Q1: If (a + b√n) is the positive square root of (29 - 12√5), where a and b are integers, and n is a natural number, then the maximum possible value of (a + b + n) is
(a) 4
(b) 18
(c) 6
(d) 22
Ans: b
Sol: So 29 - 12√5 is the positive square root of (29 - 12√5).
So 29 - 12√5 = (a + b√n)²
29 - 12√5 = a² + b²n + 2ab√n
a² + b²n = 29
ab√n = -6√5
a²b²n² = 180
b²n = 180 / a²
Substituting this in the above equation:
a² + b²n = 29
a² + (180 / a²) = 29
a⁴ - 29a² + 180 = 0
a² = (29 ± √(29² - 4*180)) / 2
a² = 9 or 20
That means one of a² or b²n is 9 or 20.
We also have ab√n = -6√5; that means one of a or b should be negative.
And also the fact that this is a positive root, and we need to maximize the value of a, b, and n.
We can have a = -3, b = 1, and n = 20.
This satisfies all the above equations, and the value of a + b + n = 18
Q2: For any natural number n, let an be the largest integer not exceeding √n. Then the value of a1 + a2 + a3 + ... + a50 is
Ans: 217
Sol:
Q1: If then
is equal to
(a) 4√5
(b) 2√7
(c) 3√31
(d) 3√7
Ans: d
Sol:
Q2: For some positive and distinct real numbers x,y and z, if is the arithmetic mean of
and
then the relationship which will always hold true, is
(a) √x, √y and √z are in arithmetic progression
(b) √x, √z and √y are in arithmetic progression
(c) y, x and z are in arithmetic progression
(d) x, y and z are in arithmetic progression
Ans: c
Sol:
x is the Arithmetic Mean of z & y, therefore, z, x, y form an A.P.
It goes without saying that y, x, z also forms an A.P.
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