ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: Solve the quadratic equation: x² - 2x - 8 = 0

Solution:
x² - 2x - 8 = 0
⇒ x² - 4x + 2x - 8 = 0
⇒ x(x - 4) + 2(x - 4) = 0
⇒ (x - 4)(x + 2) = 0
∴ x - 4 = 0
⇒ x = 4

or,
x + 2 = 0
⇒ x = - 2

Question: If y = 5, then what is the value of 10y√(y³ - y²)?

Solution:
Given,
y = 5

∴ 10y√(y³ - y²)
= 10. 5. √(5³ - 5²)
= 50√(125 - 25)
= 50√(100)
= 50 × 10
= 500

Question: Find the value of x, when x³ - x = 0

Solution:
Given,
x³ - x = 0
⇒ x(x² - 1) = 0
⇒ x(x + 1)(x - 1) = 0
∴ x = 0, 1, - 1

Question: If x = a + (1/a) and y = a - (1/a) then x⁴ + y⁴ - 2x²y² = ?

Solution:
Given,
x = a + 1/a
y = a - 1/a

x + y = a + (1/a) + a - (1/a) = 2a
x - y = a + (1/a) - a + (1/a) = 2/a

Now,
x² + y²- 2. x². y² = (x²)² + (y²)² - 2. x² .y²
= (x² - y²)²
= {(x + y)(x - y)}²
= {(2a)(2/a)}²
= 4²
= 16

Question: If (4 - x)/(2 + x) = x, what is the value of x² + 3x - 4?

Solution:
Given,
(4 - x)/(2 + x) = x
⇒ x(2 + x) = (4 - x)
⇒ 2x + x² = 4 - x
⇒ x² + 2x - 4 + x = 0
∴ x² + 3x - 4 = 0

Question: The polynomial equation x(x + 3) + 8 = (x + 2)(x - 2) is-

Solution:
We have
x(x + 1) + 8 = (x + 2)(x - 2)
⇒ x² + 3x + 8 = x² - 4
⇒ x² + 3x + 8 - x² + 4 = 0
⇒ 3x + 12 = 0 which is a linear equation.

Question: Find the solution to the equation (x² - x - 6)/(x + 2) = 0

Solution:
(x² - x - 6)/(x + 2) = 
⇒ (x² - 3x + 2x - 6) )/(x + 2)=0
⇒ {x(x - 3) + 2(x - 3)}/(x + 2) = 0
⇒ (x - 3)(x + 2)/(x + 2) = 0
⇒ x - 3 = 0
∴ x = 3

Question: If x + 1/x = 2 then, find the value of x⁷ - 1/x⁷

Solution:
Given,
x + 1/x = 2
⇒ x²  + 1 = 2x
⇒ x² - 2x + 1 = 0
⇒ x² - x - x + 1 = 0
⇒ x(x - 1)- 1(x - 1)  = 0
⇒ (x - 1)(x - 1) = 0
∴ x - 1 = 0
⇒ x = 1

∴ x⁷ - 1/x⁷ = 1⁷ - 1/1⁷
= 1 - 1
= 0

Question: If (x - 1) is a factor of 4x³ + 3x² - 4x + k, then find the value of k?

Solution:
Given,
(x - 1) is a factor of 4x³ + 3x² - 4x + k
So x - 1 = 0
⇒ x = 1

∴ f(x) = 4x³ + 3x² - 4x + k
∴ f(1) = 4(1)³ + 3(1)² - 4. 1 + k
= 4 + 3 - 4 + k
= k + 3

Now,
k + 3 = 0
∴ k = - 3

Question: If a + b = √5 and a - b = √3 the, 8ab(a² + b²) = ?

Solution:
Given,
a + b = √5
and a - b = √3

∴ 8ab(a² + b²)
= 4ab × 2(a² + b²)
= {(a + b)² - (a - b)²}{(a + b)² + (a - b)²}
= {(√5)² - (√3)²}{(√5)² + (√3)²}
= (5 - 3)(5 + 3)
= 2 × 8
= 16

Question: α and β are the roots of 2x² + 5x + 2 = 0 then, the value of (1/α) + (1/β) is-

Solution:
Here,
2x² + 5x + 2 = 0
where, a = 2, b = 5 and c = 2

∴ α + β = - (b/a) = - 5/2
and αβ = c/a = 2/2 = 1

∴ (1/α) + (1/β)
= (β + α)/αβ
= {- (5/2)}/1
= - 5/2

Question: The expression (3x² + x - 12) - 2(x² + 4x + 9) is equivalent to which of the following:

Solution:
Here,
(3x² + x - 12) - 2(x² + 4x + 9)
= 3x² + x - 12 - 2x² - 8x - 18
= x² - 7x - 30

Question: If a³ - b³ = 117 and a - b = 3 What is the value of ab?

Solution:
Given,
a³ - b³ = 117
a - b = 3

We know,
⇒ (a - b)³ + 3ab(a - b) = a³ - b³
⇒ 3³ + 3ab(3) = 117
⇒ 27 + 9ab = 117
⇒ 9ab = 117 - 27
⇒ 9ab = 90
⇒ ab = 90/9
∴ ab = 10

সঠিক উত্তর: খ) ± 2, ± 1 ও ঘ) ± 1, ± 2
অপশনে দ্বৈত উত্তর থাকায় প্রশ্নটি বাতিল করা হলো। 
---------------------------------- 

Question: Solve: x⁴ - 5x² + 4 = 0

Solution:
Given,
x⁴ − 5x² + 4 = 0
(x²)² - 5x² + 4 = 0
Let
x² = y

∴ y² − 5y + 4 = 0
⇒ y² − 4y - y + 4 = 0
⇒ y(y − 4) - 1(y - 4) = 0
⇒ (y − 4)(y − 1) = 0

∴ y − 4 = 0
⇒ y = 4
⇒ x² = 4
∴ x = ± 2

and,
y − 1 = 0
⇒ y = 1
⇒ x² = 1
∴ x = ± 1

Question: If f(x) = x² - 5x + 6 and f(x) = 0 then, x = ?

Solution:
Given,
f(x) = x² - 5x + 6
and f(x) = 0

∴ x² - 5x + 6 = 0
⇒ x² - 2x - 3x + 6 = 0
⇒ x(x - 2) - 3(x - 2) = 0
⇒ (x - 2)(x - 3) = 0

Here,
x - 2 = 0
⇒ x = 2

or,
x - 3 = 0
⇒ x = 3

Question: If dividing P(x) = 5x³ + 6x² - ax + 6 by x - 2 results the remainder 6 then find the value of a.

Solution:
Dividing P(x) by x - 2 we will get the remainder
∴  P(2) = 5 × 2³ + 6 × 2² - a × 2 + 6
= 40 + 24 - 2a + 6
= 70 - 2a

ATQ,
70 - 2a = 6
⇒ 2a = 70 - 6
⇒ 2a = 64
∴ a = 32

Question: Which is the correct factor analysis of x² - 2xy - z² + 2yz:

 Solution:
x² - 2xy - z² + 2yz
= x² - z² - 2xy + 2yz
= (x + z)(x - z) - 2y(x - z)
= (x - z)(x - 2y + z)

Question: Solve the following equation: x/3 - 1/12 = 1/6 + x/4

Solution:
Given,
x/3 - 1/12 = 1/6 + x/4
⇒ x/3 - x/4 = 1/6 + 1/12
⇒ (4x - 3x)/12 = (2 + 1)/12
⇒ x/12 = 3/12
⇒ 12x = 3 × 12
∴ x = 3

Question: The equation x² - kx + 36 = 0 has two equal roots, then the value of k is-

Solution:
Here
a = 1, b = - k and c = 36
Since the equation has two equal roots
: b² - 4ac = 0
⇒ (- k)² - 4 × 1 × 36 = 0
⇒ k² = 144
⇒ k = ± √(144)
k = ± 12