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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 x² – (α + β) x + αβ = 0 

So the required quadratic equation is 

x² – (4) x + (-21) = 0 

(i.e.,) x² – 4x – 21 = 0

Given α = 1 + √5 

So, β = 1 – √5 

α + β = 2; αβ = 1² - (-√5)² = 1 - 5 = -4

The quadratic polynomial is

p(x) = x² – (α + β)x + αβ 

p(x) = k(x² – 2x – 4) 

p(1) = k(1 – 2 – 4) = -5k

Given p (1) = 2

-5k = 2

k = -2/5

∴ p(x) = (-2/5)(x² - 2x - 4)

-1/4 is a rational number (i.e.) 1-/4 ∈ Q,

3.14 ∈ Q 

0, 4 are integers and 0 ∈ Z, 4 ∈ N, Z, Q

22/7 ∈ Q

Let 3x³ + 8x² + 8x + a = (x² + x + 1) (3x + a). 

Equating coefficient of x 

8 = a + 3 

8 – 3 = a 

a = 5

-1/4 is a rational number (i.e.) (-1/4) Q, 

3.14 ∈ Q 

0, 4 are integers and 0 ∈ Z, 4 ∈ N, Z, Q

(22/7) ∈ Q

log_{5 - x} (x² – 6x + 65) = 2 

⇒ x² – 6x + 65 = (5 – x)² 

x² – 6x + 65 = 25 + x² – 10x 

⇒ x² – 6x + 65 – 25 – x² + 10x = 0

4x + 40 = 0 ⇒ 4x = -40 

x = -10

Taking two irrational numbers as 3 + √2 and 1 + √2 

Their difference is a rational number. But if we take two irrational numbers as 2 – √3 and 4 + √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.

(b) (2, ∞) 

(|x - 2|/(x - 2)) ≥ 0

∴ x > 2

x ∈ (2, ∞)

|x – 1| ≤ 5 and |x| ≥ 2 

⇒ -5 ≤ x – 1 ≤ 5 and x ≤ -2 (or) x > 2 

⇒ – 5 + 1 ≤ x ≤ 5 + 1 

⇒ -4 ≤ x ≤ 6 and x ≤ -2 (or) x ≥ 2 

Hence x < [-4, -2] ∪ [2, 6]

23x < 100

23x/23 < 100/23

(i.e.,) x > 4.3 

(i) x = 1, 2, 3, 4 (x ∈ N) 

(ii) x = …. -3, -2, -1, 0, 1, 2, 3, 4 (x ∈ Z)

(a) xb < yb

x < y, (b ≥ 0) 

x/y < y/b

(i.e.,) xb < yb

-2x > 9 ⇒ 2x ≤ -9

(i) 2x/2 ≤ -9/2

x ≤ -9/2 (= -4.5)

(ii) x = … -3, -2, -1, 0, 1, 2, 3, 4 

(iii) x = 1, 2, 3, 4

Answer is (b) (2, ∞)

|x| < -3 + 10 (= 7) 

|x| < 7 ⇒ -7 < x < 7

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

(b) 7

⇒ log_x 343 = 3 ⇒ 343 = x³ 

(.i.e.,) 7³ = x³ ⇒ x = 7 

⇒ x = 7

343 = 7³ 

So (343)^2/3 = (7³)^2/3 = 7^{3 x (2/3)} = 7² = 49

(b) 1 

log_b a log_c b log_a c = log_a a = 1

(c) 9, 1

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

(a) 4

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

(x + 3)⁴ + (x + 5)⁴ = 2⁴ 

This is biquadratic equation. It has 4 roots.

Let x be the smaller of the two consecutive odd positive integers, then the other is x + 2. 

According to the given conditions. 

x < 10, x + 2 < 10 and x + (x + 2) > 11

⇒ x < 10, x < 8

and 2x > 9 

x < 10, (∵ x < 8 automatically smallest of the lesser than) ... (1)

and x > (9/2) ... (2) 

From (1) and (2), we get 9

(9/2) < x < 8 

Also, x is an odd positive integer. x can take values 5 and 7.

So, the required possible pairs will be (x, x + 2) = (5, 7), (7, 9)

h(t) = -5t² + 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.

12% solution of acid in 600 l ⇒ 600 × (12/100) = 72 l of acid 

15% of 600 l ⇒ 600 × (15/100) = 90 l 

18% of 600 l ⇒ 600 × (18/100) = 108 l 

Let x litres of 18% acid solution be added

Given, (600 + x) (15/100) ≥ 72 + (30x/100)

(600 + x)15 ≥ 7200 + 30x 

9000+ 15x ≥ 7200 + 30x 

1800 ≥ 15x 

x ≤ 120 

Let x litres of 18% acid solution be added

Given, (600 + x)(18/100) ≤ 72 + (30/100)x 

(600 + x)18 ≤ 7200 + 30x

10800 + 18 ≤ 7200 + 30x 

3600 ≤ 12x 

x > 300 

The solution is 120 ≤ x > 300

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

50x/50 < 430/50

x < 8.6 (i.e.) when x is less than 9 hrs the first scheme gives better wages.

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

A’s monthly salary = Rs x 

B’s monthly salary = Rs 27000 

Their annual salaries differ by Rs 6000 

A’s salary – 27000 > 6000

A’s salary > Rs 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

-x² + 3x – 2 ≥ 0 ⇒ x² – 3x + 2 ≤ 0 

(x – 1) (x – 2) ≤ 0 

[(x – 1) (x – 2) = 0 ⇒ x = 1 or 2. Here α = 1 and β = 2. Note that α < β] 

So for the inequality (x – 1) (x – 2) ≤ 2 

x lies between 1 and 2 

(i.e.) x ≥ 1 and x ≤ 2 or x ∈ [1, 2] or 1 ≤ x ≤ 2

(i) -x² + 3x + 1 = 0 

⇒ comparing with ax² + bx + c = 0

∆ = b² – 4ac = (3)² – 4(1)(-1) = 9 + 4 = 13 > 0 

⇒ The roots are real and distinct

(ii) 4x² – x – 2 = 0 

a = 4, b = -1, c = -2 

∆ = b² – 4ac = (-1)² – 4(4)(-2) = 1 + 32 = 33 > 0 

⇒ The roots are real and distinct

(iii) 9x² + 5x = 0 

a = 9, b = 5, c = 0 

∆ = b² – 4ac = 5² – 4(9)(0) = 25 > 0 

⇒ The roots are real and distinct

⇒ -7 < 3 – x < 7 3 – x > -7 

-x > -7 -3 (= -10) 

-x > -10 ⇒ x < 10 ..... (1)

3 – x < 7 

– x < 7 – 3 (= 4) 

– x < 4x > -4 ..... (2) 

From (2) and (1) ⇒ x > -4 and x < 10

⇒ -4 < x < 10

Let x = 0.33333…. 

10x = 3.3333 …. 

10x – x 

= 9x = 3

x = 3/9 = 1/3

Let the two irrational numbers as 2 + √3 and 3 – √3 

Their sum is 2 + √3 + 3 – 3√3 which is a rational number. 

But the sum of 3 + √5 and 4 – √7 is not a rational number. So the sum of two irrational numbers is either rational or irrational.

Again taking two irrational numbers as π and 3/π their product is √3 and √2 = √3 x √2 which is irrational, So the product of two irrational numbers is either rational or irrational.