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JEE Mathematics 2025 Question with Solution

Let f(x)f(x) be a real differentiable function such that f(0)=1f(0) = 1 and f(x+y)=f(x)f′(y)+f′(x)f(y)f(x + y) = f(x)f'(y) + f'(x)f(y) for all x,y∈Rx, y \in \mathbb{R}. Then ∑n=1100log⁡ef(n)\sum_{n=1}^{100} \log_e f(n) is equal to :

  • A

    23842384

  • B

    25252525

  • C

    52205220

  • D

    24062406

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: f(0)=1f(0)=1 and

f(x+y)=f(x)f′(y)+f′(x)f(y)f(x+y)=f(x)f'(y)+f'(x)f(y)

for all x,y∈Rx,y\in\mathbb{R}.

Find: ∑n=1100log⁡ef(n)\sum_{n=1}^{100}\log_e f(n).

Substitute x=y=0x=y=0 in the functional equation:

f(0)=f(0)f′(0)+f′(0)f(0)f(0)=f(0)f'(0)+f'(0)f(0)

Using f(0)=1f(0)=1,

1=2f′(0)1=2f'(0)

Hence,

f′(0)=12f'(0)=\frac{1}{2}

Now substitute y=0y=0 in the original equation:

f(x)=f(x)f′(0)+f′(x)f(0)f(x)=f(x)f'(0)+f'(x)f(0)

Using f(0)=1f(0)=1 and f′(0)=12f'(0)=\frac{1}{2},

f(x)=12f(x)+f′(x)f(x)=\frac{1}{2}f(x)+f'(x)

So,

f′(x)=12f(x)f'(x)=\frac{1}{2}f(x)

Solve the differential equation with the initial condition f(0)=1f(0)=1:

f′(x)=12f(x)f'(x)=\frac{1}{2}f(x)

Therefore,

f(x)=ex/2f(x)=e^{x/2}

Now,

log⁡ef(n)=log⁡e(en/2)=n2\log_e f(n)=\log_e\left(e^{n/2}\right)=\frac{n}{2}

Hence,

∑n=1100log⁡ef(n)=∑n=1100n2=12∑n=1100n\sum_{n=1}^{100}\log_e f(n)=\sum_{n=1}^{100}\frac{n}{2}=\frac{1}{2}\sum_{n=1}^{100}n

Using

∑n=1100n=100⋅1012=5050\sum_{n=1}^{100}n=\frac{100\cdot 101}{2}=5050

we get

∑n=1100log⁡ef(n)=50502=2525\sum_{n=1}^{100}\log_e f(n)=\frac{5050}{2}=2525

Therefore, the required sum is 25252525. The correct option is B.

Shortcut via constant coefficient form

Given: f(0)=1f(0)=1 and

f(x+y)=f(x)f′(y)+f′(x)f(y)f(x+y)=f(x)f'(y)+f'(x)f(y)

Find: ∑n=1100log⁡ef(n)\sum_{n=1}^{100}\log_e f(n).

Put y=0y=0:

f(x)=f(x)f′(0)+f′(x)f(0)f(x)=f(x)f'(0)+f'(x)f(0)

Since f(0)=1f(0)=1,

f′(x)=(1−f′(0))f(x)f'(x)=(1-f'(0))f(x)

So ff satisfies a linear differential equation of the form

f′(x)=cf(x)f'(x)=cf(x)

where c=1−f′(0)c=1-f'(0).

Hence,

f(x)=Aecxf(x)=Ae^{cx}

Using f(0)=1f(0)=1 gives A=1A=1, so

f(x)=ecxf(x)=e^{cx}

Then

f′(0)=cf'(0)=c

But also c=1−f′(0)=1−cc=1-f'(0)=1-c. Therefore,

2c=12c=1

which gives

c=12c=\frac{1}{2}

Thus,

f(x)=ex/2f(x)=e^{x/2}

Now,

ln⁡f(n)=n2\ln f(n)=\frac{n}{2}

Therefore,

∑n=1100ln⁡f(n)=12⋅100⋅1012=2525\sum_{n=1}^{100}\ln f(n)=\frac{1}{2}\cdot\frac{100\cdot101}{2}=2525

Therefore, the required sum is 25252525. The correct option is B.

Common mistakes

  • Substituting only y=0y=0 and stopping at f′(x)=(1−f′(0))f(x)f'(x)=(1-f'(0))f(x) without finding f′(0)f'(0). This leaves an unknown constant unresolved. You must use another substitution such as x=y=0x=y=0 to determine f′(0)=12f'(0)=\frac{1}{2}.

  • Solving f′(x)=12f(x)f'(x)=\frac{1}{2}f(x) as f(x)=x2f(x)=\frac{x}{2} or another non-exponential form. This is incorrect because the differential equation is proportional to the function itself. The correct general solution is f(x)=Cex/2f(x)=Ce^{x/2}, and then use f(0)=1f(0)=1.

  • Computing ∑n=1100log⁡ef(n)\sum_{n=1}^{100} \log_e f(n) incorrectly by treating it as log⁡e(∑f(n))\log_e\left(\sum f(n)\right). Logarithms do not distribute over sums in that way. First evaluate each term using f(n)=en/2f(n)=e^{n/2}, so log⁡ef(n)=n2\log_e f(n)=\frac{n}{2}, and then sum the arithmetic series.

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