Test Level 3: Functions - CAT MCQ

f(x) = 
Let f(x) be y.

For x = 2 and a = 3
We have,

Going with the options, the second option comes out to be the answer.
f(t) = … (1)
f(-t) = 
Subtracting (2) from (1), we get
f(t) - f(-t) = (1 - e^t) + t = -t + t = 0

f(4, 100) = f[3, f(3, 100)] …………(1)
f(3, 100) = f[2, f(2, 100)] ……….. (2)
f(2, 100) = f[1, f(1, 100)] …………(3)
f(1, 100) = f[0, f(0, 100)] ……….. (4)
f(0, 100) = 100 + 1 = 101.
f(1, 100) = f(0, 101) = 102
f(2, 100) = f(1, 102) …………. (5)
f(1, 102) = f[0, f(0, 102)] = f(0, 103) = 104
Put in (5) = f(2, 100) = 104
Put this value in (2),
f(3, 100) = f(2, 104) …………… (6)
Now, f(2, 104) = f[1, f(1, 104)] …………… (7)
f(1, 104) = f[0, f(0, 104)] = f(0, 105) = 106
Put in (7)... and so on
f(4, 100) = 116

The given equation 2f(x) + f(1 – x) = x² holds for all x.
2f(x) + f(1 – x) = x².......(i)
In particular, the equation holds, if we replace x with 1 – x.
Thus, we deduce that 2f(1 – x) + f(x) = (1 – x)².
2f(1 – x) + f(x) = (1 – x)²..............(ii)
Multiply equation (i) by 2 and then subtract equation (ii) from equation (i).
Thus, we get:
3f(x) = 2x² – (1 – x)² = x² + 2x – 1
f(x) = (x² + 2x – 1)/3

The given information implies that f(x) = (x – 3) (x – 13) g(x), where g(x) is a polynomial with integer coefficients. Hence, f(10) = – 21g(10), so that 21 must be a divisor of f(10). The only choice divisible by 21 is 42, so the correct answer is 42. To check if f(x) exists, consider f(x) = – 2(x – 3) (x – 13). 

Statement (i): Let F(x) be an even function and G(x) be odd.
H(x) = Product of an even function and an odd function
H(x) = F(x) G(x)
Now, H(-x) = F(-x) G(-x) = F(x) (-G(x)) = -F(x) G(x) = -H(x)
Hence, H(x) is an odd function. Statement (i) is not true.
Statement (ii): Let F(x) and G(x) be even functions.
H(x) = F(x) + G(x)
H(-x) = F(-x) + G(-x) = F(x) + G(x) = H(x)
It is an even function; hence, statement (ii) is true.
Statement (iii): Let F(x) and G(x) be odd functions.
H(x) = F(x) G(x)
H(-x) = F(-x) G(-x) = (-F(x))(-G(x)) = F(x) G(x) = H(x)
It is an even function. Hence, statement (iii) is true.
So, statements (ii) and (iii) are true.

Given: 2f(x) + f(1 - x) = x² ... (A)
Put x = 5 in (i).
2f(5) + f(-4) = 25
Multiply by 2
4f(5) + 2f(-4) = 50 ... (1)
Put x = -4 in (A).
2f(-4) + f(5) = 16 ... (2)Subtract (2) from (1):
4f(5) - f(5) = 50 - 16
3f(5) = 34
f(5) = 34/3

 f(x) = 1/(x - 2)
When x = 1, f(x) = -1
When x = 2, f(x) = ∞
When x = 3, f(x) = 1, and so on
Therefore, option d is correct.

 I  = I (-1) = -1