B(0 + 1) + C(0 2 + 1)
1 = 0 + B + C
1 = B + 4
B = 1 – 4 = – 3
Equating the coefficient of x 2 in equation (1) we have
6 = A + C
6 = A + 4
⇒ A = 6 – 4
⇒ A = 2
∴ The required partial fraction is

Answer:

Q.2.9.11\(\frac{2 x^{2}+5 x-11}{x^{2}+2 x-3}\)v

Solution

\(\frac{2 x^{2}+5 x-11}{x^{2}+2 x-3}\)
Since the degree of the numerator is equal to the degree of the denominator divide the numerator by the denominator
Put x = 1 in equation (3)
1 – 5 = A(1 + 3) + B(1 – 1)
– 4 = 4A + 0
⇒ A = – 1
Put x = – 3 in equation (3)
– 3 – 5 = A (- 3 + 3) + B(- 3 – 1)
– 8 = 0 – 4B
⇒ B = 2
Substituting the values of A and B in equation (2) we have
∴ The required partial fraction is

Answer:

Q.2.9.12\(\frac{7+x}{(1+x)\left(1+x^{2}\right)}\)v

Solution

7 + x = A( 1 + x 2 ) + Bx (1 + x) + C(1 + x) ——- (1)
Put x = -1, in equation (1)
7 – 1 = A(1 + (-1) 2 ) + B (- 1) (1 – 1) + C(1 – 1)
6 = A(1 + 1) + 0 + 0
A = \(\frac{6}{2}\) = 3
⇒ A = 3
Put x = 0, in equation (1)
7 + 0 = A(1 + 0 2 ) + B × 0 (1 + 0) + C(1 + 0)
7 = A + 0 + C
7 = 3 + C
⇒ C = 4
Equating the coefficient of x 2 in equation (I) we have
0 = A + B
0 = 3 + B
⇒ B = – 3
∴ The required partial fraction is

Answer:

10Exercise 2.107 questions

Q.2.10.1x ≤ 3y, x ≥ yv

Solution

Consider the line x = 3y
When y = 0 ⇒ x = 0
When y = 1 ⇒ x = 3
When y = – 1 ⇒ x = – 3
Consider the line x = y
When x = 0 ⇒ y = o
When x = 1 ⇒ y = 1
When x = – 1 ⇒ y = – 1
To find the region of x ≤ 3y: The line x = 3y divides the cartesian plane into two half planes. Consider the point (1,1) this point satisfies the inequality x < 3y. This point (1,1) lies above the line x = 3y.
Hence all points satisfying the inequality lie above the line x = 3y.
Therefore x < 3y represents the upper half plane of the Cartesian plane bounded by the line x = 3y. Since x ≤ 3y, this region also contains all the points on the straight line x = 3y.
To find the region x ≥ y: The line x = y divides the cartesian plane into two half planes. Consider the point (2,1). This point (2,1) satisfies the inequality x > y and this point (2,1) lies below the line x = y. ∴ All points satisfying the inequality x > y will lie below the line x = y. Therefore x > y represents the lower half plane of the cartesian plane bounded by the line x = y. Since x > y this region also contains all the points on the straight line x = y.
The required region is the region common to the regions x ≤ 3y and x ≥ y

Answer:

Q.2.10.2y ≥ 2x, – 2x + 3y ≤ 6v

Solution

Consider the straight line y = 2x
When x = 0 ⇒ y = 2 × 0 = 0
When x = 0 ⇒ y = 2 × 1 = 2
When x = – 1 ⇒ y = 2 × – 1 = – 2
Consider the line – 2x + 3y = 6
When x = 0 ⇒ – 2 × 0 + 3y = 6 ⇒ 3y = 6 ⇒ y = 2
When y = 0 ⇒ – 2x + 3 × 0 = 6 ⇒ -2x = 6 ⇒ x = – 3
To find the region of y ≥ 2x: The straight line y = 2x divides the cartesian plane into two half planes. Consider the point (1, 3) satisfying the inequality y > 2x. This point (1, 3) lies above the straight line y = 2x. ∴ All points satisfying the inequality y > 2x will lie above the line y = 2x.
To find the region of -2x + 3y ≤ 6:
The straight line – 2x + 3y = 6 divides the cartesian plane into two half planes one half plane contains the origin and the other half plane does not contain the origin.
Substitute the origin (0, 0) in the inequality – 2x + 3y < 6
we get – 2 × 0 + 3 × 0 < 6 ⇒ 0 < 6
∴ (0,0) satisfies the inequality – 2x + 3y < 6
∴ The inequality – 2x + 3y < 6 represents the half plane containing the origin. Since – 2x + 3y ≤ 6, this region contains all the points on the straight line – 2x + 3y = 6.
The required region is the region common to y ≥ 2x and – 2x + 3y ≤ 6

Answer:

Q.2.10.33x + 5y ≥ 45, x ≥ 0, y ≥ 0v

Solution

Consider the line 3x + 5y = 45
when x = 0, 3 × 0 + 5y = 45 ⇒ y = \(\frac{45}{5}\) = 9
when y = 0, 3x + 5 × 0 = 45 ⇒ x = \(\frac{45}{3}\) = 15
To find the region of x ≥ 0, y ≥ 0:
x ≥ 0 and y ≥ 0 denote the first quadrant of the Cartesian plane.
Since x ≥ 0 and y ≥ 0 this region contains all the points on the lines x = 0 and y = 0.
To find the region of 3x + 5y ≥ 45: The straight-line 3x + 5y = 45 divides the cartesian plane into two half-planes, one-half plane containing the origin and the other half-plane does not contain the origin.
Substitute the origin (0,0) in the inequality 3x + 5y > 45 we get 3 × 0 + 5 × 0 > 45 ⇒ 0 > 45 which is impossible. Therefore, (0, 0) does not satisfy the inequality 3x + 5y > 45 represents the half plane that does not contain the origin bounded by the straight line 3x + 5y = 45. Since 3x + 5y ≥ 45, x ≥ 0, y ≥ 0, this region contains all the points on the straight lines 3x + 5y = 45.
∴ The required region is the common region bounded by x ≥ 0, y ≥ 0, and 3x + 5y ≥ 45.

Answer:

Q.2.10.42x + 3y ≤ 35, y ≥ 2, x ≥ 5.v

Solution

Consider the straight line 2x + 3y = 35
When x = 0, 2 × 0 + 3y = 35 ⇒ \(\frac{35}{3}\)
When y = 0, 2x + 3 × 0 = 35 ⇒ \(\frac{35}{2}\)
To find the region of 2x + 3y ≤ 35: The straight-line 2x + 3y = 35 divides the cartesian plane into two half-planes, one-half plane contains the origin and the other half-plane does not contain the origin.
Substitute the origin (0, 0) in the inequality 2x + 3y < 35, we get 2 × 0 + 3 × 0 < 35 ⇒ 0 < 35. (0, 0) satisfies the inequality.
Therefore, the inequality 2x + 3y < 35 represents the half plane that contains the origin (0, 0) bounded by the straight line 2x + 3y = 35. Since 2x + 3y ≤ 35, this region contains all the points on the straight line 2x + 3y = 35.
To find the region y ≥ 2: The straight line y = 2 divides the cartesian plane into two half-planes, one-half plane contains the origin and the other half-plane does not contain the origin. Substitute the point (0, 0) in the inequality y ≥ 2 we get 0 >2 which is impossible. Hence (0, 0) does not satisfy the inequality y > 2.
∴ The inequality y > 2 represents the half-plane that does not contain the origin bounded by the straight line y = 2. Since y ≥ 2, this region contains all the points on the straight line y = 2.
To find the region x ≥ 5: The straight line x – 5 divides the cartesian plane into two half-planes, one half-plane containing the origin and the other half-plane does not contain the origin.
Substitute the origin (0,0) in the inequality x > 5 we get 0 > 5 which is impossible. Hence (0, 0) does not satisfy the inequality x > 5.
∴ The inequality x > 5 represents the half-plane that does not contain the origin bounded the straight line x = 5.
Since x ≥ 5, this region contains all the points on the straight line x = 5
∴ The required region is the common region bounded by 2x + 3y ≤ 35, y ≥ 2, x ≥ 5.

Answer:

Q.2.10.52x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0v

Solution

Consider the straight line 2x + 3y = 6
When x = 0, 2 × 0 + 3y = 6 ⇒ y = \(\frac{6}{3}\) = 2
When y = 0, 2x + 3 × 0 = 6 ⇒ x = \(\frac{6}{2}\) = 3
Consider the straight line x + 4y = 4
When x = 0 ⇒ 0 + 4y = 4 ⇒ y = \(\frac{4}{4}\) = 1
When y = 0 ⇒ x + 4 × 0 = 4 ⇒ x = 4
To find the region of 2x + 3y ≤ 6: The straight line 2x + 3y = 6 divides the cartesian plane into two half planes, one half plane contains the origin and the other half plane does not contain the origin. Substitute the origin (0, 0) in the inequality 2x + 3y < 6 we get 2 × 0 + 3 × 0 < 6 ⇒ 0 < 6
∴ The origin (0,0) satisfies the inequality 2x + 3y < 6. Hence the inequality 2x + 3y < 6 represents the half plane that contains the origin (0, 0). Since 2x + 3y ≤ 6, this region contains all the points on the straight line 2x + 3y = 6.
To find the region of x + 4y ≤ 4: The straight line x + 4y = 4 divides the cartesian plane into two half planes, one half plane contains the origin and the other half plane does not contain the origin. Substitute the origin (0, 0) in the inequality x + 4y < 4, we get 0 + 4 × 0 < 4 ⇒ 0 < 4.
Therefore the origin (0, 0) satisfies the inequality x + 4y < 4. Therefore the inequality x + 4y < 4 represents the half-plane that contains the origin bounded by the line x + 4y = 4. Since x + 4y ≤ 4, this region contains all the points on the straight line.
To find the region x ≥ 0, y ≥ 0: x ≥ 0 and y ≥ 0 denote the first quadrant of the cartesian plane.
Since x ≥ 0 and y ≥ 0 this region contains all the points on the lines x = 0 and y = 0.
The required region is the common region bounded by 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0 and y ≥ 0.

Answer:

Q.2.10.6x – 2y ≥ 0, 2x – y ≤ – 2, x ≥ 0, y ≥ 0v

Solution

Consider the straight line x – 2y = 0
When x = 0
⇒ -2y = 0
⇒ y = 0
When x = 2
⇒ 2 – 2y = 0
⇒ 2y = 2
⇒ y = 1
When x = -2
⇒ – 2 – 2y = 0
⇒ -2y = 2
⇒ y = – 1
Consider the line 2x – y = – 2
When x = 0 ⇒ – y = – 2 ⇒ y = 2
When y = 0 ⇒ 2x – 0 = – 2 ⇒ x = – 1
To find the region of x – 2y ≥ 0: The straight line x – 2y = 0 divides the cartesian plane into two half-planes. Consider the point (3,1) satisfying the inequality x > 2y. The point (3, 1) lies below the straight line x = 2y in the cartesian plane.
∴ All the points satisfying the inequality x > 2y will lie in the half-plane below the straight line x = 2y. Since x ≥ 2y this region contains all the points on the straight line x = 2y.
To find the region of 2x – y ≤ – 2:
-(2x – y) ≥ 2 ⇒ – 2x + y ≥ 2
Consider the straight line – 2x + y = 2. This line divides the cartesian plane in to two half planes, one half plane contains the origin and the other half plane does not contain the origin. Substitute the origin (0, 0) in the inequality – 2x + y > 2 we get- 2 × 0 + 0 > 2 ⇒ 0 > 2 which is impossible. Therefore (0, 0) does not satisfy the inequality -2x + y > 2. Hence the inequality -2x + y > 2 represents the half plane that does not contain the origin bounded by the straight line. Since -2x + y > 2, this region contains all the points on the straight line -2x + y = 2
To find the region of x ≥ 0, y ≥ 0: x ≥ 0 and y ≥ 0 denote the first quadrant of the cartesian plane. Since x ≥ 0 and y ≥ 0 this region contains all the points on the lines x = 0 and y = 0.
∴ The required region is the region common to x – 2y ≥ 0, 2x – y ≤ – 2, x ≥ 0, y ≥ 0.

Answer:

Q.2.10.72x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6v

Solution

Consider the straight line 2x + y = 8
When x = 0 ⇒ 2 × 0 + y = 8 ⇒ y = 8
When y = 0 ⇒ 2x + 0 = 8 ⇒ x = \(\frac{8}{2}\) = 4
Consider the straight line x + 2y = 8
When x = 0 ⇒ 0 + 2y = 8 ⇒ y = \(\frac{8}{2}\) = 4
When y = 0 ⇒ x + 2 × 0 = 8 ⇒ x = 8
Consider the straight line x + y = 6
When x = 0 ⇒ 0 + y = 6 ⇒ y = 6
When y = 0 ⇒ x + 0 = 6 ⇒ x = 6
To find the region of 2x + y ≥ 8: The straight-line 2x + y = 8 divides the cartesian plane into two half-planes one half plane contains the origin and the other half plane does not contain the origin. Substitute the origin (0, 0) in the inequality 2x + y > 8 we get 2 × 0 + 0 > 8 ⇒ 0 > 8 which is impossible.
∴ The origin (0, 0) does not satisfy the inequality 2x + y > 8. Hence the inequality 2x + y > 8 represents the half plane that does not contain the origin bounded by the straight line 2x + y = 8. Since 2x + y ≥ 8, this region contains all the points on the straight line 2x + y = 8.
To find the region of x + 2y ≥ 8: The straight line x + 2y = 8 divides the cartesian plane into two half planes, one half plane contains the origin and the other half plane doesnot contain the origin. Substitute the origin (0, 0) in the inequality x + 2y > 8 we get 0 + 2 × 0 > 8 ⇒ 0 > 8 which is impossible. ∴ The origin (0, 0) does not satisfy the inequality x + 2y > 8. Therefore, the inequality x + 2y > 8 represents the half plane that does not contain the origin bounded by the straight line x + 2y = 8. Since x + 2y ≥ 8 this region contains all the points on the straight line x + 2y = 8.
To find the region of x + y ≤ 6: The straight line x + y = 6 divides the cartesian plane into two half-planes, one half-plane contains the origin and the other half-plane does not contain the origin. Substitute the origin (0, 0) in the inequality x + y < 6 we get 0 + 0 < 6 ⇒ 0 < 6 which is true.
∴ The origin (0, 0) satisfies the inequality x + y < 6. Hence, the inequality x + y < 6 represents the half plane that contains the origin bounded by the straight line x + y = 6. Since x + y ≤ 6, this region contains all the points on the straight line x + y = 6.
Thus the required region is the region common to 2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6

Answer:

11Exercise 2.114 questions

Q.2.11.1Simplify (a) (125) 2/3 (b) 16 -3/4 (c) (- 1000) -2/3 (d) (3 -6 ) 1/3 (e) \(\frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}}\)v

Solution

(a) (125) 2/3
(b) 16 -3/4
(c) (- 1000) -2/3
(d) (3 -6 ) 1/3
(e) \(\frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}}\)

Answer:

(a) (125) 2/3
(b) 16 -3/4
(c) (- 1000) -2/3
(d) (3 -6 ) 1/3
(e) \(\frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}}\)

Q.2.11.4Simplify and hence find the value of n: \(\frac{3^{2 n} \cdot 9^{2} \cdot 3^{-n}}{3^{3 n}}\) = 27v

Solution

Given
4 – 2n = 3
2n = 4 – 3
2n = 1
n = \(\frac { 1 }{ 2 }\)

Answer:

Given
4 – 2n = 3
2n = 4 – 3
2n = 1
n = \(\frac { 1 }{ 2 }\)

Q.2.11.5Find the radius of the spherical tank whose volume is \(\frac{32 \pi}{3}\) units.v

Solution

Let r be the radius of the spherical tank.
Given volume of the spherical tank = \(\frac{32 \pi}{3}\)
\(\frac{4}{3}\)πr 3 = \(\frac{32 \pi}{3}\)
4r 3 = 32
r 3 = \(\frac{32}{4}\) = 8
r 3 = 2 3
r = 2
∴ Radius of the spherical tank r = 2 units.

Answer:

Q.2.11.6Simplify by rationalizing the denominator \(\frac{7+\sqrt{6}}{3-\sqrt{2}}\)v

Solution

\(\frac{7+\sqrt{6}}{3-\sqrt{2}}\)
Multiply the numerator and denominator by 3 + √2

Answer:

\(\frac{7+\sqrt{6}}{3-\sqrt{2}}\)
Multiply the numerator and denominator by 3 + √2

12Exercise 2.127 questions

Q.2.12.1Let b > 0 and b ≠ 1. Express y = b x in logarithmic form. Also, state the domain and range of the logarithmic function.v

Solution

Given y = b x ⇒ log b y = x, x ∈ R with range (0, ∞) (-∞, ∞)

Answer:

Given y = b x ⇒ log b y = x, x ∈ R with range (0, ∞) (-∞, ∞)

Q.2.12.2Compute log 9 27 – log 27 9v

Solution

log 9 27 – log 27 9 = log 9 3 3 – log 27 3 2
= 3 log 9 3 – 2 log 27 3 —— (1)
By change of base rule [log b a = \(\frac{1}{\log _{a} b}\)]

Answer:

log 9 27 – log 27 9 = log 9 3 3 – log 27 3 2
= 3 log 9 3 – 2 log 27 3 —— (1)
By change of base rule [log b a = \(\frac{1}{\log _{a} b}\)]

Q.2.12.4Solve log 4 2 8x = 2 log 2 8v

Solution

Given log 4 2 8x = 2 log 2 8
log 4 2 8x = 2 log 2 2 3
log 4 2 8x = 2 3 log 2 2
log 4 2 8x = 2 3 = 8
2 8x = 4 8
(2 2 ) 4x = 4 8
⇒ (4) 4x = 4 8
⇒ 4x = 8
⇒ x = \(\frac { 8 }{ 4 }\) = 2

Answer:

Given log 4 2 8x = 2 log 2 8
log 4 2 8x = 2 log 2 2 3
log 4 2 8x = 2 3 log 2 2
log 4 2 8x = 2 3 = 8
2 8x = 4 8
(2 2 ) 4x = 4 8
⇒ (4) 4x = 4 8
⇒ 4x = 8
⇒ x = \(\frac { 8 }{ 4 }\) = 2

Q.2.12.5If a 2 + b 2 = 7ab, show that log \(\left(\frac{a+b}{3}\right)\) = \(\frac{1}{2}\) (log a + log b)v

Solution

Given
a 2 + b 2 = 7ab
Adding both sides 2ab we get
a 2 + b 2 + 2ab = 7ab + 2ab
(a + b) 2 = 9ab
Taking square root on both sides
Taking logarithm on both sides

Answer:

Given
a 2 + b 2 = 7ab
Adding both sides 2ab we get
a 2 + b 2 + 2ab = 7ab + 2ab
(a + b) 2 = 9ab
Taking square root on both sides
Taking logarithm on both sides

Q.2.12.9Prove log a + log a 2 + log a 3 + ……… + log a n = \(\frac{n(n+1)}{2}\) log av

Solution

log a + log a 2 + log a 3 + ……… + log a n
= log a + 2 log a + 3 log a + ………….. + n log a
= log a (1 + 2 + 3 + ………….. + n)
= log a × \(\frac{n(n+1)}{2}\)
= \(\frac{n(n+1)}{2}\) log a

Answer:

Q.2.12.10If, then prove that xyz = 1v

Solution

Let \(\frac{\log x}{y-z}\) = k
log x = k(y – z)
log x = ky – kz ——— (1)
Similarly log y = k(z – x) = kz – kx ——(2)
log z = k(x – y) = kx – ky ——- (3)
Adding (1), (2) and (3)
log x + log y + log z = ky – kz + kz – kx + kx – ky
log (xyz) = 0 = log 1
xyz = 1

Answer:

Q.2.12.12Solve log 5 – x (x 2 – 6x + 65) = 2v

Solution

log 5 – x (x 2 – 6x + 65) = 2
⇒ x 2 – 6x + 65 = (5 – x) 2
⇒ x 2 – 6x + 65 = 25 + x 2 – 10x
⇒ x 2 – 6x + 65 – 25 – x 2 + 10x = 0
⇒ 4x + 40 = 0
⇒ 4x = -40
⇒ x = -10

Answer:

log 5 – x (x 2 – 6x + 65) = 2
⇒ x 2 – 6x + 65 = (5 – x) 2
⇒ x 2 – 6x + 65 = 25 + x 2 – 10x
⇒ x 2 – 6x + 65 – 25 – x 2 + 10x = 0
⇒ 4x + 40 = 0
⇒ 4x = -40
⇒ x = -10

13Exercise 2.1320 questions

Q.2.13.1If |x + 2| ≤ 9 then x belongs to (1) (- ∞,- 7) (2) [- 11, 7] (3) (-∞, – 7) ∪ [11, ∞] (4) (-11, 7)v

Solution

(2) [- 11, 7]
Explanation:
-x – 2 ≤ 9 x + 2 ≤ 9
-x < 9 + 2 = 11 x ≤ 9 – 2 = 7
⇒ x ≥ -11
so x ∈ [-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]

Q.2.13.2Given that x, y and b are real numbers x < y, b > 0 then (1) xb < yb (2) xb > yb (3) xb < yb (4) \(\frac{x}{b}\) ≥ \(\frac{y}{b}\)v

Solution

(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

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

Q.2.13.3If \(\frac{|x-2|}{x-2}\) ≥ 0, then x belongs to (1) [2, ∞) (2) (2, ∞) (3) (-∞, 2) (4) (-2, ∞)v

Solution

(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

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

Q.2.13.4The solution of 5x – 1 < 24 and 5x + 1 > – 24 is (1) (4, 5) (2) (- 5, – 4) (3) (- 5, 5) (4) (- 5, 4)v

Solution

(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)

Answer:

Q.2.13.5The solution set of the following inequality |x – 1| ≥ |x – 3| is (1) [0, 2] (2) [2, ∞) (3) (0, 2) (4) (-∞, 2)v

Solution

(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, ∞)

Answer:

Q.2.13.6The value of \(\log _{\sqrt{2}} 512\) is (1) 16 (2) 18 (3) 9 (4) 12v

Solution

(2) 18
Explanation:

Answer:

(2) 18
Explanation:

Q.2.13.7The value of \(\log _{3} \frac{1}{81}\) is (1) – 2 (2) – 8 (3) – 4 (4) – 9v

Solution

(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

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

Q.2.13.8If log √x 0.25 = 4, then the value of x is (1) 0.5 (2) 2.5 (3) 1.5 (4) 1.25v

Solution

(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

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

Q.2.13.9The value of log a b. log b c. log c a is (1) 2 (2) 1 (3) 3 (4) 4v

Solution

(2) 1
Explanation:
log a b. log b c. log c a = log a c. log c a
= log a a = 1

Answer:

(2) 1
Explanation:
log a b. log b c. log c a = log a c. log c a
= log a a = 1

Q.2.13.10If 3 is the logarithm of 343 then, the base is (1) 5 (2) 7 (3) 6 (4) 9v

Solution

(2) 7
Explanation:
⇒ log x 343 = 3 ⇒ 343 = x 3
(.i.e.,) 7 3 = x 3 ⇒ x = 7
⇒ x = 7

Answer:

(2) 7
Explanation:
⇒ log x 343 = 3 ⇒ 343 = x 3
(.i.e.,) 7 3 = x 3 ⇒ x = 7
⇒ x = 7

Q.2.13.11Find 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) 4v

Solution

(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

Answer:

Q.2.13.12If 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) 6v

Solution

(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

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

Q.2.13.13The number of solutions of x 2 + |x – 1| = 1 is (1) 1 (2) 0 (3) 2 (4) 3v

Solution

(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

Answer:

Q.2.13.14The 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 = 0v

Solution

(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.

Answer:

Q.2.13.15If 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, 2v

Solution

(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

Answer:

Q.2.13.16If 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) \(\sqrt{\mathbf{k}-8 \mathbf{c}}\)v

Solution

(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

Answer:

Q.2.13.17If then the value of k is (1) 1 (2) 2 (3) 3 (4) 4v

Solution

(3) 3
Explanation:
kx = 2x – 2 + x + 2
kx = 3x ⇒ k = 3

Answer:

(3) 3
Explanation:
kx = 2x – 2 + x + 2
kx = 3x ⇒ k = 3

Q.2.13.18If, then the value of A + B is (1) – \(\frac{1}{2}\) (2) – \(\frac{2}{3}\) (3) \(\frac{1}{2}\) (4) \(\frac{2}{3}\)v

Solution

(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

Answer:

Q.2.13.19The number of roots of (x + 3) 4 + (x + 5) 4 = 16 is (1) 4 (2) 2 (3) 3 (4) 0v

Solution

(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.

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.

Q.2.13.20The 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) 4v

Solution

(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

Answer:

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