CAT Previous Year Questions: Functions (July 9) - CAT MCQ

let a = 3x + 2y; b= 2x − 5y

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

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Let, f(x) = ax² + bx + c

Since, f(x) ≥ 0,

This means that graph of f(x) is U shaped and above the x-axis.
Since f(x) = 0 at x = 2, the graph is centered at x = 2.
This means f(2 - k) = f(2 + k).
f(2 - 2) = f(2 + 2)
f(0) = f(4)
f(0) = 6
f(0) = 6
c = 6
f(4) = 6
16 a + 4 b + 6 = 6
f(2) = 0
4 a + 2 b + 6 = 0
Solving the two equations…

If we closely observe the coefficients of the terms in the numerator and denominator, we see that the coefficients of the x² and x in the numerators are in ratios 1:2. This gives us a hint that we might need to adjust the numerator to decrease the number of variables.

Now, we only have terms of x in the denominator.
The maximum value of the expression is achieved when the quadratic expression 2x²
 + 4x + 9 achieves its highest value, that is infinity.
In that case, the second term becomes zero and the expression becomes 1/2. However, at infinity, there is always an open bracket ')'.
To obtain the minimum value, we need to find the minimum possible value of the quadratic expression.
The minimum value is obtained when 4x + 4 = 0 [d/dx = 0]
x = -1.
The expression comes as 7. 
The entire expression becomes 3/7.
Hence, 

Now we have :
f(g(x)) − 3x
so we get f(x + 3) - 3x
= 
= x² − 4x − 12
Now minimum value of expression = 
We get - (16 + 48)/4
= -16

Given f ( x + y ) = f ( x) f (y)
f ( x) = a^x ( where a is a constant ) Given, f (5) = 4 ⇒ a⁵ = 4 ⇒ a = 2^{2/ 5}

= 

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)