ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: If x/y = 1/3,  then (x² + y²)/(x² - y²) = ?

Solution:
(x² + y²)/(x² - y²) 
= {(x² + y²)/y²}/{x² - y²)/y² [Dividing the numerator and denominator by y²]
= {(x²/y²) + 1}/{(x²/y²) - 1}
= {(x/y)² + 1}/{(x/y)² - 1)}
= {(1/3)² + 1}/{(1/3)² - 1} [given, x/y = 1/3]
= {(1/9) + 1}/{(1/9} - 1}
= (10/9)/(-8/9)
= (10/9) × (9/-8)
= - 5/4

Shortcut:
Take x = 1, y = 3 (because 1/3 = 1/3)
∴ (x² + y²)/(x² - y²) = (1² + 3² )/(1² - 3²)
= (1 + 9)/(1 - 9) = - 5/4

Question: (a² - b² - 2bc - c²)/(a² + b² + 2ab - c²) is equivalent to?

Solution:
(a² - b² - 2bc - c²)/(a² + b² + 2ab - c²)
= {a² - (b² + 2bc + c²)}/{(a² + b² + 2ab) - c²}
= {a² - (b + c)²}/{(a + b)² - c²}
= (a + b + c)(a - b - c)/(a + b + c)(a + b - c)
= (a - b - c)/(a + b - c)

Question: If a + b + c = 13 and a² + b² + c² = 69, then what is the value of ab + bc + ca?

Solution: 
We know,
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
⇒ 2(ab + bc + ca) = (a + b + c)² - ( a² + b² + c²)
⇒ 2(ab + bc + ca) = 13² - 69 [given, a + b + c = 13 and a² + b² + c² = 69]
⇒ 2(ab + bc + ca) = 169 - 69 = 100
⇒ (ab + bc + ca) = 100/2
∴ (ab + bc + ca) = 50

Question: 

Solution: 

Question: (p² - 7p + 10)/(p² - 8p + 15) = ?

Solution: 
(p² - 7p + 10)/(p² - 8p + 15) 
= (p² - 2p - 5p + 10)/(p² - 3p - 5p + 15) 
= {p(p - 2) - 5(p - 2)}/{p(p - 3) - 5(p - 3)} 
= (p - 2)(p - 5)/(p - 3)(p - 5) 
= (p - 2)/(p - 3)

Question: If x + (2/x) = 4, what is the value of x³ + (8/x³)?

Solution: 
Here, x + (2/x) = 4

Now, 
x³ + (8/x³)
= (x)³ + (2/x)³
= {(x + (2/x)}³ - 3 . x . 2/x {x + (2/x)}
= 4³ - 3 . 2 . 4
= 64 - 24
= 40

Question: If x = 1 + √2 and y = 1 - √2, find the value of (x² + y²)².

Solution: 
Given that, x = 1 + √2 and y = 1 - √2
∴ x + y = 1 + √2 + 1 - √2
= 2

And, xy = (1 + √2)(1 - √2)
= 1² - (√2)²
= 1 - 2
= - 1

Now,
x² + y² = (x + y)² - 2xy
= (2)² - 2(- 1)
= 4 + 2
= 6

∴ (x² + y²)² = 6²
= 36

Question: If a + 1/a = √3, then what is the value of a³⁰ + a²⁴ + a⁶ + 1?

Solution:
Given, a + 1/a = √3
Now,
a³ + 1/a³ = (a + 1/a)³ - 3 . a . (1/a)(a + 1/a)
⇒ a³ + 1/a³ = (√3)³ - 3(√3) [∵ a + 1/a = √3]
⇒ a³ + 1/a³ = 3(√3) - 3(√3)
⇒ a³ + 1/a³ = 0 
⇒ a⁶ + 1 = 0 [Multiplying both sides by a³]

Then,
a³⁰ + a²⁴ + a⁶ + 1
= a²⁴ (a⁶ + 1) + (a⁶ + 1)
= (a²⁴ × 0) + 0
= 0

Question: The sum of the squares of three numbers is 123, and the sum of their products taken two at a time is 119. What is the sum of the three numbers?

Solution:
Let the three numbers be a, b, and c
Then,
a² + b² + c² = 123
ab + bc + ca = 119

Now,
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
⇒ (a + b + c)² = 123 + (2 × 119) 
⇒ (a + b + c)² = 361
⇒ (a + b + c) = √361
∴ (a + b + c) = 19

Question: If x = 7 - 4√3, then 

Solution: 
Given,
x = 7 - 4√3
⇒ x = 4 + 3 - 4√3
⇒ x = 2² + (√3)² - 2 × 2√3
⇒ x = (2 - √3)²  [ a² - 2ab + b² = (a - b)² ]
∴ √x = 2 - √3

Again,
√x = 2 - √3
⇒ 1/√x = 1/(2 - √3)
⇒ 1/√x = (2 + √3)/{(2 - √3) (2 + √3)}
⇒ 1/√x = (2 + √3)/(4 - 3)
∴ 1/√x = 2 + √3

∴ √x + (1/√x) = 2 - √3 + 2 + √3
∴ √x + (1/√x) = 4