Let f(x) be a real differentiable function such that f(0)=1 and f(x+y)=f(x) f^\prime(y)+f^\prime(x) f(y) for all x, y \in R . Then \sum_n=1^100 \log _e f(n) is equal to :

Differential Equations PYQ — JEE Main 2025 · 22 Jan · Shift 1 · Q36 with Solution

View written solutionFree

Correct answer: C

Given functional equation

We have $f(0)=1$ and for all real $x,y$ , $f(x+y)=f(x)f'(y)+f'(x)f(y).$

We need to find $\sum_{n=1}^{100} \ln f(n).$

Use special values to determine $f$

We are given $f(x+y)=f(x)f'(y)+f'(x)f(y).$

We now substitute convenient values.

Step 1: Put $x=0$

Then $f(y)=f(0)f'(y)+f'(0)f(y).$ Since $f(0)=1$ , this becomes $f(y)=f'(y)+f'(0)f(y).$ So, $f'(y)=(1-f'(0))f(y).$

Thus $f$ satisfies the differential equation $f'(x)=cf(x),$ where $c=1-f'(0).$

Hence $f(x)=Ae^{cx}.$ Using $f(0)=1$ , we get $A=1$ . Therefore $f(x)=e^{cx}.$

Determine $c$ using the original equation

Now $f(x)=e^{cx}, \qquad f'(x)=ce^{cx}.$ Substitute into the functional equation:

Left side: $f(x+y)=e^{c(x+y)}.$

Right side: $f(x)f'(y)+f'(x)f(y)=e^{cx}(ce^{cy})+(ce^{cx})e^{cy}=2ce^{c(x+y)}.$

So $e^{c(x+y)}=2ce^{c(x+y)}.$ Since $e^{c(x+y)}\neq 0$ , we get $1=2c \implies c=\frac12.$

Therefore $f(x)=e^{x/2}.$

Compute the required sum

We need $\sum_{n=1}^{100} \ln f(n).$ Since $f(n)=e^{n/2},$ we have $\ln f(n)=\ln\left(e^{n/2}\right)=\frac n2.$

Therefore

=\frac12\sum_{n=1}^{100} n.$$ Now $$\sum_{n=1}^{100} n=\frac{100\cdot 101}{2}=5050.$$ Hence $$\sum_{n=1}^{100} \ln f(n)=\frac12\cdot 5050=2525.$$ --- 5. **Check options** The value is $$2525,$$ which corresponds to **Option C**. --- 6. **Comparison with stored correct answer** Stored correct answer: **C** Our derived answer: **C** So the answer agrees with the stored correct answer.

More from Differential Equations

View written solutionFree

Correct answer: C

Given functional equation

We have $f(0)=1$ and for all real $x,y$ , $f(x+y)=f(x)f'(y)+f'(x)f(y).$

We need to find $\sum_{n=1}^{100} \ln f(n).$

Use special values to determine $f$

We are given $f(x+y)=f(x)f'(y)+f'(x)f(y).$

We now substitute convenient values.

Step 1: Put $x=0$

Then $f(y)=f(0)f'(y)+f'(0)f(y).$ Since $f(0)=1$ , this becomes $f(y)=f'(y)+f'(0)f(y).$ So, $f'(y)=(1-f'(0))f(y).$

Thus $f$ satisfies the differential equation $f'(x)=cf(x),$ where $c=1-f'(0).$

Hence $f(x)=Ae^{cx}.$ Using $f(0)=1$ , we get $A=1$ . Therefore $f(x)=e^{cx}.$

Determine $c$ using the original equation

Now $f(x)=e^{cx}, \qquad f'(x)=ce^{cx}.$ Substitute into the functional equation:

Left side: $f(x+y)=e^{c(x+y)}.$

Right side: $f(x)f'(y)+f'(x)f(y)=e^{cx}(ce^{cy})+(ce^{cx})e^{cy}=2ce^{c(x+y)}.$

So $e^{c(x+y)}=2ce^{c(x+y)}.$ Since $e^{c(x+y)}\neq 0$ , we get $1=2c \implies c=\frac12.$

Therefore $f(x)=e^{x/2}.$

Compute the required sum

We need $\sum_{n=1}^{100} \ln f(n).$ Since $f(n)=e^{n/2},$ we have $\ln f(n)=\ln\left(e^{n/2}\right)=\frac n2.$

Therefore

=\frac12\sum_{n=1}^{100} n.$$ Now $$\sum_{n=1}^{100} n=\frac{100\cdot 101}{2}=5050.$$ Hence $$\sum_{n=1}^{100} \ln f(n)=\frac12\cdot 5050=2525.$$ --- 5. **Check options** The value is $$2525,$$ which corresponds to **Option C**. --- 6. **Comparison with stored correct answer** Stored correct answer: **C** Our derived answer: **C** So the answer agrees with the stored correct answer.

Differential Equations question

Differential Equations question

Step 1: Put $x=0$

More from Differential Equations

Differential Equations question

Differential Equations question

Step 1: Put $x=0$

More from Differential Equations

Step 1: Put x=0x=0x=0

More from Differential Equations

Step 1: Put x=0x=0x=0

More from Differential Equations

Step 1: Put $x=0$

Step 1: Put $x=0$