Daily Practice Questions CAT Previous Year Questions with Answer PDF

let a = 3x + 2y; b= 2x − 5y

Let's try to write x in terms of a and b

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If n is even, f(n) = n (n + 1)
So, f (2) = 2(2 + 1) = 2 (3) = 6
If n is odd, f(n) = n + 3
f (1) = 1 + 3 = 4
It is given that, 8 x f(m + 1) - f(m) =2
So, m can either be even or odd
Case-1: If m were even and m + 1 odd
So, 8 x f(m + 1) - f(m) =2
8(m + 4) - m (m + 1) = 2
8m + 32 - m² - m = 2
m² - 7m - 30 = 0
(m-10) (m + 3) = 0
m = 10 or -3
m = 10, since m is positive
Case-2: If m were even and m + 1 odd
8 x f(m + 1) - f(m) =2
8 (m + 1) (m + 2) - (m + 3) = 2
Now, let us substitute m = 1 which is the minimum possible value
8 (1 + 1) (1 + 2) - (1 + 3) ≠ 3
Case 2 does not work

We know that f(x + y) = f(x) f(y)
Let x = a and y = 1,
f (a + 1) = f(a). f (1)
f(a + 1) = f(a). 2
f(a + 2) = f(a + 1) f(1)
f(a+2) = f(a) x 2 x 2 = f(a) x 2²
Similarly, f(a + n) = f(a) x 2ⁿ
(a + 1) + f (a + 2) + ..... + f(a + n) = f(a) {2¹ + 2² + 2³ +...+2ⁿ}
(a + 1) + f (a + 2) + ..... + f(a + n) = 2f(a) {1 + 2 + 2² + 2³ +...+ 2^n-1}, which is an infinite GP series
On solving,
(a + 1) + f (a + 2) + ..... + f(a + n) = 2f(a) {2ⁿ - 1}
We know that, (a + 1) + f (a + 2) + ..... + f(a + n) = 16 (2ⁿ - 1)
So, 2f(a) {2ⁿ - 1} = 16 (2ⁿ - 1)
2 f(a) = 16
f(a) = 8
We know that, f (1) = 2 , f(2) = 4 and f(3) = 8
So, a = 3

It is given that f (2) > 1 and f(mn) = f(m) f(n)
So, f (2) = f (2) f (1), as m and n are positive integers.
Only possible value for f (1) = 1
f (2) > 1
Now we know, f (24) = 54
So, f (2) f (3) f (4) = 54
f (3) f (2)³ = 54
Now we know, 54 = 2 x 3³
Therefore, f (2) = 3, f (3) = 2 and f (1) = 1
Now, we need to find the value of f (18)
f (18) = f (3) x f (2) x (3)
f (18) = 2 x 3 x 2 = 12
f (18) = 12

Given, f(x+2) = f(x) + f(x + 1) f(11) = 91, f(15) = 617
We get 91 + f(12) = f(13)
Let f(12) be equal to some value ‘a’
So, 91 + a = f(13).
f(12) + f(13) = f(14)
a + 91 + a = f(14)
So, f(14) = 2a + 91 f(13) + f(14) = 617
So, 91 + a + 2a + 91 = 617
3a + 182 = 617
a = 145
Substituting the value of a and f(11), we get
f(10) + 91 = 145
f(10) = 54

Given f(x) = min {2x², 52 − 5x}
From graph, we see that f(x) increases initially and then decreases after intersection
So, maximum value occurs when 2x² = 52 − 5x
2x² + 5x – 52 = 0
2x² - 8x + 13x - 52 = 0
2x(x - 4) + 13(x - 4) = 0
(2x + 13) (x - 4) = 0
Since x is a Positive real number, x = 4
min{2x², 52 - 5x} = min {32, 32} = 32 = max f(x)

Given that f(x) = max{5x , 52 - 2x²}, where x is any positive real number.
The minimum possible value of f(x) has to found.
The minimum possible value is obtained if the two curves intersect or
5x = 52 - 2x²
2x² + 5x – 52 = 0
2x² – 8x + 13x – 52 = 0
2x(x - 4) + 13(x - 4) = 0
(2x + 13)(x - 4) = 0
x = −13 / 2 or 4
One of these would be the minimum possible value
It is given that x is a positive real number. So, we have to consider only when x = 4
When x = 4
5x = 5 × 4 = 20
52 - 2x² = 52 – 2(4)² = 20
The minimum possible value of f(x) is 20.

Given that f₁(x) = x² + 11x + n and f₂(x) = x,
Then we have to find the largest positive integer n for which the equation f₁(x) = f₂(x) has two distinct real roots.
⟹ x² + 11x + n = x
⟹ x² + 11x – x + n = 0
⟹ x² + 10n + n = 0
The discriminant should be greater than zero, D > 0.
So, b² – 4ac > 0.
⟹ 10² – 4n > 0
⟹ 100 – 4n > 0
⟹ 25 – n > 0
⟹ n = 24
Hence if f₁(x) = x² + 11x + n and f₂(x) = x, then the largest positive integer n for which the equation
f₁(x) = f₂ (x) has two distinct real roots, is 24.

Given that f(ab) = f(a) × f(b) for all possible integers a and b.
We have to find the largest possible value of f(1). We know that,
f(ab) = f(a) × f(b)
⟹ f(a × 1) = f(a) × f(1)
⟹ [f(a) × f(1)] – f(a) = 0
⟹ f(a)[f(1) – 1] = 0
There are two possibilities here, either f(a) = 0 or f(1) – 1 = 0.
If f(a) = 0, then it is a constant function. Hence, f(1) = 0.
But, if f(1) – 1 = 0, f(1) = 1.
Hence, f(1) = 0 or f(1) = 1.
Out of these, f(1) = 1 is the largest possible value.