Continuity & DifferentiabilitymediumFree

A Differentiable Function with f(x+y) = f(x-y) Must Be Constant | JEE

JEE Maths question with a full step-by-step solution.

Question
Let f(x)f(x) be a non-zero differentiable function for all x∈Rx \in \mathbb{R} satisfying
f(x+y)=f(x−y)∀ x,y∈R.f(x+y) = f(x-y) \quad \forall\ x, y \in \mathbb{R} .
 Then
Af(1)+f(0)=f′(0)f(1)+f(0) = f'(0)
Bf(1)−f(0)=f′(0)f(1)-f(0) = f'(0)correct
Cf(0)+f(1)=f(2)f(0)+f(1) = f(2)
D3f(2)=2f(3)3f(2) = 2f(3)
Solution
Step 1: Write the derivative from first principles, using y=hy = h in the given relation with the ++ sign.
f′(x)=lim⁡h→0f(x+h)−f(x)h.f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}{h} .
 Step 2: Now use the relation itself: f(x+h)=f(x−h)f(x+h) = f(x-h). Substituting,
f′(x)=lim⁡h→0f(x−h)−f(x)h.f'(x) = \lim_{h\to0}\frac{f(x-h)-f(x)}{h} .
 Step 3: Recognise the right-hand side. Multiplying and dividing by −1-1,
lim⁡h→0f(x−h)−f(x)h=−lim⁡h→0f(x−h)−f(x)−h=−f′(x).\lim_{h\to0}\frac{f(x-h)-f(x)}{h} = -\lim_{h\to0}\frac{f(x-h)-f(x)}{-h} = -f'(x) .
 Step 4: So f′(x)=−f′(x)f'(x) = -f'(x), giving
f′(x)=0for every x  ⟹  f(x)=k (a constant).f'(x) = 0 \quad \text{for every } x \implies f(x) = k \ \text{(a constant)} .
 Step 5: Note k≠0k \ne 0, since ff is given to be non-zero. Step 6: Test each option with f≡kf \equiv k and f′(0)=0f'(0) = 0.
(1) k+k=2k=0? only if k=0. ×\text{(1)}\ k+k = 2k = 0 ? \ \text{only if } k = 0 . \ \times
(2) k−k=0=f′(0).\text{(2)}\ k-k = 0 = f'(0) .
(3 k+k=2k versus f(2)=k. ×\text{(3}\ k+k = 2k \ \text{versus}\ f(2) = k . \ \times
(4) 3k=2k? only if k=0. ×\text{(4)}\ 3k = 2k ? \ \text{only if } k = 0 . \ \times
 Answer: (2).
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