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Maths Question 2 – AP-EAMCET 2026

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If f satisfies the relation f(x+y)+f(x−y)=2f(x)f(y) for all x,y∈R and f(0)≠0, then f(10)−f(−10)=

Start by evaluating the function at specific points, such as x=0 or y=0, to discover fundamental properties of the function f.

Step 1: Determine the value of f(0)✦ Active

Substitute x=0 and y=0 into the given functional equation f(x+y)+f(x−y)=2f(x)f(y).

f(0+0)+f(0−0)=2f(0)f(0) f(0)+f(0)=2(f(0))2 2f(0)=2(f(0))2

Since it is given that f(0)≠0, we can divide both sides by 2f(0):

1=f(0)

So, f(0)=1.

💡 Teacher's Secret Hint

Always check for special values like f(0) or f(1) in functional equations, as they often reveal key properties or constants.

Step 2: Determine the parity of the function f(x)○ Expand

Substitute x=0 into the original functional equation:

f(0+y)+f(0−y)=2f(0)f(y) f(y)+f(−y)=2f(0)f(y)

Now, substitute the value of f(0)=1 found in Step 1:

f(y)+f(−y)=2(1)f(y) f(y)+f(−y)=2f(y)

Subtract f(y) from both sides:

f(−y)=2f(y)−f(y) f(−y)=f(y)

This shows that f is an even function.

💡 Teacher's Secret Hint

Recognizing if a function is even or odd can significantly simplify expressions involving f(x) and f(−x). An even function is symmetric about the y-axis.

Step 3: Calculate the required expression○ Expand

Since f is an even function, we know that f(−x)=f(x) for all x∈R.

Therefore, for x=10, we have f(−10)=f(10).

The expression to be evaluated is f(10)−f(−10).

Substitute f(−10)=f(10) into the expression:

f(10)−f(−10)=f(10)−f(10)=0

The value of f(10)−f(−10) is 0.

💡 Teacher's Secret Hint

This type of functional equation is known as d'Alembert's functional equation, whose continuous solutions are f(x)=cos⁡(ax) or f(x)=cosh⁡(ax). Both are even functions, confirming our result.

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