SAT (Stage-2) Mock Test - 17

Since all are prime numbers and N itself is an even number, so there should be a prime factor which is even and we have only one even prime number which is 2. Also 2 is smallest prime number.

Let the roots of the polynomial be R and r.
Now, rR = 127
This implies that either both are positive or both are negative.
Also, r + R = -99
This implies that both must be negative.
Hence, both the roots are negative.

We compare the highest powers of a and b.
a = x³y²
b = xy³
HCF will have common terms and lowest exponent = xy²

Product of the zeros taken all at a time is -c.

Let α and β be the other roots.

Then,

α × β × (- 1) = –c

α × β = c

Also, y = x³ + ax² + bx + c

-1 is the root of the equation.

=> -1 + a - b + c = 0

=> c = b - a + 1

A pair of simultaneous equations with 2 unknowns will always have a finite solution.

To get a dependent linear equation, you have to multiply the given equation by any integer. If the given equation is multiplied by -2, we get 10x - 14y = -4

AP₁ = -1, (-1 + d), .....
AP₂ = -8, (-8 + d), .....a4 = a + 3d
Difference between their 4^th term = (-8 + 3d) - (-1 + 3d)
= -8 + 3d + 1 - 3d
= -8 + 1
= -7

Let the 1^st term of an arithmetic progression be 'a' and the common difference be 'd'.
Now, 2^nd term = T₂ = a + d = 13
5^th term = T₅ = a + 4d = 25
Now, T₅ - T₂ = 3d = 25 - 13 = 12
On solving, we get d = 4
7^th term = T₇ = a + 6d = T₅ + 2d = 25 + 2 x 4 = 33

n^th term of AP = a + (n - 1)d, where 'a' is the 1^st term and d is the common difference

1^st term of AP a₁ = -3
2^nd term of AP a₂ = 4
d = a₂ - a₁ = 4 - (-3) = 7
21^st term = -3 + (21 - 1)7 = 137

d₁² = 10² + 10² = 200 m²
Similarly, d₂² = 200 m²
So, d₁² + d₂² = 400 m²