ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: If a + b = -1 and a² + b² = 25, then find the value of (a - b)².

Solution:
Given that 
(a + b) = -1
a² + b² = 25

We know
(a + b)² = a² + b² + 2ab
(- 1)² = 25 + 2ab
1 = 25 + 2ab
1 - 25 = 2ab
2ab = - 24
ab = -  12

Now
(a - b)² = (a + b)²  - 4ab
            =(- 1)² - 4 × (- 12)
            = 1 + 48 
            = 49

Question: The product of the roots of the equation 2a² - 5a + p = 10 is - 6. Find the value of p.

Solution: 
Rearranging the given equation we have 2a² - 5a + (p - 10) = 0

We know that,
if ax² + bx + c = 0 is a quadratic equation, then the product of their roots = c/a

Given the product of the roots = - 6
⇒ (p - 10)/2 = - 6
⇒ (p - 10) = - 12
⇒ p = - 12 + 10
   p = - 2

Question: If (3 + √3)z + 2 = 5 + 3√3, then the value of z is-

Solution: 
(3 + √3)z + 2 = 5 + 3√3
⇒ (3 + √3)z + 2 = 5 + 3√3
⇒ (3 + √3)z = 5 + 3√3 - 2
⇒ (3 + √3)z = 3 + 3√3
⇒ z = (3 + 3√3)/(3 + √3)
⇒ z =√3(3 + √3)/(3 + √3)
      z = √3

Question: The factors of the expression 2a² - a - 3 is-

Solution: 
    2a² - a - 3
= 2a² - 3a + 2a - 3
= a(2a - 3) + 1(2a - 3)
= (2a - 3)(a + 1)

Question: What are the solutions to the equation x² - 2x - 2 = 0

Solution: 
Given that 
x² - 2x - 2 = 0.........(1)
Comparing ax² + bx + c = 0 with (1) get, a = 1, b = - 2 and c = - 2

We know
x = {(- b) ± √(b² - 4ac)}/2a
   = [{- (- 2)} ± √{(- 2)² - 4.1(- 2)]/2.1
    = (2 ± √12)/2
    =(2 ± 2√3)/2
    = 1 ± √3
    = (1 + √3), (1 - √3)

Question: The value of (x - y)³ + (x + y)³ + 6x(x² - y²)

Solution: 
(x - y)³ + (x + y)³ + 6x(x² - y²)
= (x - y)³ + (x + y)³ + 3.2x(x - y)(x + y)
Let
a = x - y
b = x + y
a + b = x - y + x + y = 2x

Given expression 
= (x - y)³ + (x + y)³ + 3.2x(x - y)(x + y)
= a³ +b³ + 3(a + b)ab
= a³ + b³ + 3ab(a + b)
= (a + b)³
= (2x)³
= 8x³

Question: What is the sum of the reciprocals of the values of zeroes of the polynomial 6x² + 3x² - 5x + 1?

Solution: 
6x² + 3x² - 5x + 1
⇒ 9x² - 5x + 1
Let α and β are two roots of the equations

As we know,
Sum of roots (α + β) = 5/9
Product of roots (αβ) = 1/9
According to the question :
⇒ 1/α + 1/β = (α + β)/αβ
                     = (5/9)/(1/9)
                     = (5/9) × (9/1)
                     = 5

Question: If p/q = 5, then the value of (p + q)/(p - q) is-

Solution: 
Given that 
 p/q = 5/1
Now
(p + q)/(p - q) = (5 + 1)/(5 - 1)
                        = 6/4
                        = 3/2

Question: One of the roots of the equation x² - 13x + k = 0 is x = 4. The other root is:

Solution:
Let us put x = 4 in the equation x² - 13x + k = 0,
⇒ 16 - 52 + k = 0
⇒ k = 36

Putting the value of k in the equation,
we get:
x² - 13x + 36 = 0
⇒ x² - 9x - 4x + 36 = 0
⇒ x(x - 9) - 4 (x - 9) = 0
⇒ (x - 4)(x - 9) = 0
⇒ x = 4 and 9

∴ Other root of the equation is 9

Question: If a + b = 2, ab = 1 what is the value of (a, b)?

Solution:
Given that 
a + b = 2...........(1)
ab = 1..................(2)

From (1)
b = 2 - a

Putting the value in (2), we get 
a(2 - a) = 1
2a - a² = 1
2a = 1 + a² 
a² - 2a + 1 = 0
(a - 1)² = 0
a - 1 = 0
a = 1

a = 1 Putting the value in (2), we get 
1.b = 1
b = 1

(a, b) = (1,1)

Question: If (x + 3) is a factor of 3x² + ax + b, then find the value of 3a - b.

Solution:
According to the question,
⇒ (x + 3) = 0
⇒ x = - 3
Now, substitute the value of x = - 3 in the function 3x² + ax + b and then equate to 0.
⇒ 3(- 3)² - 3a + b = 0
⇒ 27 - 3a + b = 0
⇒ - 3a + b = - 27 
⇒ - (3a - b) = - 27
⇒ 3a - b = 27

Question: If 2x + y = 15, 2y + z = 25 and 2z + x = 26, what is the value of z? 

Solution:
2x + y = 15..............(1)
2y + z = 25 ..............(2)
2z + x = 26..............(3)

(1) + (2) + (3)⇒
2x + y  + 2y + z + 2z + x = 15 + 25 + 26
3x + 3y + 3z = 66
3(x + y + z) = 66
x + y + z = 22..................(4)

From (2)
2y + z = 25
2y = 25 - z
y = (25 - z)/2

From (3)
2z + x = 26
x = 26 - 2z

From (4)
x + y + z = 22
26 - 2z + {(25 - z)/2} + z = 22
52 - 4z + 25 - z + 2z = 44
77 - 3z = 44
- 3z = 44 - 77
- 3z = - 33
z = 11

Question:  If x = √10 + 3 then find the value of x - 1/x

Solution:
Given that 
x = √10 + 3
1/x = 1/(√10 + 3)
      =(√10 - 3) /(√10 + 3)(√10 - 3)
      = (√10 - 3)/{(√10)² - (3)²}
      = (√10 - 3)/(10 - 9)
     = (√10 - 3)

x - 1/x = √10 + 3 -  (√10 - 3)
            = √10 + 3 - √10 + 3
             = 6

Question: If √32 +√128 = √x then find the value of x.

Solution:
√32 +√128 =√x
⇒ 4√2 + 8√2 = √x
⇒ 12√2 = √x
⇒ x = (12√2)²
⇒ x = (144× 2)
⇒ x = 288

∴ The required value of x is 288

Question: If α and β are roots of the equation x² + x - 1 = 0, then the equation whose roots are α/β and β/α is:

Solution:
As α and β are roots of x² + x - 1 = 0,
then
⇒ α + β = - ( + 1) = - 1
⇒ αβ = - 1

Now, if (α/β) and (β/α) are roots then,
⇒ Sum of roots = (α/β) + (β/α)
= (α² + β²)/αβ
= [(α + β)² - 2αβ]/αβ
= (- 1)² - 2(-1)]/(-1)
= (1+ 2)/(- 1)
= - 3

Product of roots = (α/β) × (β/α) = 1
Now, then the equation is,
⇒ x² - (Sum of roots)x + Product of roots = 0
⇒ x² - (- 3)x + (1) = 0
⇒ x² + 3x + 1 = 0

Question: Find the product of two consecutive numbers where four times the first number is 10 more than thrice the second number.
Solution:
Let the numbers are ‘a' and ‘a + 1’.
According to the question :
4a = 3 × (a + 1) + 10
4a = 3a + 3 + 10
4a - 3a = 13
⇒ a = 13
Hence, the numbers are 13 and 14.

∴ Product = 13 × 14 = 182

Question: One of the factors of the expression: a³ - 6a² + 12a - 9

Solution:
Given that 
a³ - 6a² + 12a - 9

Let 
f(a) = a³ - 6a² + 12a - 9
f(3) = 3³ - 6 × 3² + 12 × 3 - 9
      = 27 - 54 + 36 - 9
      = 63 - 63
      = 0
a - 3 one of the factors of the expression a³ - 6a² + 12a - 9

Question: If a² + 2a/5 + 1/25 = 0, then (a - 2/5)² = ?

Solution:
a² + 2a/5 + 1/25 = 0
⇒ a² + (1/5)² + 2 × a × 1/5 = 0
⇒ (a + 1/5)² = 0
⇒ a = - 1/5

{(- 1/5) - (2/5)}² =  {(- 1 - 2)/5}²
                           =  9/25

Question: One factor of x² - y² + 2y - 1 is (x + y - 1) then another factor is-

Solution:
x² - y² + 2y - 1
= x² - (y² - 2.y.1 + 1²)
= x² - (y - 1)²
= {x + (y - 1)}{x - (y - 1)}
= (x + y - 1)(x - y + 1)

Question:  If a/3 = b/4 = c/7, the value of (a + b + c)/c  is-

Solution:
Let
a/3 = b/4 = c/7 = x
a = 3x, b = 4x, c = 7x. 

Now 
 (a + b + c)/c = (3x + 4x + 7x)/7x
                       = 14x/7x
                      = 2

Question: If x + y = 2a then the value of {a/(x - a)} + {a/(y - a)} is 

Solution:
a/(x - a) + a/(y - a)
= a(y - a) + a(x - a)}/(x - a)(y - a)
= (ay - a² + ax - a²)/(x - a)(y - a)
= a(x +y) - 2a²/(x - a)(y - a)
= a. 2a - 2a²/(x - a)(y - a)
= 2a² - 2a²/(x - a)(y - a)
= 0/(x - a)(y - a)
= 0

Question: If x⁶ + x⁵ + x⁴ + x³ + x² + x + 1 = 0, then, find the value of x³⁵ + x⁶³.

Solution:
Given that
x⁶ + x⁵ + x⁴ + x³ + x² + x + 1 = 0
x⁶ + x⁵ + x⁴ + x³ + x² + x = - 1..........(1)

Multiplying by x,
we have
 x⁷ + x⁶ + x⁵ + x⁴ + x³ + x² + x = 0 ........(2) 
⇒ x⁷ -  1 = 0
⇒ x⁷ = 1

According to question 
x³⁵ + x⁶³ = (x⁷)⁵ + (x⁷)⁹ = 1⁵ + 1⁹ = 1 + 1 = 2

Question:  If a + b = √7 and a - b = √5, what is the value of 8ab(a² + b²)?

Solution:
Given that,
 a + b = √7 
a - b = √5

8ab(a² + b²) = 4ab. 2(a² + b²)
                      = {(a + b)² - (a - b)²}{(a + b)² + (a - b)²}
                        = {(√7)² - (√5)²}{(√7)² + (√5)²}
                       = (7 - 5)(7 + 5)
                        = 2 × 12
                         = 24

Question: 3x² + kx + 4 is divisible by x - 1. The expression is also divisible by-

Solution:
Given that 
The expression 3x² + kx + 4 divisible by x - 1,
⇒ 3x² + kx + 4 = 0............(1)

Put x = 1 in equation (1)
⇒ 3 × 1² + k × 1 + 4 = 0
⇒ 3 + k + 4 = 0
⇒ k = - 7

Put the value k in equation (1) 
⇒ 3x² - 7x + 4 = 0
⇒ 3x² - 3x - 4x + 4 = 0
⇒ 3x (x - 1) - 4(x – 1) = 0
⇒ (x - 1) (3x - 4) = 0

So, expression divisible by (x - 1) and (3x - 4).