Suppose for all integers x, there are two functions f and g such that f(x)+f(x-1)-1=0 and g(x)=x². If f(x²-x)=5, then the value of the sum f(g(5))+g(f(5)) is

Question

UPSC Quizroom · Accepted Answer

Step 1: Understand the problem

We are given:

f(x) + f(x-1) - 1 = 0 for all integers x
This can be rewritten as: f(x) = 1 - f(x-1)

g(x) = x^2

f(x^2 - x) = 5

We are asked: f(g(5)) + g(f(5)) = ?

Step 2: Solve the recurrence for f(x)

The functional equation: f(x) = 1 - f(x-1)

Notice the pattern:

f(x) = 1 - f(x-1)
f(x-1) = 1 - f(x-2)
=> f(x) = 1 - (1 - f(x-2)) = f(x-2)

So f is periodic with period 2:

f(even) = A, f(odd) = B

And the recurrence gives:
A = 1 - B
B = 1 - A
Step 3: Use the given value f(x^2 - x) = 5

x^2 - x = x(x-1) is always even.
So f(even) = 5

Then f(odd) = 1 - f(even) = 1 - 5 = -4

So:
f(even) = 5, f(odd) = -4

Step 4: Compute f(g(5)) + g(f(5))

g(5) = 5^2 = 25

f(g(5)) = f(25) → 25 is odd → f(25) = -4

f(5) = -4 (5 is odd)

g(f(5)) = g(-4) = (-4)^2 = 16

Step 5: Sum

f(g(5)) + g(f(5)) = -4 + 16 = 12

Answer:
12

🎯 UPSC Quiz Room

Suppose for all integers x, there are two functions f and g such that f(x)+f(x-1)-1=0 and g(x)=x². If f(x²-x)=5, then the value of the sum f(g(5))+g(f(5)) is

Options

Recommended articles

Suppose for all integers x, there are two functions f and g such that f(x)+f(x-1)-1=0 and g(x)=x². If f(x²-x)=5, then the value of the sum f(g(5))+g(f(5)) is

Options

Recommended articles

📱 Follow Us