Test: SSC CGL Previous Year Questions: Algebra (2023-18) - 1 - SSC CGL MCQ

And, 
= 23 × 5√21 = 115√21 ----- (2)
Now, 
= 527 × 115√21 = 60605√21

Since, α, β are the roots of 6x² + 13x + 7 = 0. then, α + β = (-13)/6 and αβ = 7/6

So, the equation whose r oots are α² β² , is given by :
 ⇒ 36x² - 85 + 49 = 0

+ (223 − 224)²  + (223 − 224)² + (224 − 222)²]
= (669/2)[(−1)² + (−1)² + (2)²]
= (669/2) x 6 = 2007

Dividing and multiplying by (k + (1/k)), we get

Statement 1.
Sum of squares of n even natural no.
= (2/3) x (n)(n + 1 )(2n + 1)
Sum of squares of n odd natural no.
= (1/3) x n x (2n + 1) (2n − 1)
= (100² + 98² + 96² ......... + 2²) - (99² + 97² + 95² ... 1²)
× 50 × (2 × 50 + 1) × (2 × 50 − 1)

Statement 1 is correct .
Statement 2 :-

Now, x¹⁹⁷ + x^-197 = -2
Statement 2 is incorrect.

Given , x = and y = 

x² + y² = 

Given,
x² - 3x - 1 = 0 ⇒ x² - 1 = 3x ⇒ x - (1/x) = 3
Now, (x² + 8x - 1)


 = k² + 2)

x = 2 - p ⇒ x + p = 2
Cubing both sides
(x + p)³ = 2³
⇒ x³ + p³ + 3xp (x + p) = 8
⇒ x³ + p³ + 3xp(2) = 8
⇒ x³ + 6xp + p³ = 8

2x² + y² + 6x - 2xy + 9 = 0
⇒ x² + 6x + 9 + x² + y² - 2xy = 0
⇒ (x + 3)² + (x - y)² = 0
Sum of the squares of two numbers is equal to zero only if the numbers themselves are zero.
⇒ (x + 3) = 0 and (x - y) = 0 ⇒ x = y = - 3
Now,
4x³ - y³ + x²y² ⇒ 3x³ + x⁴ ...(x = y)
⇒ x³ (3 + x) ⇒ (-3)³ (3 - 3) = 0

(a + b + c)² - a² - b² - c²
= a² + b² + c² + 2(ab + bc + ac) - a² - b² - c² = 2(ab + bc + ac)

According to the question,

13X − 65Y = 7X + 35Y
6X = 100Y

a³ + b³ + c³ - 3abc = (a + b + c)/2
{3(a² + b² + c²) - (a + b + c)²}
= 6 {3 × 50 − 144} = 36

Then, 
Now, 

x + (1/x) = 2 ⇒ x = 1
Then,  1 + 1 = 2

[(1 + P) (1 + P²) (1 + P⁴) (1 + P⁸) (1 + P¹⁶)(1 − P) − 1]
⇒ [(1 − P²) (1 + P²) (1 + P⁴) (1 + P⁸) (1 + P¹⁶) − 1]
⇒ [(1 − P⁴) (1 + P⁴) (1 + P⁸) (1 + P¹⁶) − 1]
⇒ [(1 − P⁸) (1 + P⁸) (1 + P¹⁶) − 1]
⇒ [(1 − P¹⁶) (1 + P¹⁶) − 1]
⇒ [(1 − P³²) − 1] = − P³²

a³ - b³ = (a - b)(a² + b² + ab)
According to the question,
a − b = 1
Hence, a³ - b³ = (a² + b² + ab)

x³ - y³ = 6272 and x - y = 8
x³ - y³ = (x - y){(x - y)² + 3xy}

3xy = 784 − 64 = 720 ⇒ xy = 240

As per question, 
So,  

According to question,

 

x² + 1 = 15x
Squaring both side
x⁴ + 1 + 2x² = 225x²
x⁴ - 223x² + 1 = 0
x⁴ - 223x² + 6 = 5

x³ + y³ = (x + y)[ (x + y)² - 3xy]
= 25[25² - 3 x 20] = 25[625 - 60]
= 25 × 565 = 14125

Two lines have no solution, if these two lines are parallel to each other. The lines are parallel to each other means that the slopes of the lines are equal.
18x − 72y + 13 = 0 ⇒ y = 
Slope (m) = 1/4 and 7x − my − 17 = 0 ⇒ y = 
Value of (m) for no solution ⇒ 7/m = 1/4 
m = 28.

k + k^-1 = -2
k + 1/k = -2
Clearly, K = -1
Putting the value in given equation:-

Given , x + y = 36 then , x = 36 - y
= (x − 27)³ + (y − 9)³
= (36 − y − 27)³ + (y - 9)³
= (9 − y)³ + (y - 9)³
= (9 − y)³ + (9 - y)³ = 0

Formula :-
(a + b − c)² = a² + b² + c² + 2ab - 2bc - 2ca
Given,
a² + b² + c² - 1024 = -2ab + 2bc + 2ca
a² + b² + c² + 2ab − 2bc − 2ca = 1024
(a + b − c)² = 1024
(5c − c)² = 1024
 ...(Given, a + b = 5c) ⇒ c = 8.

x⁴ + 1/(x⁴) = 2207 ...eq .(1)
Adding 2 in eq .(1), we get ,

Adding 2 in eq .(2), we get ,
⇒ x + 1/x = 7
Then , 
Now, using Identities :-

= 322 × 2207 - 7 = 710647

x = 3 + 2√2

Taking reciprocal of eq .(1), we get

Now, From eq .(1) and eq .(2), we get,