Test Level 1: Quadratic Equations & Linear Equations - 2 - CAT MCQ

The given equation can be written as x² - 5x + (2k - 1) = 0.
If p and q are the roots, it is given that p - q = 3.
⇒ (p + q)2 - 4pq = 9. (∵ (p - q)2 = (p + q)2 - 4pq)
⇒ 25 - 4(2k - 1) = 9 (∵ p + q = 5, pq = 2k - 1)
∴ k = 2.5

Given: x = 3 + √8
1/x = 1/(3 + √8) = (3 - √8)/[3² - (√8)²] = 3 - √8/(9 - 8) = 3 - √8
x + 1/x = 3 +√8 + 3 - √8 = 6
Squaring both sides, we get:
(x + 1/x)² = 62
x² + 1/x² + 2 = 36
x² + 1/x² = 36 - 2 = 34

Product of all the roots = e/a
Sum of the roots taking two at a time = c/a

Given, (c - 3)² = c² + kc + 9
We know the formula: (a - b)² = a² - 2ab + b²
Applying the above formula, we get
c² - 6c + 9 = c² + kc + 9
Comparing both sides, we get
k = -6

5x + 9y + 13z = 50 and 24x + 20y + 16z = 66
Adding both, we get
29x + 29y + 29z = 116
Thus, x + y + z = 4

(i) x² + 9x + 27 = 0
D = 92 - 4(27)
D < 0; therefore, roots are imaginary.
(ii) 6x² - 13x - 5 = 0
D = (-13)² - 4(-5)(6)
D > 0; therefore, roots are real.
So, option (3) is the correct answer.

Since the roots are in the ratio of 1 : 2, they can be assumed as α and 2α.
Sum of the roots

Product of the roots = α (2α) = k/3 ⇒ k = 6

The 24th term in the series will be (x – x).
Now, (x – x) = 0
So, (x – a)(x – b) … (x – z) = 0.

Let Ralph's present age be x years.
After 15 years, his age will be (x + 15) years. His age 5 years back was (x - 5) years.
According to the condition: x + 15 = 5(x - 5)
On solving, we get x = 10.
Therefore, Ralph's present age = 10 years

Since one of a, b, c is equal to zero, we can put the values of a, b and c as a = 2, b = -1, c = 0 in the following equation

On solving we get x = 2/3.

Since, x, y, z are distinct positive real numbers, use AM > GM

Hence, option (3) is correct.

For the given expression to be a maximum, the denominator should be minimized. (Since, the function in the denominator has imaginary roots and is always positive). x² + 5x + 10 will be minimized at x = – 2.5 and its minimum values at x = –2.5 is 3.75. Hence, required answer = 1/3.75 = 4/15.

The minimum value of ( p + 1/p) is at p = 1. The value is 2.

To solve this take any expression whose roots differ by 2.
Thus, (x – 3) (x – 5) = 0
⇒ x² – 8x + 15 = 0
In this case, a = 1, b = –8 and c = 15.
We can see that b² = 4(c + 1).

h (x) = 4x2 + 13x + 10.
Sum of roots -13/4.